Amazon Interview Question for SDE-2s


Country: United States
Interview Type: In-Person




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

public class StringReverse {

    public static void main(String[] args) {
        String input = "aabdceaaabbbcd";
        reverseWithOneCharacterOnce(input);
    }

    private static void reverseWithOneCharacterOnce(String input) {

        Set<Character> alreadyPrintedCharacter = new HashSet<Character>();
        String reversed = "";

        for (int index = input.length() - 1; index >= 0; index--) {
            Character ch = input.charAt(index);
            if (!alreadyPrintedCharacter.contains(ch)) {
                alreadyPrintedCharacter.add(ch);
                reversed = reversed + ch;
            }
        }

        System.out.println(reversed);

    }

}

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

Can be implemented with DS - combination of hash map & stack.
1. put data in above DS
2. While putting data in step-1, assign head with new node in HashMap which was inserted in step-1
3. after completing the iteration on the input string & putting all data in above DS using step-1& step-2, start reading data from above DS with stack.pop() operation

Run-time complexity - O(N)
Memory complexity - O(N)

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

suppose k = "aabdceaaabbbcd"
temp = []
string_list = list(reversed(k))
def get_unique_with_reverse(string_list):
for i in string_list:
if not i in temp:
temp.append(i)
print "".join(temp)

- amit.barnwal09 June 02, 2016 | Flag Reply
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0
of 0 vote

void solve(string s) {
	reverse(s.begin(), s.end());
	set<char> seen;
	string ans;
	for (auto c : s) {
		if (!seen.count(c)) {
			ans += c;
			seen.insert(c);	
		}
	}
	cout << c;
}

- Anonymous May 31, 2016 | Flag Reply
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0
of 0 vote

public class StringReverse {

    public static void main(String[] args) {
        String input = "aabdceaaabbbcd";
        reverseWithOneCharacterOnce(input);
    }

    private static void reverseWithOneCharacterOnce(String input) {

        Set<Character> alreadyPrintedCharacter = new HashSet<Character>();
        String reversed = "";

        for (int index = input.length() - 1; index >= 0; index--) {
            Character ch = input.charAt(index);
            if (!alreadyPrintedCharacter.contains(ch)) {
                alreadyPrintedCharacter.add(ch);
                reversed = reversed + ch;
            }
        }

        System.out.println(reversed);

    }

}

- Hi May 31, 2016 | Flag Reply
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0
of 0 vote

private static String reversedUnique(String input)
	{
		StringBuilder sb = new StringBuilder();
		
		for(int i=input.length()-1;i>0;i--)
		{
			if(!sb.toString().contains(input.substring(i, i+1))){
				
				sb.append(input.substring(i, i+1));
			}
		}
		return sb.toString();
	}

- harsh May 31, 2016 | Flag Reply
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0
of 0 vote

private static String reversedUnique(String input)
	{
		StringBuilder sb = new StringBuilder();
		
		for(int i=input.length()-1;i>0;i--)
		{
			if(!sb.toString().contains(input.substring(i, i+1))){
				
				sb.append(input.substring(i, i+1));
			}
		}
		return sb.toString();
	}

- harshmighlani May 31, 2016 | Flag Reply
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0
of 0 vote

void solution(String in)
{
	HashMap<Character> hm = new HashMap<~> ();
	for(int i=in.length-1;i>0;i--)
	{
	
		Character c = in.charAt(i);
		if(!hm.containsKey(c))
		{
			hm.add(c);
			System.out.print(c);
		}	

	}
}

- skumaringuva May 31, 2016 | Flag Reply
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0
of 0 vote

1. Create data-structure - hybrid of HashMap & Stack. Head = pointing to the top of the stack. Put() - put value in hash map & changes head to point to new value inserted.
2. Start iterating over the input of string
3. After completing the iteration in step-2, start stack.pop(head) & print till stack is empty.

- thakur.871 May 31, 2016 | Flag Reply
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0
of 0 vote

#include <bits/stdc++.h>
#include<queue>
#include<vector>
#define ll long long int
using namespace std;
class Word
{
public:
int index;
char w;
};
bool comp(Word a,Word b)
{

return (a.index > b.index);
}
int main()
{
vector<Word> v1;
unordered_map<char,int> u1;
string str="aabdceaaabbbcd";
for(int i= str.length() - 1;i>=0;i--)
{
Word obj;
obj.index=i;
obj.w=str[i];
if(u1.count(str[i])==0)
{
u1[str[i]]=1;
v1.push_back(obj);
}
}

sort(v1.begin(),v1.end(),comp);
for (int i=0;i<v1.size();i++)
{
cout<<v1[i].w;
}
return 0;
};

