# 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

##### Complexity:

time complexity is O(N*M)

space complexity is O(N*M)

##### Execution:

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

##### Solution:
#!/usr/bin/py

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
else:
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__':
main()


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