Interview Question for SDE-2s


Country: United States




Comment hidden because of low score. Click to expand.
1
of 1 vote

the answer is to use boyer-moore or KMP, which ar eboth O(n) algorithms. I'd probably study boyer-moore since it's easier to implement if some crazy interviewer actually wants you to implement an O(n) substring matching.

- Anonymous December 11, 2014 | Flag Reply
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0
of 0 votes

Here's a KMP implementation:

public static boolean isIn(String str1, String str2){
	char[] arr1 = str1.getChars();
	char[] arr2 = str2.getChars();
	//compute table for str2:
	int[] t = computeTable(arr2);

	int arr1Index = 0, arr2Index = 0;
	while(arr1Index + arr2Index < arr1.length){
		if(arr1[arr1Index + arr2Index] == arr2[arr2Index]){
			arr2Index++;
			if(arr2Index == arr2.length){
				return true;
			}
		}
		else{
			if(t[arr2Index] > -1){
				arr1Index += (arr2Index - t[arr2Index]);
				arr2Index = t[arr2Index];
			}
			else{
				arr1Index++;
				arr2Index = 0;
			}
		}
	}
	return false;
}

private static int[] computeTable(char[] arr){
	int[] table = new int[arr.length];
	int i = 1; m = 0;
	table[0] = -1;
	while(i < arr.length){
		if(arr[i] == arr[m]){
			table[i] = m;
			m++;
			i++;
		}
		else if (m > 0){
			m = t[m];
		else {
			table[i] = 0;
			i++;
		}
	}
	return table;
}

- zortlord December 11, 2014 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

Objective C solution

NSString *stringOne = @"aaaaaabbb";
    NSString *stringTwo = @"aaaaaabbbbbbb";
    BOOL containsString = NO;
    if (stringOne.length>stringTwo.length) {
         //NSLog(@"Contains Substring: %d",[stringOne containsString:stringTwo]);
        for (int i = 0; i<=(stringOne.length-stringTwo.length); i++) {
            NSRange mainRange = NSMakeRange(i, stringTwo.length);
            NSString *subString = [stringOne substringWithRange:mainRange];
            if ([subString isEqualToString:stringTwo]) {
                containsString = YES;
                break;
            }
        }
        NSLog(@"%@",containsString?@"Contains String":@"Does not contain String");
    }else{
         //NSLog(@"Contains Substring: %d",[stringTwo containsString:stringOne]);
        for (int i = 0; i<=(stringTwo.length-stringOne.length); i++) {
            NSRange mainRange = NSMakeRange(i, stringOne.length);
            NSString *subString = [stringTwo substringWithRange:mainRange];
            if ([subString isEqualToString:stringOne]) {
                containsString = YES;
                break;
            }
        }
        NSLog(@"%@",containsString?@"Contains String":@"Does not contain String");
    }

- deepak.hebbar December 11, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 2 vote

def SS(str1, str2):
    if(len(str1)>len(str2)):
        return False
    
    for i in range(len(str2)-len(str1)+1):
        found = True
        for j in range(len(str1)):    
            if(str2[i+j]!=str1[j]):
                found = False
                break
        if (found):
            return True
    return False
    
A = 'aaaaaabbb'
B = 'aaaabbb'
print SS(A,B) or SS(B,A)

- Aerofoil.kite December 11, 2014 | Flag Reply
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0
of 0 vote

not sure if would be ok to use charAt(), if not convert the two string to arrays won't change much in the actual implementation but it would affect time.

private static boolean evaluate(String stringOne, String stringTwo) {
	  for(int i=0; i<stringOne.length(); i++){
			int startIndex = i, counter = 0;
			for(int j=0;j<stringTwo.length(); j++){
				if(stringOne.charAt(startIndex) == stringTwo.charAt(j)){					
					startIndex++;
					counter++;
				}
				if(counter == 2){
					return true;
				}
			}
		}
		return false;
	}

- Moody123 December 11, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public boolean strstr (String src, String target){
		if (src.length() < target.length()) return strstr(target,src);
		for (int i = 0 ; src.length() - i >= target.length() ; ++i) {
			int j = 0 ;
			int k = i ; 			
			while (j < target.length() && src.charAt(k) == target.charAt(j)) {
				k++;
				j++;
			}
			if (j == target.length()) return true ;
		}		
		return false;