- Anonymous May 31, 2016 | Flag Reply
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0
of 0 vote

#include <bits/stdc++.h>
#include<queue>
#include<vector>
#define ll long long int
using namespace std;
class Word
{
public:
int index;
char w;
};
bool comp(Word a,Word b)
{

return (a.index > b.index);
}
int main()
{
vector<Word> v1;
unordered_map<char,int> u1;
string str="aabdceaaabbbcd";
for(int i= str.length() - 1;i>=0;i--)
{
Word obj;
obj.index=i;
obj.w=str[i];
if(u1.count(str[i])==0)
{
u1[str[i]]=1;
v1.push_back(obj);
}
}

sort(v1.begin(),v1.end(),comp);
for (int i=0;i<v1.size();i++)
{
cout<<v1[i].w;
}
return 0;
};

- Anonymous May 31, 2016 | Flag Reply
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0
of 0 vote

#include <bits/stdc++.h>
#include<queue>
#include<vector>
#define ll long long int
using namespace std;
class Word
{
public:
int index;
char w;
};
bool comp(Word a,Word b)
{

return (a.index > b.index);
}
int main()
{
vector<Word> v1;
unordered_map<char,int> u1;
string str="aabdceaaabbbcd";
for(int i= str.length() - 1;i>=0;i--)
{
Word obj;
obj.index=i;
obj.w=str[i];
if(u1.count(str[i])==0)
{
u1[str[i]]=1;
v1.push_back(obj);
}
}

sort(v1.begin(),v1.end(),comp);
for (int i=0;i<v1.size();i++)
{
cout<<v1[i].w;
}
return 0;
};

- bhanu870 May 31, 2016 | Flag Reply
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0
of 0 vote

Traverse the characters of the string in reverse order and use java.util.LinkedHashSet.

- RoBa May 31, 2016 | Flag Reply
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0
of 0 vote

public static String printReversedOnlyOnce(String s) {
		if (s == null || s.equals("")) {
			return null;
		}		
		StringBuilder sb = new StringBuilder();
		char[] c = s.toCharArray();
		Map<Character, Boolean> mapChars = new HashMap<>();
		for (int i = c.length - 1; i >= 0; i--) {
			if (!mapChars.containsKey(c[i])) {
				sb.append(c[i]);
				mapChars.put(c[i], Boolean.TRUE);
			}
		}
		return sb.toString();
	}

- guilhebl May 31, 2016 | Flag Reply
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0
of 0 vote

#include <bits/stdc++.h>
#include<queue>
#include<vector>
#define ll long long int
using namespace std;

int main()
{
unordered_map<char,int> u1;
string str="aabdceaaabbbcd";
for(int i= str.length() - 1;i>=0;i--)
{

if(u1.count(str[i])==0)
		{
		u1[str[i]]=1;
		cout<<str[i];

}

return 0;
};

- bhanu870 May 31, 2016 | Flag Reply
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0
of 0 vote

#include <bits/stdc++.h>
#include<queue>
#include<vector>
#define ll long long int
using namespace std;

int main()
{
unordered_map<char,int> u1;
string str="aabdceaaabbbcd";
for(int i= str.length() - 1;i>=0;i--)
{
		if(u1.count(str[i])==0)
		{
		u1[str[i]]=1;
		cout<<str[i];
	     }
}

return 0;	
}

- bhanu870 May 31, 2016 | Flag Reply
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0
of 0 vote

/* print all characters present in the given string only once in a revers order*/

import java.io.*;
import java.util.*;

public class SringReverseOnce {

public static void main(String args[])
{
Scanner sc = new Scanner(System.in);

System.out.print("Enter String input: ");
String in = sc.next();

int n = in.length();
char buf[] = in.toCharArray();


for(int i=n-1;i>=0;i--)
{ int flag;
 flag =0;
	for(int j=n-1;j>=i;j--)
	{ 
		if(buf[i]==buf[j] && j!=i)
		{ flag =1;
			break;
		}
	
	}	
	
	if(flag==0)
	{System.out.print(buf[i]);}
}

}
}

- Devesh June 01, 2016 | Flag Reply
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0
of 0 vote

define boolean array of character set with false value.
if characters ascii value location of character set is false, set it true and print character,
otherwise ignore the character.

