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Amazon Interview Question


Country: United States



Comment hidden because of low score. Click to expand.
3
of 3 vote

I would prose the following solution:
(i) Extract k-th digits from the end of the strings (if index exceeds dimension return '0').
(ii) Sum these digits and the carry
(iii) Compute resulting digit as modulo 2
(iv) Compute new carry as inreger division by 2
(v) Do until digits and carry are all zero.

A sample code is shown below:

public static String sumBinary(String a, String b) {
	int carry = 0;
	StringBuffer sb = new StringBuffer();
	int k = 0;
	while (true) {
		int da = getChar(a, k) - '0'; 	// extract digit
		int db = getChar(b, k) - '0';
		int aux = da + db + carry;
		if ( aux == 0 && !(k<a.length() && k<b.length()) )			
				break;
		sb.append(aux%2);
		carry = aux/2;
		k++;
	}

	return sb.reverse().toString();
}
private static char getChar(String s, int k) {
	int idx = s.length()-1-k;
	return (idx >= 0)?s.charAt(idx):'0';
}

- autoboli April 22, 2015 | Flag Reply
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0
of 0 votes

why if (aux == 0) break; ??? This will fail for "111100" and "1100"

- Rahul April 22, 2015 | Flag
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0
of 0 votes

Good point Rahul, not only 'aux' must be zero but also 'k' must exceeds length of at least one string. Bug fixed, thank you!

- autoboli April 23, 2015 | Flag
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0
of 0 votes

I think the correct condition to break would be:

if ( aux == 0 && (k>a.length() && k>b.length()) )

The above fails for ("100001", "1"). The output comes 10 instead of 100010

- devdeep May 03, 2015 | Flag
Comment hidden because of low score. Click to expand.
0
of 2 vote

public static String sumBinary(String binary1, String binary2)
{
char[] binary1Array = binary1.toCharArray();
char[] binary2Array = binary2.toCharArray();

int length = binary1.length() > binary2.length() ? binary1.length() + 1 : binary2.length() +1;
char[] sum = new char[length];
int i=binary1Array.length-1, j=binary2Array.length-1, k= length-1;
char carry = '0';
while (i >= 0 && j >= 0)
{
if (binary1Array[i] == '0' && binary2Array[j] == '0' && carry == '0') { sum[k] = '0'; carry = '0'; } else
if (binary1Array[i] == '1' && binary2Array[j] == '0' && carry == '0') { sum[k] = '1'; carry = '0'; } else
if (binary1Array[i] == '0' && binary2Array[j] == '1' && carry == '0') { sum[k] = '1'; carry = '0'; } else
if (binary1Array[i] == '0' && binary2Array[j] == '0' && carry == '1') { sum[k] = '1'; carry = '0'; } else
if (binary1Array[i] == '1' && binary2Array[j] == '1' && carry == '0') { sum[k] = '0'; carry = '1'; } else
if (binary1Array[i] == '1' && binary2Array[j] == '0' && carry == '1') { sum[k] = '0'; carry = '1'; } else
if (binary1Array[i] == '0' && binary2Array[j] == '1' && carry == '1') { sum[k] = '0'; carry = '1'; } else
if (binary1Array[i] == '1' && binary2Array[j] == '1' && carry == '1') { sum[k] = '1'; carry = '1'; }
i--;
j--;
k--;
}
if (i < 0 && j >=0)
{
while (j >=0)
{
if (binary2Array[j] == '1' && carry == '0') {sum[k] = '1'; carry = '0';} else
if (binary2Array[j] == '0' && carry == '1') {sum[k] = '1'; carry = '0';} else
if (binary2Array[j] == '1' && carry == '1') {sum[k] = '0'; carry = '1';}
j--;
k--;
}
}
if ( j < 0 && i >= 0)
{
while (i >=0)
{
if (binary1Array[i] == '1' && carry == '0') {sum[k] = '1'; carry = '0';} else
if (binary1Array[i] == '0' && carry == '1') {sum[k] = '1'; carry = '0';} else
if (binary1Array[i] == '1' && carry == '1') {sum[k] = '0'; carry = '1';}
i--;
k--;
}
}

if (carry == '1') sum[k] = carry;
return new String(sum);
}

- Javacode April 21, 2015 | Flag Reply
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0
of 0 votes

This is too heavy code, it can be done with just few lines. It would be a red flag during interview.

- autoboli April 22, 2015 | Flag
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0
of 0 vote

I have created a solution with code clearance/quality in mind. So, I divided it into two methods that could be reused:

/**
 * Created by sky on 21/04/15.
 */
public class SumBinary {
    public static int binaryStringToInteger(String binary) {
        int a = 0;
        int mul = 1;

        for (int i = binary.length() - 1; i >= 0; i--) {
            Character chr = binary.charAt(i);
            int bit = chr == '1' ? 1 : 0;
            a += bit * mul;
            mul <<= 1;
        }
        return a;
    }

    public static String integerToBinaryString(int v) {
        int mask = 1;
        StringBuilder sb = new StringBuilder();

        for (int i = 0; i < 32; i++) {
            if ((v & mask) > 0) {
                sb.append('1');
            } else {
                sb.append('0');
            }

            mask <<= 1;
        }

        String bin = sb.reverse().toString();
        int idx = bin.indexOf('1');
        return bin.substring(idx);
    }

    public static String sumBinary(String binary1, String binary2) {
        int a = 0;
        int b = 0;
        a = binaryStringToInteger(binary1);
        b = binaryStringToInteger(binary2);

        int res = a + b;
        return integerToBinaryString(res);
    }

    public static void main(String[] args) {
        System.out.println(sumBinary("0111101", "1101"));
    }
}

- Felipe Cerqueira April 21, 2015 | Flag Reply
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0
of 0 votes

Already these function are available?
Integer.tobinaryString(integer);
Integer.parseInt(String,2);

- Ramesh May 12, 2015 | Flag
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0
of 0 vote

