Google Interview Question for Software Engineers


Country: United States




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If B matches repeated A that means B can be written as <S><A...><P> where <S> is any suffix of A including empty suffix, <A...> is A repeated some number of times including zero times and <P> is any prefix of A including empty prefix. Let len(A) = N, len(B) = M then ans is 1 (if non-empty S exists) + number of times A is repeated + 1 (if non-empty P exists):

<S> and <P> can be checked to exist in O(N*N) while individual <A> can be checked in N time and there might be at max ceil(M/N) times <A> in B.

If this pattern is not matched, return -1.

- ashu1.220791 July 07, 2019 | Flag Reply
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0
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public static void main(String[] args) {

		String A = "abcd";
		String B = "cdabcdabcdabcdab";
		String temp = "";
		
		int len = B.length() / A.length();
		
		for(int i=0; i<len; i++) {
			temp = temp + A;
		}

		while(len<=(B.length() / A.length() + 1)) {
			if(temp.contains(B)) {
				System.out.println(len);
				break;
			}else {
				len++;
				temp = temp+A;
			}
		}
	}

- iamthem July 12, 2019 | Flag Reply
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0
of 0 vote

count=0
while True:
    count+=1
    if B in A*count:
        print(count)
        break

    if count>1000:
        print(-1)
        break

- Root July 22, 2019 | Flag Reply
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0
of 0 vote

int getCount(String A, String B) {
int startIndex = B.indexOf(A);

int count = B.length() / A.length();
if (startIndex > 0) {
count++;
}
return count;
}

- Martee July 29, 2019 | Flag Reply
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0
of 0 vote

int getCount(String A, String B) {
        int startIndex = B.indexOf(A);

        int count = B.length() / A.length();
        if (startIndex > 0) {
            count++;
        }
        return count;
    }

- Martee July 29, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int getCount(String A, String B) {
        int startIndex = B.indexOf(A);

        int count = B.length() / A.length();
        if (startIndex > 0) {
            count++;
        }
        return count;

}

- Martee July 29, 2019 | Flag Reply
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0
of 0 vote

Runtime O(m+n), Space O(1)

public static int NumOfRepeatsToGetSubstring(string i_A, string i_B)
{
    int count = 1;
    int a = 0, b = 0;

    while (b < i_B.Length)
    {
        if (i_A[a] == i_B[b])
        {
            ++b;
        }
        else if (count > 1)
        {
            count = -1;
            break;
        }

        ++a;
        if (a == i_A.Length)
        {
            a = 0;
            ++count;
        }
    }

    return count;
}

- Navro August 01, 2019 | Flag Reply
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0
of 0 vote

The test string never needs to be longer than A + B + A.

function getRepetitions(A, B) {
    var test = "", repetitions = 0, max;
    if (A.length < B.length)
        max = B.length + (A.length * 2);
    else
        max = A.length * 2;
    while (test.length <= max) {
        test += A;
        repetitions++;
        if (test.indexOf(B) != -1) { // I'm not sure if including test.length >= B.length would improve performance here
            return repetitions;
	}
    }
    return -1;
}

- someone September 05, 2019 | Flag Reply
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-1
of 1 vote

I think we can use KMP algorithm to solve it, and it takes linear time to solve it.

- lixx3527 July 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

public static void main(String[] args) {

		String A = "abcd";
		String B = "cdabcdab";
		String temp = "";
		
		int len = B.length() / A.length();
		
		for(int i=0; i<len; i++) {
			temp = temp + A;
		}

		while(true) {
			if(temp.contains(B)) {
				System.out.println(len);
				break;
			}else {
				len++;
				temp = temp+A;
			}
		}
	}

- iamthem July 12, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

public static void main(String[] args) {

		String A = "abcd";
		String B = "cdabcdab";
		String temp = "";
		
		int len = B.length() / A.length();
		
		for(int i=0; i<len; i++) {
			temp = temp + A;
		}

		while(true) {
			if(temp.contains(B)) {
				System.out.println(len);
				break;
			}else {
				len++;
				temp = temp+A;
			}
		}
	}

- iamthem July 12, 2019 | Flag Reply


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