/**
     * Print String in reverse order but without any duplicate characters 
     * Assume character set is asci characters 0 - 255
     */
    private static boolean printStringRevNoDuplicate(String str){
        int setSize = 255;
        boolean[] charSet = new boolean[setSize];
        for(int index = str.length() - 1; index >= 0; --index){
            if(!charSet[str.charAt(index)]){
                charSet[str.charAt(index)] = true;
                System.out.print(str.charAt(index));
            }
        }
        System.out.println();
        return true;

}

- Anonymous June 01, 2016 | Flag Reply
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0
of 0 vote

def solution(s):
    result = []
    table = [0] * 256
    for x in reversed(s):
        if not table[ord(x)]:
           result.append(x)
        table[ord(x)] += 1
    print(result)
    
    # Extra feature: Print the char appears only once in reversed order
    print('Print the char shows only once')
    reduced = [x for x in result if table[ord(x)] == 1]
    print(reduced)

Time complexity: O(n)

- cnhnyu June 01, 2016 | Flag Reply
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0
of 0 vote

def rev_char_set(word):
	rev_word = list(reversed(word))
	tem_list = rev_word

	result = []

	for i in rev_word:
		if i in tem_list:
			result.append(i)
			tem_list = [x for x in tem_list if x != i]

	return ''.join(result)

- Anonymous June 01, 2016 | Flag Reply
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0
of 0 vote

def rev_char_set(word):
	rev_word = list(reversed(word))
	tem_list = rev_word

	result = []

	for i in rev_word:
		if i in tem_list:
			result.append(i)
			tem_list = [x for x in tem_list if x != i]

	return ''.join(result)

- Anonymous June 01, 2016 | Flag Reply
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0
of 0 vote

private static String reverse(String string) {
StringBuffer s1 = new StringBuffer(string);
s1.reverse();
System.out.println(s1);
for (int i = 0; i < s1.length(); i++) {
for (int j = i + 1; j < s1.length(); j++) {
if (s1.charAt(i) == s1.charAt(j)) {
s1.deleteCharAt(j);
} else {
continue;
}
}
}
return s1.toString();
}

- Anonymous June 01, 2016 | Flag Reply
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0
of 0 vote

private static String reverse(String string) {
StringBuffer s1 = new StringBuffer(string);
s1.reverse();
System.out.println(s1);
for (int i = 0; i < s1.length(); i++) {
for (int j = i + 1; j < s1.length(); j++) {
if (s1.charAt(i) == s1.charAt(j)) {
s1.deleteCharAt(j);
} else {
continue;
}
}
}
return s1.toString();
}

- razo June 01, 2016 | Flag Reply
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0
of 0 vote

private static String reverse(String string) {
        StringBuffer s1 = new StringBuffer(string);
        s1.reverse();
        System.out.println(s1);
        for (int i = 0; i < s1.length(); i++) {
            for (int j = i + 1; j < s1.length(); j++) {
                if (s1.charAt(i) == s1.charAt(j)) {
                    s1.deleteCharAt(j);
                } else {
                    continue;
                }
            }
        }
        return s1.toString();
    }

- razo June 01, 2016 | Flag Reply
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0
of 0 vote

private static String reverse(String string) {
StringBuffer s1 = new StringBuffer(string);
s1.reverse();
System.out.println(s1);
for (int i = 0; i < s1.length(); i++) {
for (int j = i + 1; j < s1.length(); j++) {
if (s1.charAt(i) == s1.charAt(j)) {
s1.deleteCharAt(j);
} else {
continue;
}
}
}
return s1.toString();
}

- razo June 01, 2016 | Flag Reply
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0
of 0 vote

private char array[] = new char[26];
	private void reverseString(String str, int index) {
		if (index < 0) {
			return;
		} else {
			char sym = str.charAt(index);
			index = index - 1;
			if (array[sym - 97] == '\u0000') {
				System.out.print(sym);
				array[sym - 97] = sym;
			}
			reverseString(str, index);
		}

	}

- verma82 June 01, 2016 | Flag Reply
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0
of 0 vote

def reversed_letters(string):
	outstring = ''
	for c in string[::-1]:
		if c in outstring:
			pass
		else:
			outstring += c

	print oustring

- Nate June 01, 2016 | Flag Reply
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0
of 0 vote

$value = 'aaafffcccddaabbeeddhhhaaabbccddaaaa';
$final_array = array();
$len = strlen($value) - 1;
for ($i = $len; $i >= 0; $i--) {
$a = $value[$i];
$final_array[$a] = $value[$i];
}
echo implode('', $final_array);

- amitinfo86 June 01, 2016 | Flag Reply
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0
of 0 vote