Summing the binary bits and carry bit and based on total sum choosing possibilities,
sum=3 or 2 ==> carry=1 else carry=0

public class BinarySum {
	public static void main(String[] args) {
		String str1="0111101",str2="1101";
		int i=str1.length()-1,j=str2.length()-1;
		char[] str1Array=str1.toCharArray();
		char[] str2Array=str2.toCharArray();
		int max=(i>=j)?i:j;
		char[] sumArray=new char[max+1];
		int b1,b2,sum,carry=0;
		for(int k=0;k<=max;k++,i--,j--){
			b1=(i>=0)?Character.getNumericValue(str1Array[i]):0;
			b2=(j>=0)?Character.getNumericValue(str2Array[j]):0;
			sum=b1+b2+carry;
			if(sum==3){
				sumArray[max-k]=Character.forDigit(1, 10);
				carry=1;
			}else if(sum==2){
				sumArray[max-k]=Character.forDigit(0, 10);
				carry=1;
			}
			else{
				sumArray[max-k]=Character.forDigit(sum, 10);
				carry=0;
			}
		}
		String s=new String(sumArray);
		System.out.println(s);
	}

}

- achyuth April 22, 2015 | Flag Reply
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0
of 0 vote

Convert to int then sum then convert it back

public void SumString(string s1, string s2)
        {
            Console.WriteLine(ConvertDecimalToBinaryString((ConvertBinaryToDecimal(s1) + ConvertBinaryToDecimal(s2))));
        }

        private int ConvertBinaryToDecimal(string str)
        {
            int num = 0;
            int length = str.Length - 1;
            for (int i = str.Length - 1; i >= 0; i--)
            {
                num += int.Parse(str[i].ToString()) * (int)Math.Pow(2, length - i);
            }
            return num;
        }

        private string ConvertDecimalToBinaryString(int num)
        {
            StringBuilder sb = new StringBuilder();
            while (num != 0)
            {
                sb.Insert(0, num % 2);
                num = num / 2;
            }
            return sb.ToString();
        }

- Tu Nguyen April 22, 2015 | Flag Reply
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0
of 0 vote

it.... is not good code but its work.

public class SumBinary {

    public static String sumBinary(String s1, String s2) {
	
	char[] a, b;

	if(s1.length()>s2.length()){
	    a = s1.toCharArray();
	    b = s2.toCharArray();
	}else {
	    b = s1.toCharArray();
	    a = s2.toCharArray();
	}
	
	int length = s1.length() > s2.length() ? s1.length() + 1 : s2.length() + 1;
	char[] sum = new char[length];
	
	char c = '0';

	int i = s1.length() - 1, j = s2.length() - 1; //for looks code easy

	
	while( i>=0 ) {
	    if(a[i] == '0' && c == '0') {
		if(j<0) sum[i+1] = '0';
		else
		    sum[i+1] = b[j];
		
	    }else if(a[i] == '0' && c == '1') {
		if(j<0) {
		    sum[i+1] = '1';
		    c = '0';
		}else if(b[j] == '0') {
		    sum[i+1] = '1';
		    c = '0';
		}else{
		    sum[i+1] = '0';
		    c = '1';
		}
		
	    }else if(a[i] == '1' && c == '1') {
		if(j<0) {
		    sum[i+1] = '0';
		    c = '1';
		}else if(b[j] == '0') {
		    sum[i+1] = '0';
		    c = '1';
		}else{
		    sum[i+1] = '1';
		    c = '1';
		}
		
		
	    }else if(a[i] == '1' && c == '0') {
		if(j<0) {
		    sum[i+1] = '1';
		    //c == '0';
		} else if(b[j] == '0') {
		    sum[i+1] = '1';
		}else {
		    sum[i+1] = '0';
		    c='1';
		}
		
	    }

	    
	    

	    i--;
	    j--;
	}
	
	if(c == '1') sum[0] = '1';
	//	System.out.println(sum);

	return new String(sum); // or return sum?
    }

    public static void main(String[] args) {


        System.out.println(sumBinary("", ""));
        System.out.println(sumBinary("0", "0"));
        System.out.println(sumBinary("1", "1"));	
        System.out.println(sumBinary("0111101", "0001101"));
	System.out.println(sumBinary("1111", "0000"));
	System.out.println(sumBinary("1010", "0101"));
	System.out.println(sumBinary("1111", "1111"));

	System.out.println(sumBinary("1111", ""));
	
	System.out.println(sumBinary("1111", "1"));
	
	System.out.println(sumBinary("1111", "00"));
	
	System.out.println(sumBinary("1111111111111111", "111"));

	
    }
    
}

- tk0221 April 22, 2015 | Flag Reply
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0
of 0 vote

public class SumBinary {

    public static String sumBinary(String s1, String s2) {
	
	char[] a, b;

	if(s1.length()>s2.length()){
	    a = s1.toCharArray();
	    b = s2.toCharArray();
	}else {
	    b = s1.toCharArray();
	    a = s2.toCharArray();
	}
	
	int length = s1.length() > s2.length() ? s1.length() + 1 : s2.length() + 1;
	char[] sum = new char[length];
	
	char c = '0';

	int i = s1.length() - 1, j = s2.length() - 1; //for looks code easy

	
	while( i>=0 ) {
	    if(a[i] == '0' && c == '0') {
		if(j<0) sum[i+1] = '0';
		else
		    sum[i+1] = b[j];
		
	    }else if(a[i] == '0' && c == '1') {
		if(j<0) {
		    sum[i+1] = '1';
		    c = '0';
		}else if(b[j] == '0') {
		    sum[i+1] = '1';
		    c = '0';
		}else{
		    sum[i+1] = '0';
		    c = '1';
		}
		
	    }else if(a[i] == '1' && c == '1') {
		if(j<0) {
		    sum[i+1] = '0';
		    c = '1';
		}else if(b[j] == '0') {
		    sum[i+1] = '0';
		    c = '1';
		}else{
		    sum[i+1] = '1';
		    c = '1';
		}
		
		
	    }else if(a[i] == '1' && c == '0') {
		if(j<0) {
		    sum[i+1] = '1';
		    //c == '0';
		} else if(b[j] == '0') {
		    sum[i+1] = '1';
		}else {
		    sum[i+1] = '0';
		    c='1';
		}
		
	    }

	    
	    

	    i--;
	    j--;
	}
	
	if(c == '1') sum[0] = '1';
	//	System.out.println(sum);

	return new String(sum); // or return sum?
    }

    public static void main(String[] args) {


        System.out.println(sumBinary("", ""));
        System.out.println(sumBinary("0", "0"));
        System.out.println(sumBinary("1", "1"));	
        System.out.println(sumBinary("0111101", "0001101"));
	System.out.println(sumBinary("1111", "0000"));
	System.out.println(sumBinary("1010", "0101"));
	System.out.println(sumBinary("1111", "1111"));

