Hacker Rank ‘Common Child’ Solution

Short Problem Definition:

Given two strings a and b of equal length, what’s the longest string (S) that can be constructed such that it is a child of both?

A string x is said to be a child of a string y if x can be formed by deleting 0 or more characters from y.


Common Child


time complexity is O(N*M)

space complexity is O(N*M)


This is a longest common subsequence problem in disguise. I encourage you to look at a good explanation here.


def lcs(a, b):
    lengths = [[0 for j in range(len(b)+1)] for i in range(len(a)+1)]
    for i, x in enumerate(a):
        for j, y in enumerate(b):
            if x == y:
                lengths[i+1][j+1] = lengths[i][j] + 1
                lengths[i+1][j+1] = \
                    max(lengths[i+1][j], lengths[i][j+1])
    return lengths[-1][-1]

def main():
    s1 = raw_input()
    s2 = raw_input()
    print lcs(s1,s2)
if __name__ == '__main__':

If you enjoyed this post, then make sure you subscribe to my Newsletter and/or Feed.