Here is a C++ implementation

#include <iostream>
#include <unordered_set>
#include <string>

void printReverseUniqueString(std::string inputStr)
{
    std::unordered_set<char> uniqueChars;
    std::string uniqueString;
    for(std::string::iterator i = inputStr.end(); i!= inputStr.begin(); i--)
    {
        if(uniqueChars.count(*i) == 0)
        {
            uniqueString.push_back(*i);
            uniqueChars.insert(*i);
        }
    }
    std::cout << uniqueString << std::endl;
}

- Anonymous June 01, 2016 | Flag Reply
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0
of 0 vote

Here is a C++ implementation

#include <iostream>
#include <unordered_set>
#include <string>

void printReverseUniqueString(std::string inputStr)
{
    std::unordered_set<char> uniqueChars;
    std::string uniqueString;
    for(std::string::iterator i = inputStr.end(); i!= inputStr.begin(); i--)
    {
        if(uniqueChars.count(*i) == 0)
        {
            uniqueString.push_back(*i);
            uniqueChars.insert(*i);
        }
    }
    std::cout << uniqueString << std::endl;
}

- Anirudh June 01, 2016 | Flag Reply
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0
of 0 vote

Steps:
1.Go to the last of that string
2. Check a bit in int of he bit is set
3. If not set, print and set the bit.
4. repeat until the beginning.

- Rsl June 01, 2016 | Flag Reply
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0
of 0 vote

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

using namespace std;

int main(int argc, char **argv)
{
   if(argc != 2)
   {
      printf("usage: %s [test_chars]\n", argv[0]);
      return 0;
   }

   char *test_chars = argv[1];
   char temp[256];

   memset(temp, 0, sizeof(temp));

   for(int i = strlen(test_chars); i >= 0; i--)
   {
      if(temp[test_chars[i]] == 0)
      {
         printf("%c", test_chars[i]);
         temp[test_chars[i]]++;
      }
   }

   printf("\n");

   return 0;
}

- sears1234 June 02, 2016 | Flag Reply
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0
of 0 vote

{{ static string StringManipulatorReverseUnique(string s)
{
char[] charArray = s.ToCharArray();
return new string(charArray.Reverse().Distinct().ToArray());
}}}

- Random Guy June 03, 2016 | Flag Reply
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0
of 0 vote

public class Solution {
    public String removeDuplicateLetters(String s) {
        boolean[] validate=new boolean[300];
        StringBuilder sb = new StringBuilder();
        for(char c:s.toCharArray()){
            if(validate[c]){
                continue;
            }
            validate[c]=true;
        }
        for(int i=299;i>=0;i++){
            if(validate[i]){
                sb.append((char)i);
            }
        }
        return sb.toString();
    }

}

- Anonymous June 03, 2016 | Flag Reply
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0
of 0 vote

/* We can solve this by using HashMap or HashSet. Hope this is also one of the effective solution */

public class ReverseArrayOnlyOneCharacter {
	public static void main(String[] args) {
		String str = "aabdceaaabbbcd";
		int[] tmparr = new int[str.length()];
		for (int i = str.length() - 1; i >= 0; i--) {
			int ch = (int) str.charAt(i);
			int mod = (((ch - 97) % 26) % (str.length() - 1));
			if (tmparr[mod] == ch) {
				continue;
			} else {
				System.out.println((char)ch);
				tmparr[mod] = ch;
			}
		}
	}
}

- anand June 03, 2016 | Flag Reply
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0
of 0 vote

String str = "aaaabbcddddccbbdaaeee";    
        int end = str.length()-1;
        String output="";

        while(str.length() !=0 && end>=0){
            if(output.indexOf(str.charAt(end)) == -1) output +=str.charAt(end);
            end--;
        }
        System.out.println(output);

- Painter Mathavan June 03, 2016 | Flag Reply
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0
of 0 vote

int visited[255];
void print_rev(char *str, int index, int len)
{
if(index < 0)
return;
visited[(str[index])%len] = 1;
printf("%c", str[index]);
index--;
while(visited[(str[index])%len])
{

index = index-1;

}
print_rev(str, index, len);

}


int main()
{
char *str1 = "aabdceaaabbbcd";
int len = strlen(str1);
int i = 0;
for(i = 0; i<len; i++)
visited[i] = 0;

print_rev(str1, len-1, len);
return 0;
}

- Swati June 04, 2016 | Flag Reply
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0
of 0 vote

/* I am using visited array which is just like hash table */

void print_rev(char *str, int index, int len)
{
	if(index < 0)
		return;
	visited[(str[index])%len] = 1;
		printf("%c", str[index]);
	index--;
	while(visited[(str[index])%len])
	{
		
		index = index-1;		
		
	}
	print_rev(str, index, len);