	System.out.println(sumBinary("1111", ""));
	
	System.out.println(sumBinary("1111", "1"));
	
	System.out.println(sumBinary("1111", "00"));
	
	System.out.println(sumBinary("1111111111111111", "111"));

	
    }
    
}

- kim.taehun0221 April 22, 2015 | Flag Reply
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0
of 0 vote

I hope my Java solution is easy to read and understand:

String binarySum(String n1, String n2){
	if (n2.length() > n1.length()){
		String tmp = n1;
		n1 = n2;
		n2 = tmp;
	}
		
	StringBuilder sum = new StringBuilder();
	int carry = 0;
		
	for (int i = 0; i < n1.length(); i++){
		int idx1 = n1.length() - 1 - i;
		int bit1 = n1.charAt(idx1) - '0';
			
		int bit2 = 0;
		int idx2 = n2.length() - 1 - i;
			
		if (idx2 >= 0){
			bit2 = n2.charAt(idx2) - '0';
		}
			
		int bitSum = bit1 + bit2 + carry;
			
		if (bitSum >= 2){
			carry = 1;
		}
		else {
			carry = 0;
		}
			
		bitSum %= 2;
		sum.append((char)(bitSum + '0'));
	}
		
	if (carry == 1){
		sum.append('1');
	}
		
	return sum.reverse().toString();
}

- inucoder April 22, 2015 | Flag Reply
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0
of 0 vote

I see a lot of brute force like methods. Although my answer is probably not what the interviewer is looking for it can be solved much simpler.

string sumBinary(string s1,string s2)
{
	string sum;
	int result = (std::stoull(s1.c_str(),NULL,2) + std::stoull(s2.c_str(),NULL,2));	 
	while(result != 0)
	{
		if(result & 0x1)
		{
			sum = "1" + sum; 
		}else
		{
		    sum = "0" + sum; 
		}

		result >>= 1;
	}
	 
	return sum;
}

- Alex April 22, 2015 | Flag Reply
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0
of 0 votes

That looks interesting. Could you pls describe in a nut shell how the code works?

- spagaty April 22, 2015 | Flag
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0
of 0 votes

This code converts the strings into numbers, adds them, then converts the sum into a string.

Converting the sum into a string is done by reading the least significant digit and shifting right until the sum equals 0.

- JW April 26, 2015 | Flag
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0
of 0 vote

My clean java code

public String sumBinary (String a, String b){
		int carrier = 0 ;
		StringBuilder sb = new StringBuilder ();
		int i = a.length() - 1 , j = b.length() - 1 ; 
		while (i >= 0 || j >= 0) {
			int v1 = i < 0 ? 0 : a.charAt(i) - '0' ;
			int v2 = j <0 ? 0 : b.charAt(j) - '0' ;
			int sum = v1 + v2 + carrier ;
			sb.append(sum % 2) ;
			carrier = sum / 2 ;
			i--;
			j--;
		}
		if (carrier == 1) {
			sb.append(carrier) ;
		}
		sb.reverse() ;
		return sb.toString() ;
	}

- Scott April 22, 2015 | Flag Reply
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0
of 0 votes

Exactly the same solution I ended up with, line for line.

Except...I used (sum ^ 1) >> 1 to calculate the carry. But seems everyone is using sum / 2, which is simpler.

- JW April 26, 2015 | Flag
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0
of 0 vote

private static String sumBinary(String binary1, String binary2)
{
String binary_sum = "";


int carry = 0;

for(int cnt1 = binary1.length()-1,cnt2 = binary2.length()-1; cnt1 >= 0 || cnt2 >=0; cnt1--,cnt2--)
{

int digit1 = 0;
int digit2 = 0;

if(cnt1 >= 0)
digit1 = new Integer(binary1.charAt(cnt1)) - '0';

if(cnt2 >=0)
digit2 = new Integer(binary2.charAt(cnt2)) - '0';

int sum = digit1 + digit2 + carry;

switch(sum)
{
case 0:
binary_sum = "0"+binary_sum;
carry = 0;
break;
case 1:
binary_sum = "1"+binary_sum;
carry = 0;
break;
case 2:
binary_sum = "0"+binary_sum;
carry = 1;
break;
case 3:
binary_sum = "1"+binary_sum;
carry = 1;
break;
}
}

if(carry > 0)
binary_sum = "1"+binary_sum;

return binary_sum;
}

- Satyawan April 22, 2015 | Flag Reply
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0
of 0 vote

#include<iostream>
#include<string.h>

using namespace std;

#define GET_VAL(a) (a=='1'?1:0)
#define GET_STR(a) (a?'1':'0')

int add(const char a, const char b)
{
	if(GET_VAL(a) && GET_VAL(b))
		return 10;
	else
		return (GET_VAL(a) | GET_VAL(b));
}

main()
{
	std::string a = "0100101";//"0111101";
	std::string b = "1101110";//"1101";
	std::string exp = "10010011";//"1001010";

	std::string smallNum, bigNum, result;
	int diff = 0;
	
	if(a.length() < b.length())
	{
		bigNum = b;
		smallNum = a;
		diff = b.length() - a.length();
	}
	else
	{
		bigNum = a;
		smallNum = b;
		diff = a.length() - b.length();
	}
	for(int i=0;i<diff;i++)
	{
		smallNum = '0' + smallNum;
	}
	int carry = 0, j=0;
	for(j=smallNum.length()-1;j>=0;j--)
	{
		int val = add(bigNum[j],smallNum[j]);
		if(carry < 10 && val < 10)
			carry = add(GET_STR(carry),GET_STR(val));
		else
			carry += val;
		int rem = (carry % 10);
		result = GET_STR(rem) + result;
		carry = carry / 10;
	}
	if(carry)
		result = GET_STR(carry) + result;
	cout<<"\nstring 1: "<<bigNum<<"\nstring 2: "<<smallNum<<"\nexpected: "<<exp;
	if(exp == result)
		cout<<"\nPerfect result  : "<<result<<endl;
	else
		cout<<"\nwrong result  : "<<result<<endl;
}

- Madhuri sripada April 22, 2015 | Flag Reply
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0
of 0 vote

#include<iostream>
#include<string.h>

using namespace std;