}

- Anonymous December 12, 2014 | Flag Reply
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0
of 0 vote

def is_sub(txt, txt2):
    uno, dos = set(txt), set(txt2)
    return uno.issuperset(dos) or dos.issuperset(uno)

print(is_sub('aaaaaabbb', 'aaabbb'))  # True
print(is_sub('asdasdf', 'cxvbxcvb'))  # False

- Anonymous December 12, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

class SubstringPresent{
private String largerString = "aaabcdedfd";
private String smallerString = "cdf";

public boolean isSubString(String largerString, String smallerString){
int temp=0;
for(int k = 0; k < largerString.length(); k++){
if(largerString.charAt(k)==smallerString.charAt(temp)){
temp = k;
break;
}
}

int temp_1 = 0;
for(int m = temp; m<largerString.length(); m++){
if(largerString.charAt(m)!=smallerString.charAt(temp_1))
return false;
temp_1++;
if(temp_1 == smallerString.length()){
return true;
}
}


return false;
}

public static void main(String[] args){
SubstringPresent test = new SubstringPresent();
System.out.println(test.isSubString(test.largerString, test.smallerString));
}

}

- vipinsharma1388 December 12, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

class SubstringPresent{
private String largerString = "aaabcdedfd";
private String smallerString = "cdf";

public boolean isSubString(String largerString, String smallerString){
int temp=0;
for(int k = 0; k < largerString.length(); k++){
if(largerString.charAt(k)==smallerString.charAt(temp)){
temp = k;
break;
}
}

int temp_1 = 0;
for(int m = temp; m<largerString.length(); m++){
if(largerString.charAt(m)!=smallerString.charAt(temp_1))
return false;
temp_1++;
if(temp_1 == smallerString.length()){
return true;
}
}


return false;
}

public static void main(String[] args){
SubstringPresent test = new SubstringPresent();
System.out.println(test.isSubString(test.largerString, test.smallerString));
}

}

- vipinsharma1388 December 12, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

class SubstringPresent{
private String largerString = "aaabcdedfd";
private String smallerString = "cdf";

public boolean isSubString(String largerString, String smallerString){
int temp=0;
for(int k = 0; k < largerString.length(); k++){
if(largerString.charAt(k)==smallerString.charAt(temp)){
temp = k;
break;
}
}
int temp_1 = 0;
for(int m = temp; m<largerString.length(); m++){
if(largerString.charAt(m)!=smallerString.charAt(temp_1))
return false;
temp_1++;
if(temp_1 == smallerString.length()){
return true;
}
}
return false;
}
public static void main(String[] args){
SubstringPresent test = new SubstringPresent();
System.out.println(test.isSubString(test.largerString, test.smallerString));
}
}

- vipinsharma1388 December 12, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

class SubstringPresent{
    private String largerString = "aaabcdedfd";
    private String smallerString = "cdf";
    public boolean isSubString(String largerString, String smallerString){
        int temp=0;
        for(int k = 0; k < largerString.length(); k++){
            if(largerString.charAt(k)==smallerString.charAt(temp)){
                temp = k;
                break;
            }
        }
        int temp_1 = 0;
        for(int m = temp; m<largerString.length(); m++){
        	if(largerString.charAt(m)!=smallerString.charAt(temp_1))
        		return false;
        	temp_1++;
        if(temp_1 == smallerString.length()){
                return true;
            }	
        }  
        return false;
    }
    public static void main(String[] args){
        SubstringPresent test = new SubstringPresent();
        System.out.println(test.isSubString(test.largerString, test.smallerString));
    }
}

- vipinsharma1388 December 12, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public boolean strstr (String src, String target){
		if (src.length() < target.length()) return strstr(target,src);
		for (int i = 0 ; src.length() - i >= target.length() ; ++i) {
			int j = 0 ; 			
			while (j < target.length() && src.charAt(i + j) == target.charAt(j)) j++;
			if (j == target.length()) return true ;
		}		
		return false;

}

- Scott December 13, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 3 vote

In Java

//aaaabbb aaaaaabbb
//Checking if s1 is a substring of s2 (s1 < s2)
// O(n) - letters in the larger string
public boolean isSubstring(String smallerString, String largerString){
  for(int i = 0; i < largerString.length(); ++i){
    int endIndex = i + smallerString.length();
    if(endIndex > largerString.length())
      break;
    if(largerString.substring(i, endIndex).equals(smallerString))
      return true;
    }
    return false;
  }

  public boolean checkSubstring(String s1, String s2){
    if(s1.length() < s2.length())
      return isSubstring(s1, s2);
    //s2 >= s1
    return isSubstring(s2, s1);
  }
}

- jacks205 December 10, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.


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