}

- Swati June 04, 2016 | Flag Reply
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0
of 0 vote

{
void print_rev(char *str, int index, int len)
{
	if(index < 0)
		return;
	visited[(str[index])%len] = 1;
		printf("%c", str[index]);
	index--;
	while(visited[(str[index])%len])
	{
		
		index = index-1;		
		
	}
	print_rev(str, index, len);

}

- swati3458 June 04, 2016 | Flag Reply
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0
of 0 vote

import java.util.*;
import java.lang.*;

/* Print all the characters present in the given string only once in a reverse order.
 * Time and space complexity should not be more than o(N)
 */
/* Test : aabdceaaabbbcd => dcbae */

public class ReverseUniqueString {

    public String printReverse(String input) throws Exception {
        String msg = null;
        List<Character> charPresent = new ArrayList<Character>();
        String output = "";
        try {
            if(null != input) {
                for(int i = input.length()-1; i > 0; i--) {
                    Character c = input.charAt(i);
                    if(!charPresent.contains(c)) {
                        charPresent.add(c);
                    }
                }
            }

            for(Character c : charPresent) {
                output = output + c.toString();
            }
        } catch (Exception e) {
            msg = "Error while determining and printing reverse order of character found.";
            System.err.println(msg);
            throw new Exception(msg,e);
        }
        return output;
    }
}

/*----------------------Tests------------------------------*/

import org.junit.After;
import org.junit.Assert;
import org.junit.Before;
import org.junit.Test;

public class ReverseUniqueStringTest {
    @Before
    public void setUp() throws Exception {

    }

    @After
    public void tearDown() throws Exception {

    }

    @Test
    public void testGeneralScenario() throws Exception {
        ReverseUniqueString r = new ReverseUniqueString();
        String output = r.printReverse("aabdceaaabbbcd");
        Assert.assertEquals(output,"dcbae");
    }

    @Test
    public void testNullString() throws Exception {
        ReverseUniqueString r = new ReverseUniqueString();
        String output = r.printReverse(null);
        Assert.assertEquals(output,"");
    }

    @Test
    public void testAllSame() throws Exception {
        ReverseUniqueString r = new ReverseUniqueString();
        String output = r.printReverse("aaaaaa");
        Assert.assertEquals(output,"a");
    }
}

- maverick85 June 08, 2016 | Flag Reply
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of 0 vote

public class Reverse {
	public static void main(String[] args) {
		String s = "aabdceaaabbbcd";
		char[] c = s.toCharArray();
		
		Set<String> mySet = new LinkedHashSet<String>();
		
		for(char c1 : c){
			mySet.add(String.valueOf(c1));
		}
		System.out.println(mySet);
		Comparator<String> reverse = Comparator.reverseOrder();
		
		List<String> myList = new ArrayList<String>(mySet);
		
		Collections.sort(myList, reverse);
		System.out.println(myList);
	}
}

- Sakthi June 10, 2016 | Flag Reply
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0
of 0 vote

public class Reverse {
	public static void main(String[] args) {
		String s = "aabdceaaabbbcd";
		char[] c = s.toCharArray();
		
		Set<String> mySet = new LinkedHashSet<String>();
		
		for(char c1 : c){
			mySet.add(String.valueOf(c1));
		}
		System.out.println(mySet);
		Comparator<String> reverse = Comparator.reverseOrder();
		
		List<String> myList = new ArrayList<String>(mySet);
		
		Collections.sort(myList, reverse);
		System.out.println(myList);
	}

}

- Sakthi June 10, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public void reversePrint(String value)
	{
		value = value.toUpperCase();
		int MAX_CHAR_SET=26;
		int[] charIndexer = new int[MAX_CHAR_SET];
		if(value == null || value.equals("") | value.length()==0)
			return;
		int charIdx = 0;
		for(int idx = value.length() -1 ; idx >= 0; --idx)
		{
			charIdx = getCharIdx(value.charAt(idx));
			if(charIndexer[charIdx] ==0)
			{
				System.out.print(value.charAt(idx));
				++ charIndexer[charIdx];
			}
		}
	}
	
	private int getCharIdx(char c)
	{
		return ((int)c)-65;
	}

- Harsh123 June 10, 2016 | Flag Reply
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0
of 0 vote

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

using namespace std;

int main()
{
char s[10001];
int arr[27];
int t,i,j,k,n;
cin>>t;
while(t--)
{
cin>>s;
n = strlen(s);
for(i = 0;i<27;i++) arr[i] = 0;
for(j=n-1;j>=0;j--)
{
if(!arr[s[j] - 'a'])
{
cout<<s[j];
arr[s[j]-'a'] = 1;
}
}