#define GET_VAL(a) (a=='1'?1:0)
#define GET_STR(a) (a?'1':'0')

int add(const char a, const char b)
{
	if(GET_VAL(a) && GET_VAL(b))
		return 10;
	else
		return (GET_VAL(a) | GET_VAL(b));
}

main()
{
	std::string a = "0100101";//"0111101";
	std::string b = "1101110";//"1101";
	std::string exp = "10010011";//"1001010";

	std::string smallNum, bigNum, result;
	int diff = 0;
	
	if(a.length() < b.length())
	{
		bigNum = b;
		smallNum = a;
		diff = b.length() - a.length();
	}
	else
	{
		bigNum = a;
		smallNum = b;
		diff = a.length() - b.length();
	}
	for(int i=0;i<diff;i++)
	{
		smallNum = '0' + smallNum;
	}
	int carry = 0, j=0;
	for(j=smallNum.length()-1;j>=0;j--)
	{
		int val = add(bigNum[j],smallNum[j]);
		if(carry < 10 && val < 10)
			carry = add(GET_STR(carry),GET_STR(val));
		else
			carry += val;
		int rem = (carry % 10);
		result = GET_STR(rem) + result;
		carry = carry / 10;
	}
	if(carry)
		result = GET_STR(carry) + result;
	cout<<"\nstring 1: "<<bigNum<<"\nstring 2: "<<smallNum<<"\nexpected: "<<exp;
	if(exp == result)
		cout<<"\nPerfect result  : "<<result<<endl;
	else
		cout<<"\nwrong result  : "<<result<<endl;
}

- Madhuri sripada April 22, 2015 | Flag Reply
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0
of 0 vote

#define GET_VAL(a) (a=='1'?1:0)
#define GET_STR(a) (a?'1':'0')

int add(const char a, const char b)
{
	if(GET_VAL(a) && GET_VAL(b))
		return 10;
	else
		return (GET_VAL(a) | GET_VAL(b));
}

main()
{
	std::string a = "0100101";//"0111101";
	std::string b = "1101110";//"1101";
	std::string exp = "10010011";//"1001010";

	std::string smallNum, bigNum, result;
	int diff = 0;
	
	if(a.length() < b.length())
	{
		bigNum = b;
		smallNum = a;
		diff = b.length() - a.length();
	}
	else
	{
		bigNum = a;
		smallNum = b;
		diff = a.length() - b.length();
	}
	for(int i=0;i<diff;i++)
	{
		smallNum = '0' + smallNum;
	}
	int carry = 0, j=0;
	for(j=smallNum.length()-1;j>=0;j--)
	{
		int val = add(bigNum[j],smallNum[j]);
		if(carry < 10 && val < 10)
			carry = add(GET_STR(carry),GET_STR(val));
		else
			carry += val;
		int rem = (carry % 10);
		result = GET_STR(rem) + result;
		carry = carry / 10;
	}
	if(carry)
		result = GET_STR(carry) + result;
	cout<<"\nstring 1: "<<bigNum<<"\nstring 2: "<<smallNum<<"\nexpected: "<<exp;
	if(exp == result)
		cout<<"\nPerfect result  : "<<result<<endl;
	else
		cout<<"\nwrong result  : "<<result<<endl;
}

- Madhuri sripada April 22, 2015 | Flag Reply
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of 0 vote

#include<iostream>
#include<string.h>

using namespace std;

#define GET_VAL(a) (a=='1'?1:0)
#define GET_STR(a) (a?'1':'0')

int add(const char a, const char b)
{
	if(GET_VAL(a) && GET_VAL(b))
		return 10;
	else
		return (GET_VAL(a) | GET_VAL(b));
}

main()
{
	std::string a = "0100101";//"0111101";
	std::string b = "1101110";//"1101";
	std::string exp = "10010011";//"1001010";

	std::string smallNum, bigNum, result;
	int diff = 0;
	
	if(a.length() < b.length())
	{
		bigNum = b;
		smallNum = a;
		diff = b.length() - a.length();
	}
	else
	{
		bigNum = a;
		smallNum = b;
		diff = a.length() - b.length();
	}
	for(int i=0;i<diff;i++)
	{
		smallNum = '0' + smallNum;
	}
	int carry = 0, j=0;
	for(j=smallNum.length()-1;j>=0;j--)
	{
		int val = add(bigNum[j],smallNum[j]);
		if(carry < 10 && val < 10)
			carry = add(GET_STR(carry),GET_STR(val));
		else
			carry += val;
		int rem = (carry % 10);
		result = GET_STR(rem) + result;
		carry = carry / 10;
	}
	if(carry)
		result = GET_STR(carry) + result;
	cout<<"\nstring 1: "<<bigNum<<"\nstring 2: "<<smallNum<<"\nexpected: "<<exp;
	if(exp == result)
		cout<<"\nPerfect result  : "<<result<<endl;
	else
		cout<<"\nwrong result  : "<<result<<endl;

}

- Madhuri Sripada April 22, 2015 | Flag Reply
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of 0 vote

#include<iostream>
#include<string.h>

using namespace std;

#define GET_VAL(a) (a=='1'?1:0)
#define GET_STR(a) (a?'1':'0')

int add(const char a, const char b)
{
	if(GET_VAL(a) && GET_VAL(b))
		return 10;
	else
		return (GET_VAL(a) | GET_VAL(b));
}

main()
{
	std::string a = "0100101";//"0111101";
	std::string b = "1101110";//"1101";
	std::string exp = "10010011";//"1001010";

	std::string smallNum, bigNum, result;
	int diff = 0;
	
	if(a.length() < b.length())
	{
		bigNum = b;
		smallNum = a;
		diff = b.length() - a.length();
	}
	else
	{
		bigNum = a;
		smallNum = b;
		diff = a.length() - b.length();
	}
	for(int i=0;i<diff;i++)
	{
		smallNum = '0' + smallNum;
	}
	int carry = 0, j=0;
	for(j=smallNum.length()-1;j>=0;j--)
	{
		int val = add(bigNum[j],smallNum[j]);
		if(carry < 10 && val < 10)
			carry = add(GET_STR(carry),GET_STR(val));
		else
			carry += val;
		int rem = (carry % 10);
		result = GET_STR(rem) + result;
		carry = carry / 10;
	}
	if(carry)
		result = GET_STR(carry) + result;
	cout<<"\nstring 1: "<<bigNum<<"\nstring 2: "<<smallNum<<"\nexpected: "<<exp;
	if(exp == result)
		cout<<"\nPerfect result  : "<<result<<endl;
	else
		cout<<"\nwrong result  : "<<result<<endl;
}

