Hackerrank - Common Child Solution

Hackerrank - Common Child Solution

A string is said to be a child of a another string if it can be formed by deleting 0 or more characters from the other string. Given two strings of equal length, what's the longest string that can be constructed such that it is a child of both?

For example, ABCD and ABDC have two children with maximum length 3, ABC and ABD. They can be formed by eliminating either the D or C from both strings. Note that we will not consider ABCD as a common child because we can't rearrange characters and ABCD  ABDC.

Function Description

Complete the commonChild function in the editor below. It should return the longest string which is a common child of the input strings.

commonChild has the following parameter(s):

  • s1, s2: two equal length strings

Input Format

There is one line with two space-separated strings,  and .

Constraints

  • All characters are upper case in the range ascii[A-Z].

Output Format

Print the length of the longest string , such that  is a child of both  and .

Sample Input

HARRY
SALLY

Sample Output

 2

Explanation

The longest string that can be formed by deleting zero or more characters from  and  is , whose length is 2.

Sample Input 1

AA
BB

Sample Output 1

0

Explanation 1

and  have no characters in common and hence the output is 0.

Sample Input 2

SHINCHAN
NOHARAAA

Sample Output 2

3

Explanation 2

The longest string that can be formed between  and  while maintaining the order is .

Sample Input 3

ABCDEF
FBDAMN

Sample Output 3

2

Explanation 3
is the longest child of the given strings.

Solution in Pypy3

def commonChild(s1, s2):
    m = [[0]*(len(s2)+1) for _ in range(len(s1)+1)]
    for i,c in enumerate(s1,1):
        for j,d in enumerate(s2,1):
            if c == d:
                m[i][j] = m[i-1][j-1]+1
            else:
                m[i][j] = max(m[i][j-1],m[i-1][j])
                   
    return m[-1][-1]
print(commonChild(input(), input()))

Note: Time out error with python3. Only works with Pypy3

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