}
return 0;
}

- Anonymous June 16, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

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

using namespace std;

int main()
{
   char s[10001];
   int arr[27];
   int t,i,j,k,n;
   cin>>t;
   while(t--)
   {
   	cin>>s;
   	n = strlen(s);
   	for(i = 0;i<27;i++) arr[i] = 0;
   	for(j=n-1;j>=0;j--)
   	{
   		if(!arr[s[j] - 'a'])
   		{
   			cout<<s[j];
   			arr[s[j]-'a'] = 1;
   		}
   	}
   	
   }
    return 0;
}

- Anonymous June 16, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

private static void printCharactersInReverseOrder(String iInputString) {
		char[] charArray = iInputString.toCharArray();
		String alreadyEvaluatedCharacters = "";
		for (int i = charArray.length - 1; i >= 0; i--) {
			if (!alreadyEvaluatedCharacters.contains(String.valueOf(charArray[i]))) {
				alreadyEvaluatedCharacters = alreadyEvaluatedCharacters + charArray[i];
			}
		}
		System.out.println("String in Reverse Order with characters printed only once => " + alreadyEvaluatedCharacters);
	}

- Deepak Malik June 21, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

private static void printCharactersInReverseOrder(String iInputString) {
		char[] charArray = iInputString.toCharArray();
		String alreadyEvaluatedCharacters = "";
		for (int i = charArray.length - 1; i >= 0; i--) {
			if (!alreadyEvaluatedCharacters.contains(String.valueOf(charArray[i]))) {
				alreadyEvaluatedCharacters = alreadyEvaluatedCharacters + charArray[i];
			}
		}
		System.out.println("String in Reverse Order with characters printed only once => " + alreadyEvaluatedCharacters);
	}

- Deepak June 21, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

private static void printCharactersInReverseOrder(String iInputString) {
		char[] charArray = iInputString.toCharArray();
		String alreadyEvaluatedCharacters = "";
		for (int i = charArray.length - 1; i >= 0; i--) {
			if (!alreadyEvaluatedCharacters.contains(String.valueOf(charArray[i]))) {
				alreadyEvaluatedCharacters = alreadyEvaluatedCharacters + charArray[i];
			}
		}
		System.out.println("String in Reverse Order with characters printed only once => " + alreadyEvaluatedCharacters);
	}

- Deepak Malik June 21, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

C# solution...

public static void PrintUniqueCharsInReverseOrder(string s)
        {
            List<char> list = new List<char>();

            for(int i=s.Length-1; i >= 0; i--)
            {
                if(list.Contains(s[i]))
                {
                    continue;
                }
                else
                {
                    list.Add(s[i]);
                }
            }

            foreach(char c in list)
            {
                Console.Write(c);
            }
            Console.WriteLine();
        }

- Jeanclaude June 21, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class StringReversePrint {

    public static void main(String[] args){
        printUniqueReverse("aaaabbfbccaaeeaaaff"); //faecb
    }

    public static void printUniqueReverse(String word){

        char[] cword = word.toCharArray();

        int[] tab = new int[256];

        StringBuilder sb = new StringBuilder();

        for(int i=(cword.length -1); i>=0;i--){

            char c = cword[i];

            if(tab[c] == 0){
                sb.append(c);
                tab[c] = 1;
            }

        }
        System.out.println(sb.toString());
    }

}

- fabien June 22, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

{{
bool set[128] = { false };
string str = { "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbcccaabbcc" };
int length = str.length();
int endIndex = length - 1;

while (endIndex >= 0)
{
if (!(set[str.at(endIndex) - 'a']))
{
cout << str.at(endIndex);
set[str.at(endIndex) - 'a'] = true;
}
endIndex--;
}
}}

- SinghIsKing June 22, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

bool set[128] = { false };
	string str = { "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbcccaabbcc" };
	int length = str.length();
	int endIndex = length - 1;
	
	while (endIndex >= 0)
	{
		if (!(set[str.at(endIndex) - 'a']))
		{
			cout << str.at(endIndex);
			set[str.at(endIndex) - 'a'] = true;
		}
		endIndex--;
	}

- SinghIsKing June 22, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class ReverseString {

	public static void main(String[] args) {
		String input = "aabdceaaabbbcd";
		reverseWithUniqueCharacterSet(input);
	}

	public static void reverseWithUniqueCharacterSet(String input) {
		
		LinkedHashSet<Character> output = new LinkedHashSet<Character>();
		char[] inputCharArray=input.toCharArray();		
		for(int i=inputCharArray.length-1; i>=0;i--){			
			if(output.add(inputCharArray[i])){
				System.out.print(inputCharArray[i]);
			}
		}
	}