- Madhuri Sripada April 22, 2015 | Flag Reply
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class SolutionForSumOfTwoBinaryGivenInStringFormat {
    public String solution(String str1, String str2) {
        String out = "";
        String first = str1;
        String secnd = str2;
        if (str2.length() > str1.length()) {
            first = str2;
            secnd = str1;
        }
        int j = secnd.length() - 1;
        int carry = 0;
        for (int i = first.length() - 1; i >= 0; i--) {
            int temp1 = first.charAt(i) - '0';
            int temp2 = 0;
            if (j >= 0) {
                temp2 = secnd.charAt(j--) - '0';
            }
            int sum = temp1 + temp2 + carry;
            // System.out.println("TEMP1: " + temp1 + "\t TEMP2: " + temp2 + "\t CARRY: " + carry +
            // "\t SUM: " + sum);
            if (sum == 0) {
                out = '0' + out;
                carry = 0;
            } else if (sum == 1) {
                out = '1' + out;
                carry = 0;
            } else if (sum == 2) {
                out = '0' + out;
                carry = 1;
            } else if (sum == 3) {
                out = '1' + out;
                carry = 1;
            } else {
                System.out.println("Invalid Entry");
                return null;
            }
        }
        
        if(carry != 0) {
            return carry + out;
        }
        return out;
    }
}

public class SumOfTwoBinaryGivenInStringFormat {

    public static void main(String[] args) {
        SolutionForSumOfTwoBinaryGivenInStringFormat mSol = new SolutionForSumOfTwoBinaryGivenInStringFormat();
        System.out.println(mSol.solution("0111101", "1101"));
        System.out.println(mSol.solution("", ""));
        System.out.println(mSol.solution("1", ""));
        System.out.println(mSol.solution("10101", ""));
        System.out.println(mSol.solution("1111", "11"));
        System.out.println(mSol.solution("1111", "1111"));
    }
}

- Scorpion King April 23, 2015 | Flag Reply
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sum= (firstbit XOR SecondBit XOR carry)
carry = (firstbit AND secondbit) OR (firstbit AND carry) OR (firstbit AND carry)

- puncaz April 23, 2015 | Flag Reply
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String sumBinary(String num1, String num2) {
	
	// Indexes to start from
	int i1 = num1.length() - 1;
	int i2 = num2.length() - 1;

	// characters at index
	int b1 = 0;
	int b2 = 0;

	// sum character
	int c = 0;

	// sum index
	int si = i1 > i2 ? i1 : i2;
	
	// sum char array
	char[] sum = new char[si + 1];
	
	// carry
	int carry = 0;

	// iterate until both the strings are read
	while(i1 >= 0 || i2 >= 0) {
		b1 = 0;
		b2 = 0;
		if(i1 >= 0) {
			b1 = ( (int) num1.charAt(i1) ) & 0x01;
		}
		if(i2 >= 0) {
			b2 = ( (int) num2.charAt(i2) ) & 0x01;
		}

		// compute sum of bits
		c = b1 + b2 + carry;

		// compute carry
		carry = c & 0x03;
		carry = carry >> 1;
		
		// sum bit
		c = c & 0x01;
		
		// put sum bit in the sum char array
		sum[si] = (char) (c + 0x30);

		i1--;
		i2--;
		si--;
	}

	return new String(sum);
}

- blue-j April 24, 2015 | Flag Reply
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of 0 vote

!cc
cc sumBinary.c

./a.out
0111101 + 1101 = 1001010

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

// convert from string binary to integer
int atob(char *a) {
   int n = strlen(a);
   int num = 0;
   int index = 0;

   while (n-- >= 0) {
      if (a[n] == '1') {
         num = num + (1 << index);
      }
      index++;
   }
   return num;
}

// convert an int to string
// the caller needs to delete the memory return
char *itobs(int myInt) {
  char *c = malloc(sizeof (char) * 33);
  int seenOne = 0;
  int cur = 0;

  for (int i = 0; i < 32; i++) {
    if (myInt & (1 << (31 - i)))  {
      c[cur++] = '1';
      seenOne = 1;
    } else {
      if (seenOne == 1) {
         c[cur++] = '0';
      }
    }
  }
  c[cur] = '\0';
  return c;
}

char *sumBinary(char *a, char *b) {
 return itobs(atob(a) + atob(b));
}

int
main (int arc, char *argv[])  {
  printf( "0111101 + 1101 = %s\n", sumBinary("0111101", "1101"));
}

- tam.tony April 25, 2015 | Flag Reply
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of 0 vote

char *sumBinary(char *st1, char *st2)
{
char *str1, *str2, *str3;
int length,carry=0,res=0;

if(strlen(st1) > strlen(st2))
{
	srt1=st1;
	str2=st2;
	str3=(char *) malloc (strlen(st1)+1);
}
else 
{
	srt1=st2;
	str2=st1;
	str3=(char *) malloc (strlen(st2)+1);
}

length1=strlen(str1);
length2=strlen(str2);

while (length2>=0)
{
	res = (str1[length1] - '0') + (str2[length2] - '0') + carry;
	str3[length1+1] = (char)((res % 2) + '0');
	carry = res / 2;
	length1--;
	length2--;
}

while (length1>=0)
{
	res = (str1[length1] - '0') + carry;
	str3[length1+1] = (char)(res % 2 + '0');
	carry = res / 2;
	length1--;
}

str3[0]=(char)(carry + '0');

return str3;
}

- sanjay.2812 May 05, 2015 | Flag Reply
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of 0 vote

long al = Long.parseLong(a, 2);
		long bl = Long.parseLong(b, 2);
		
		long result = al + bl;
		System.out.println(Long.toBinaryString(result));

- Fabio May 15, 2015 | Flag Reply
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My approach would be just convert binary numbers to integers (simple multiplication by power of 2), then add two decimals. Let k be number length binary number. So binary to decimal conversion will take O(n) time. Convert decimal answer back to binary. (log (k) time.)