}

- teji.catia June 30, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Assuming we only have lowercase a-z (We can extend this for all ASCII easily)
Following code has O(n) time and constant space. Space is Actually O(26) which is constant.

public class Test {
	
	public static void printOnceReverse(char[] s) {
		if(s.length == 0) return;
		
		boolean[] flag = new boolean[26];
		int i;
		
		for(i=0; i<s.length; i++) {
			int index = s[i] - 'a';
			flag[index] = true;
		}
		
		for(i=s.length-1; i>=0; i--) {
			int index = s[i] - 'a';
			if(flag[index] == true) {
				System.out.print(s[i]);
				flag[index] = false;
			}
		}
		System.out.println();
		
	}


	public static void main(String[] args) {
		printOnceReverse("aabdceaaabbbcd".toCharArray());
		printOnceReverse("aaaabbcddddccbbdaaeee".toCharArray());
		printOnceReverse("aaafffcccddaabbeeddhhhaaabbccddaaaa".toCharArray());

	}

}

- CodeNameEagle July 06, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void printReverse(String x){
		
		for(int i= x.length()-1;i>=0;i--){
			if(array[(byte)x.charAt(i)-97]==0){
				array[(byte)x.charAt(i)-97]=1;
				System.out.print(x.charAt(i));
			}
		}
	}

- Akash Bhartiya July 11, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

j=[]
def rev(s):
for i in reversed(s):
if i not in j:
j.append(i)
print ''.join(j)
d=rev('fjdcbkgjgh')]

- nandini July 14, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

private static String reverse(String param) {
        
        StringBuilder buf = new StringBuilder();
        
        for(int i = param.length(); i > 0; i--) {
            char ch = param.charAt(i-1);        
            if (buf.indexOf(ch + "") == -1)
                buf.append(ch);
        }
        
        return buf.toString();
    }
    
    public static void main(String[] args) {
        System.out.println(reverse("aaaabbbbddddeeeewwww"));
    }

- Sunil July 14, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

package amazon;

public class PrintAllStringInReverseWhichOccursOnlyOnce {
    public static void main(String[] args) {
        char[]c = "aabdceaaabbbcd".toCharArray();
        int[] temp = new int[128];
        for(char x :c){
            temp[x]++;
        }
        for(int i=c.length-1;i>=0;i--){
            if(temp[c[i]]>0){
                System.out.print(c[i]);
                temp[c[i]]=0;
            }
        }
    }
}

- Abhinaw Bhagat July 28, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

import java.util.*;
public class StringAmazon {
	public static void main(String args[]){

		String inputStr;
		Scanner sc = new Scanner(System.in);
		System.out.println("Enter the String");
		inputStr=sc.nextLine();

		StringAmazon sa =new StringAmazon();
		System.out.println(sa.reverseString(inputStr));

	}

	private String reverseString(String str){
		//String output = null;
		StringBuilder sbr = new StringBuilder();
		if(str.length()==0)
			return null;

		if(str.length()==1)
			return str;

		for(int i=str.length()-1; i>0; i--){
			if(str.charAt(i)!=str.charAt(i-1)){
				sbr=sbr.append(str.charAt(i));
				
			}
			
		}
		sbr.append(str.charAt(0));
		return sbr.toString();

	}
}

- sahilchalke3 August 03, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

//============================================================================
// Name        : StringPrintReverseOnlyOnce.cpp
// Author      : Nitish Raj, Scientist, DRDO, raj.nitp@gmail.com
// Version     :
// Copyright   : Your copyright notice
// Description : StringPrintInReverseMannerWithNoDuplicate, Ansi-style
//============================================================================

#include <bits/stdc++.h>
#include <tr1/unordered_map>
using namespace std;

int main()
{

	typedef tr1::unordered_map<char,int> charmap;

	charmap _map;
	string _str="aaaabbbbgggggffffafffbuuaaiiiiaaa";

	int str_size = _str.length();
	while(str_size>0){
		if(!_map.count(_str[str_size-1])){
			_map.insert(pair<char,int>(_str[str_size-1],1));
			cout<<_str[str_size-1];
		}


		str_size--;
	}

	/*cout<<endl;

	charmap::hasher fn = _map.hash_function();

	std::cout << "a: " << fn ('a') << std::endl;
	std::cout << "i: " << fn ('i') << std::endl;
	std::cout << "u: " << fn ('u') << std::endl;
	std::cout << "b: " << fn ('b') << std::endl;
	std::cout << "f: " << fn ('f') << std::endl;
	std::cout << "g: " << fn ('g') << std::endl;


	charmap ::iterator itr = _map.begin();
	while(itr != _map.end())
	{
		cout<<itr->first;
		itr++;
	}
*/
	return 0;
}

- raj.nitp@gmail.com August 05, 2016 | Flag Reply
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0
of 0 vote

def reverse_unique_strings(a):
temp = {j:i for i,j in enumerate(a)}
t = temp.keys()
t.sort()
t.reverse()
print reduce(lambda x,y :x+y, t)

reverse_unique_strings("aaavvvcccdeaaefe")

- mkdivis07@gmail.com September 02, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def reverse_unique_strings(a):
	temp = {j:i for i,j in enumerate(a)}
	t = temp.keys()
	t.sort()
	t.reverse()
	print reduce(lambda x,y :x+y, t)

reverse_unique_strings("aaavvvcccdeaaefe")

- mkdivis07 September 02, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def reverse_unique_strings(a):
temp = {j:i for i,j in enumerate(a)}
t = temp.keys()
t.sort()
t.reverse()
print reduce(lambda x,y :x+y, t)

reverse_unique_strings("aaavvvcccdeaaefe")

- mkdivis07 September 02, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

private static void reverseString() {
		String input = "aabdceaaabbbcd";
		Set<Character> print = new LinkedHashSet<Character>();
		for (int i = input.length() - 1; i >= 0; i--) {
			print.add(input.charAt(i));
		}
		for (Iterator iterator = print.iterator(); iterator.hasNext();) {
			System.out.println((Character) iterator.next());
		}
	}

- thangarebel November 14, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

wer

- Anonymous February 10, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

void PrintAlphabetsOnce(char *str) {
static int flags = 0;
if ((NULL == str) || (*str == '\0')) {
return;
}

PrintAlphabetsOnce(str + 1);

if (!(flags & (1 << *str - 'a'))) {
flags |= (1 << *str - 'a');
cout << *str << " ";
}

cout << endl;
}

- Anonymous February 10, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

void PrintAlphabetsOnce(char *str) {
    static int flags = 0;
    if ((NULL == str) || (*str == '\0')) {
        return;
    }
    
    PrintAlphabetsOnce(str + 1);
    
    if (!(flags & (1 << *str - 'a'))) {
        flags |= (1 << *str - 'a');
        cout << *str << " ";
    }
    
    cout << endl;

- Anonymous February 10, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

private static StringBuffer printStringRevOnce(String string) {
		StringBuffer sb = new StringBuffer();
		for (int i=string.length()-1; i>=0 ; i--) {
			if (sb.toString().lastIndexOf(string.charAt(i))<0) {
				sb.append(string.charAt(i));
			}
		}
		return sb;
	}

- Onkar Ganjewar February 19, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

static String uniqueCharsReversed(String str){
		
		boolean[] chars = new boolean[26];
		StringBuilder sb = new StringBuilder();
		
		for(int i=str.length()-1;i>=0;i--){
			
			if(!chars[str.charAt(i)-'a']){	
				sb.append(str.charAt(i));
				chars[str.charAt(i)-'a']=true;
			}
		}
		return sb.toString();
	}

- Anonymous June 18, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

static String uniqueCharsReversed(String str){
		
		boolean[] chars = new boolean[26];
		StringBuilder sb = new StringBuilder();
		
		for(int i=str.length()-1;i>=0;i--){
			
			if(!chars[str.charAt(i)-'a']){	
				sb.append(str.charAt(i));
				chars[str.charAt(i)-'a']=true;
			}
		}
		return sb.toString();
	}

|

- ag91375 June 18, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Python : Used Stack and Dictionary. Complexity O(N)

s1 = 'aabdceaaabbbcd' 
#output = dcbae 

s2 = 'aaaabbcddddccbbdaaeee' 
#output = eadbc 

s3 ='aaafffcccddaabbeeddhhhaaabbccddaaaa' 
#output = adcbhef 

def check(s1):
	len1 = len(s1)
	if len1 <= 1:
		return s1

	dict1 = {}
	count = 0
	list1 = []
	for i in range(len1-1,0,-1):

		if s1[i] in dict1:

			dict1[s1[i]] = count
			count += 1
		else:
			dict1[s1[i]] = count
			count += 1
			if s1[i] not in list1:
				list1.append(s1[i])

	print dict1
	print list1


	s2 = ''

	for items in list1:
		s2 = s2 + items

	print s2


check(s3)

- koolkhush February 09, 2018 | Flag Reply


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