This will avoid most of heavy if - else loops

- AP May 21, 2015 | Flag Reply
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public class BinaryAddition {

public static void main(String[] args){
String a="0111101";
String b="1101";

int num1=Integer.parseInt(a,2);
int num2=Integer.parseInt(b,2);

System.out.println(num1+ " " +num2);

int sum=num1+num2;

String tot=Integer.toBinaryString(sum);
System.out.println("Total:" +tot);

}

}

- Vigneshwaran June 02, 2015 | Flag Reply
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of 0 vote

public class BinaryAddition {

public static void main(String[] args){
String a="0111101";
String b="1101";

int num1=Integer.parseInt(a,2);
int num2=Integer.parseInt(b,2);

System.out.println(num1+ " " +num2);

int sum=num1+num2;

String tot=Integer.toBinaryString(sum);
System.out.println("Total:" +tot);

}

}

- Vigneshwaran June 02, 2015 | Flag Reply
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of 0 vote

public class BinaryAddition {

public static void main(String[] args){
String a="0111101";
String b="1101";

int num1=Integer.parseInt(a,2);
int num2=Integer.parseInt(b,2);

System.out.println(num1+ " " +num2);

int sum=num1+num2;

String tot=Integer.toBinaryString(sum);
System.out.println("Total:" +tot);

}

}

- Vigneshwaran June 02, 2015 | Flag Reply
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of 0 vote

public class BinaryAddition {

public static void main(String[] args){
String a="0111101";
String b="1101";

int num1=Integer.parseInt(a,2);
int num2=Integer.parseInt(b,2);

System.out.println(num1+ " " +num2);

int sum=num1+num2;

String tot=Integer.toBinaryString(sum);
System.out.println("Total:" +tot);

}

}

- Vigneshwaran June 02, 2015 | Flag Reply
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public class BinaryAddition {
	
	public static void main(String[] args){
		String a="0111101";
		String b="1101";
		
		int num1=Integer.parseInt(a,2);
		int num2=Integer.parseInt(b,2);
		
		System.out.println(num1+ " " +num2);
		
		int sum=num1+num2;
		
		String tot=Integer.toBinaryString(sum);
		System.out.println("Total:" +tot);
		
	}

}

- Vigneshwaran June 02, 2015 | Flag Reply
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of 0 vote

public class BinaryAddition {
	
	public static void main(String[] args){
		String a="0111101";
		String b="1101";
		
		int num1=Integer.parseInt(a,2);
		int num2=Integer.parseInt(b,2);
		
		System.out.println(num1+ " " +num2);
		
		int sum=num1+num2;
		
		String tot=Integer.toBinaryString(sum);
		System.out.println("Total:" +tot);
		
	}

}

- Vigneshwaran June 02, 2015 | Flag Reply
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of 0 vote

public class BinaryAddition {	
	public static void main(String[] args){
		String a="0111101";
		String b="1101";
		
		int num1=Integer.parseInt(a,2);
		int num2=Integer.parseInt(b,2);
		
		System.out.println(num1+ " " +num2);
		
		int sum=num1+num2;
		
		String tot=Integer.toBinaryString(sum);
		System.out.println("Total:" +tot);
		
	}

}

- Vigneshwaran June 02, 2015 | Flag Reply
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of 0 vote

public class BinaryAddition {	
	public static void main(String[] args){
		String a="0111101";
		String b="1101";
		
		int num1=Integer.parseInt(a,2);
		int num2=Integer.parseInt(b,2);
		
		System.out.println(num1+ " " +num2);
		
		int sum=num1+num2;
		
		String tot=Integer.toBinaryString(sum);
		System.out.println("Total:" +tot);
		
	}

}

- Vigneshwaran June 02, 2015 | Flag Reply
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of 0 vote

public class BinaryAddition {public static void main(String[] args){String a="0111101";String b="1101";int num1=Integer.parseInt(a,2);int num2=Integer.parseInt(b,2);
System.out.println(num1+ " " +num2);int sum=num1+num2;String tot=Integer.toBinaryString(sum);System.out.println("Total:" +tot);}}

- Vigneshwaran June 02, 2015 | Flag Reply
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of 0 vote

public class BinaryAddition {public static void main(String[] args){String a="0111101";String b="1101";int num1=Integer.parseInt(a,2);int num2=Integer.parseInt(b,2);
System.out.println(num1+ " " +num2);int sum=num1+num2;String tot=Integer.toBinaryString(sum);System.out.println("Total:" +tot);}}

- Vigneshwaran June 02, 2015 | Flag Reply
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of 0 vote

package proj2;

public class AddBinaryStrings {

	public AddBinaryStrings(String string, String string2) {
		
		int carry = 0;
		int i = string.length()-1;
		int j = string2.length()-1;
		StringBuilder sb = new StringBuilder();
		
		while(i >=0 && j >=0){
			int a = string.charAt(i)-'0';
			int b = string2.charAt(j)-'0';
			int c = a + b + carry;
			sb.append(c%2);
			carry = c/2;
			i--;
			j--;
		}
		while(i >= 0){
			int a = string.charAt(i)-'0';
			int c = a + carry;
			sb.append(c%2);
			carry = c/2;
			i--;
		}
		while(j >= 0){
			int b = string2.charAt(j)-'0';
			int c = b + carry;
			sb.append(c%2);
			carry = c/2;
			j--;
		}
		if (carry == 1) {
			sb.append(carry) ;
		}
		sb.reverse() ;
		System.out.println(sb);
		
	}

}

- Anonymous July 06, 2015 | Flag Reply
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public static String binarySum(String s1, String s2){
        int num1 = Integer.parseInt(s1, 2);
        int num2 = Integer.parseInt(s2, 2);
        int sum = Integer.sum(num1, num2);
        return Integer.toBinaryString(sum);
    }

- Spectral May 31, 2016 | Flag Reply
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of 1 vote

public static String sumBinary(String binary1, String binary2)
	   {
		   char[] binary1Array = binary1.toCharArray();
		   char[] binary2Array = binary2.toCharArray();
		
		   int length = binary1.length() > binary2.length() ? binary1.length() + 1 : binary2.length() +1;
		   char[] sum = new char[length];
		   int i=binary1Array.length-1, j=binary2Array.length-1, k= length-1;
		   char carry = '0';
		   while (i >= 0 && j >= 0)
		   {
			   if (binary1Array[i] == '0' && binary2Array[j] == '0' && carry == '0') { sum[k] = '0'; carry = '0'; } else
			   if (binary1Array[i] == '1' && binary2Array[j] == '0' && carry == '0') { sum[k] = '1'; carry = '0'; } else
			   if (binary1Array[i] == '0' && binary2Array[j] == '1' && carry == '0') { sum[k] = '1'; carry = '0'; } else
			   if (binary1Array[i] == '0' && binary2Array[j] == '0' && carry == '1') { sum[k] = '1'; carry = '0'; } else
			   if (binary1Array[i] == '1' && binary2Array[j] == '1' && carry == '0') { sum[k] = '0'; carry = '1'; } else
			   if (binary1Array[i] == '1' && binary2Array[j] == '0' && carry == '1') { sum[k] = '0'; carry = '1'; } else
			   if (binary1Array[i] == '0' && binary2Array[j] == '1' && carry == '1') { sum[k] = '0'; carry = '1'; } else
			   if (binary1Array[i] == '1' && binary2Array[j] == '1' && carry == '1') { sum[k] = '1'; carry = '1'; }
			   i--;
			   j--;
			   k--;
		   }
		   if (i < 0 && j >=0)
		   {
			   while (j >=0)
			   {
				   if (binary2Array[j] == '1' && carry == '0') {sum[k] = '1'; carry = '0';} else
				   if (binary2Array[j] == '0' && carry == '1') {sum[k] = '1'; carry = '0';} else
				   if (binary2Array[j] == '1' && carry == '1') {sum[k] = '0'; carry = '1';} 
				   j--;
				   k--;
			   }
		   }
		   if ( j < 0 && i >= 0)
		   {
			   while (i >=0)
			   {
				   if (binary1Array[i] == '1' && carry == '0') {sum[k] = '1'; carry = '0';} else
				   if (binary1Array[i] == '0' && carry == '1') {sum[k] = '1'; carry = '0';} else
				   if (binary1Array[i] == '1' && carry == '1') {sum[k] = '0'; carry = '1';}
				   i--;
				   k--;
			   }
		   }
		   
		   if (carry == '1') sum[k] = carry;
		   return new String(sum);
	   }

- Javacode April 21, 2015 | Flag Reply
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-1
of 1 vote

public static String sumBinary(String binary1, String binary2)
	   {
		   char[] binary1Array = binary1.toCharArray();
		   char[] binary2Array = binary2.toCharArray();
		
		   int length = binary1.length() > binary2.length() ? binary1.length() + 1 : binary2.length() +1;
		   char[] sum = new char[length];
		   int i=binary1Array.length-1, j=binary2Array.length-1, k= length-1;
		   char carry = '0';
		   while (i >= 0 && j >= 0)
		   {
			   if (binary1Array[i] == '0' && binary2Array[j] == '0' && carry == '0') { sum[k] = '0'; carry = '0'; } else
			   if (binary1Array[i] == '1' && binary2Array[j] == '0' && carry == '0') { sum[k] = '1'; carry = '0'; } else
			   if (binary1Array[i] == '0' && binary2Array[j] == '1' && carry == '0') { sum[k] = '1'; carry = '0'; } else
			   if (binary1Array[i] == '0' && binary2Array[j] == '0' && carry == '1') { sum[k] = '1'; carry = '0'; } else
			   if (binary1Array[i] == '1' && binary2Array[j] == '1' && carry == '0') { sum[k] = '0'; carry = '1'; } else
			   if (binary1Array[i] == '1' && binary2Array[j] == '0' && carry == '1') { sum[k] = '0'; carry = '1'; } else
			   if (binary1Array[i] == '0' && binary2Array[j] == '1' && carry == '1') { sum[k] = '0'; carry = '1'; } else
			   if (binary1Array[i] == '1' && binary2Array[j] == '1' && carry == '1') { sum[k] = '1'; carry = '1'; }
			   i--;
			   j--;
			   k--;
		   }
		   if (i < 0 && j >=0)
		   {
			   while (j >=0)
			   {
				   if (binary2Array[j] == '1' && carry == '0') {sum[k] = '1'; carry = '0';} else
				   if (binary2Array[j] == '0' && carry == '1') {sum[k] = '1'; carry = '0';} else
				   if (binary2Array[j] == '1' && carry == '1') {sum[k] = '0'; carry = '1';} 
				   j--;
				   k--;
			   }
		   }
		   if ( j < 0 && i >= 0)
		   {
			   while (i >=0)
			   {
				   if (binary1Array[i] == '1' && carry == '0') {sum[k] = '1'; carry = '0';} else
				   if (binary1Array[i] == '0' && carry == '1') {sum[k] = '1'; carry = '0';} else
				   if (binary1Array[i] == '1' && carry == '1') {sum[k] = '0'; carry = '1';}
				   i--;
				   k--;
			   }
		   }
		   
		   if (carry == '1') sum[k] = carry;
		   return new String(sum);
	   }

- Java April 21, 2015 | Flag Reply
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-1
of 1 vote

use sum and carry technique

Sum(string s1, string s2)
{
int [] n1 = new int[s1.Length];
int [] n2 = new int[s2.Length];
int max = s1.Length > s2.Length ? s1.Length:S2.Length;

for(i=0;i<s1.Length;i++) n1[s1.Length - 1 - i] = (s1.charAt(i) == '0') ? 0 : 1;
for(i=0;i<s2.Length;i++) n2[s2.Length - 1 - i] = (s2.charAt(i) == '0') ? 0 : 1;

int carry = new int[((s1.Length > s2.Length) ? s1.Length + 1 : s2.Length + 1) * 2];
int sum = new int[(s1.Length > s2.Length) ? s1.Length + 1 : s2.Length + 1];
bool loop = false;
int i=0;
do
{
loop = false;
if(i==0)
{
for(i=0;i<max;i++)
{
sum[i] = ((i<s1.Length) ? s1[i] : 0) ^ ((i<s2.Length) ? s2[i] : 0);
carry[i+1] = ((i<s1.Length) ? s1[i] : 0) & ((i<s2.Length) ? s2[i] : 0);
loop |= (carry[i+1] == 1) ? true ? false;
}
}
else
{
for(i=0;i<max;i++)
{
sum[i] = (sum[i]) ^ (carry[i]);
carry[i+1] = (sum[i]) & (carry[i]);
loop |= (carry[i+1] == 1) ? true ? false;
}
max = i+1;
}
}while(loop);

StringBuilder sb = new StringBuilder();
for(i=max;i>=0;i--) sb.append(sum[i]);
return sb.ToString();
}

- aditya.eagle059 April 21, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

def sumbinaryNo(val1,val2):
    maxlen = len(val1)
    sumStr = ""
    addDigit=0
    char0 = "0"
    char1="0"
    extraZero = 0
    extraZeroStr = ""
    if len(val2)>len(val1):
        extraZero = len(val2)-len(val1)
        maxlen = len(val2)
        print("extraZero  "+str(extraZero))
        for i in range(0,extraZero):
            extraZeroStr = "0"+extraZeroStr
        val1 = extraZeroStr+val1
    elif len(val1)>len(val2):
        extraZero = len(val1)-len(val2)
        maxlen = len(val1)
        for i in range(0,extraZero):
            extraZeroStr = "0"+extraZeroStr
        val2 = extraZeroStr+val2
    print(val1)
    print(val2)
    for i in range(maxlen-1,-1,-1):
        # print(i)
        char0 = int(val1[i])
        char1 = int(val2[i])
        print(char0+char1+addDigit)
        if char0+char1+addDigit == 3:
            sumStr = "1"+sumStr;
            addDigit = 1
        elif char0+char1+addDigit == 2:
            sumStr = "0"+sumStr;
            addDigit = 1
        else:
            sumStr = str(char0+char1+addDigit)+sumStr;
            addDigit = 0
    print(str(addDigit)+sumStr)


sumbinaryNo('11001111','101000000')

- Kiran Patil April 22, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

using System;

namespace binary
{
    class binarySum
    {
        public static string result = string.Empty;
        public static int carryOver = 0;

        static void Main(string[] args)
        {
            Console.WriteLine(sumBinary("0111101", "1101"));
            Console.Read();
        }

        public static string sumBinary(string number1, string number2)
        {
            if (string.IsNullOrEmpty(number1))
            {
                if (string.IsNullOrEmpty(number2))
                    return string.Empty;
                return number2;
            }

            if (string.IsNullOrEmpty(number2))
                return number1;

            int upperLimit = Math.Max(number1.Length, number2.Length);
            int n1 = number1.Length;
            int n2 = number2.Length;
            int val1 = 0, val2 = 0, sum = 0;

            for (int i = 1; i < upperLimit; i++)
            {
                if (n1 - i > -1 && n2 - i > -1)
                {
                    val1 = int.Parse(number1[n1 - i].ToString());
                    val2 = int.Parse(number2[n2 - i].ToString());
                    sum = val1 + val2 + carryOver;
                    setBit(sum);
                }
                else if (n1 - i > -1)
                {
                    val1 = int.Parse(number1[n1 - i].ToString());
                    sum = val1 + carryOver;
                    setBit(sum);
                }
                else if (n2 - i > -1)
                {
                    val2 = int.Parse(number2[n2 - i].ToString());
                    sum = val2 + carryOver;
                    setBit(sum);
                }
            }
            if (carryOver == 1)
                return 1 + result;
            return result;
        }

        public static void setBit(int sum)
        {
            switch (sum)
            {
                case 0:
                    result = result.Insert(0, "0");
                    carryOver = 0;
                    break;
                case 1:
                    result = result.Insert(0, "1");
                    carryOver = 0;
                    break;
                case 2:
                    result = result.Insert(0, "0");
                    carryOver = 1;
                    break;
                case 3:
                    result = result.Insert(0, "1");
                    carryOver = 1;
                    break;
                default:
                    break;
            }

        }
    }
}

- sa-chin April 22, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

#include<iostream>
#include<string.h>

using namespace std;

#define GET_VAL(a) (a=='1'?1:0)
#define GET_STR(a) (a?'1':'0')

int add(const char a, const char b)
{
	if(GET_VAL(a) && GET_VAL(b))
		return 10;
	else
		return (GET_VAL(a) | GET_VAL(b));
}

main()
{
	std::string a = "0100101";//"0111101";
	std::string b = "1101110";//"1101";
	std::string exp = "10010011";//"1001010";

	std::string smallNum, bigNum, result;
	int diff = 0;
	
	if(a.length() < b.length())
	{
		bigNum = b;
		smallNum = a;
		diff = b.length() - a.length();
	}
	else
	{
		bigNum = a;
		smallNum = b;
		diff = a.length() - b.length();
	}
	for(int i=0;i<diff;i++)
	{
		smallNum = '0' + smallNum;
	}
	int carry = 0, j=0;
	for(j=smallNum.length()-1;j>=0;j--)
	{
		int val = add(bigNum[j],smallNum[j]);
		if(carry < 10 && val < 10)
			carry = add(GET_STR(carry),GET_STR(val));
		else
			carry += val;
		int rem = (carry % 10);
		result = GET_STR(rem) + result;
		carry = carry / 10;
	}
	if(carry)
		result = GET_STR(carry) + result;
	cout<<"\nstring 1: "<<bigNum<<"\nstring 2: "<<smallNum<<"\nexpected: "<<exp;
	if(exp == result)
		cout<<"\nPerfect result  : "<<result<<endl;
	else
		cout<<"\nwrong result  : "<<result<<endl;
}

- Madhuri Sripada April 22, 2015 | Flag
Comment hidden because of low score. Click to expand.
-1
of 1 vote
{{{ package bitpattern; public class BinarySum { public static void main(String[] args) { String no1 = Integer.toBinaryString(178); String no2 = Integer.toBinaryString(284); String sum = sumBinary(no1,no2); System.out.println("no1 "+no1); System.out.println("no2 "+no2); System.out.println("sum "+sum); } private static String sumBinary(String binary1, String binary2) { String binary_sum = ""; int carry = 0; for(int cnt1 = binary1.length()-1,cnt2 = binary2.length()-1; cnt1 >= 0 || cnt2 >=0; cnt1--,cnt2--) { int digit1 = 0; int digit2 = 0; if(cnt1 >= 0) digit1 = new Integer(binary1.charAt(cnt1)) - '0'; if(cnt2 >=0) digit2 = new Integer(binary2.charAt(cnt2)) - '0'; int sum = digit1 + digit2 + carry; switch(sum) { case 0: binary_sum = "0"+binary_sum; carry = 0; break; case 1: binary_sum = "1"+binary_sum; carry = 0; break; case 2: binary_sum = "0"+binary_sum; carry = 1; break; case 3: binary_sum = "1"+binary_sum; carry = 1; break; } } if(carry > 0) binary_sum = "1"+binary_sum; return binary_sum; } } }} - Pravin April 22, 2015 | Flag Reply


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