# HackerRank Common Child Solution

Hello Programmers, In this post, you will learn how to solve HackerRank Common Child Solution. This problem is a part of the HackerRank Algorithms Series.

One more thing to add, don’t straight away look for the solutions, first try to solve the problems by yourself. If you find any difficulty after trying several times, then look for the solutions. We are going to solve the HackerRank Algorithms problems using C, CPP, JAVA, PYTHON, JavaScript & SCALA Programming Languages.

## 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. Letters cannot be rearranged. Given two strings of equal length, what’s the longest string that can be constructed such that it is a child of both?

Example

s1 = ‘ABCD’
s2 = ‘ABDC’

These strings have two children with maximum length 3, `ABC` and `ABD`. They can be formed by eliminating either the `D` or `C` from both strings. Return 3.

Function Description

Complete the commonChild function in the editor below.

commonChild has the following parameter(s):

• string s1: a string
• string s2: another string

Returns

• int: the length of the longest string which is a common child of the input strings

Input Format

There are two lines, each with a string, s1 and s2.

Constraints

• 1 <= |s1|, |s2| <= 5000 where |s| means “the length of s
• All characters are upper case in the range ascii[A-Z].

Sample Input

HARRY
SALLY

Sample Output

2

Explanation

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

Sample Input 1

AA
BB

Sample Output 1

0

Explanation 1

AA and BB 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 SHINCHAN and NOHARAAA while maintaining the order is NHA.

Sample Input 3

ABCDEF
FBDAMN

Sample Output 3

2

Explanation 3

BD is the longest child of the given strings.

## HackerRank Common Child Solution

### HackerRank Common Child Solution in C

```#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define SORT(a,n) qsort(a,n,sizeof(int),intcmp)
#define s(n)                        scanf("%d",&n)
#define sc(n)                       scanf("%c",&n)
#define sl(n)                       scanf("%I64d",&n)
#define sf(n)                       scanf("%lf",&n)
#define ss(n)                       scanf("%s",n)
#define fill(a,v)                   memset(a, v, sizeof(a))
int intcmp(const void *f,const void *s)
{
return (*(int *)f -*(int *)s);
}
int gcd(int a,int b){ return ((b==0)?a:gcd(b,a%b));}
int max(int a,int b){ return(a>b)?a:b;}

#define MAX 8192
#define MODBY 1000000007

typedef long long int lld;
typedef long double Lf;
int preprocess()
{
return 0;
}
int lcs[MAX][MAX];
int main()
{
int cases;
int i,j,n;
char a[MAX],b[MAX];
scanf("%s%s",a+1,b+1);
for(i=1;a[i];++i)
for(j=1;b[j];++j){
if(a[i]==b[j])
lcs[i][j]=1+lcs[i-1][j-1];
else lcs[i][j]=max(lcs[i-1][j],lcs[i][j-1]);
}
printf("%d\n",lcs[strlen(a+1)][strlen(b+1)]);
return 0;
}```

### HackerRank Common Child Solution in Cpp

```#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <cstring>
#include <string>
#include <cmath>
#include <ctime>
#include <utility>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <sstream>
#define FOR(a,b,c) for (int a=b,_c=c;a<=_c;a++)
#define FORD(a,b,c) for (int a=b;a>=c;a--)
#define REP(i,a) for(int i=0,_a=(a); i<_a; ++i)
#define REPD(i,a) for(int i=(a)-1; i>=0; --i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define sz(a) int(a.size())
#define reset(a,b) memset(a,b,sizeof(a))
#define oo 1000000007

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;

const int maxn=5007;

int dp[maxn][maxn],n;
char a[maxn],b[maxn];

int main(){
//freopen("test.txt","r",stdin);
scanf("%s",a+1);
scanf("%s",b+1);
n=strlen(a+1);
reset(dp,0);
FOR(i,1,n) FOR(j,1,n){
dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
if(a[i]==b[j]) dp[i][j]=max(dp[i][j],dp[i-1][j-1]+1);
}
printf("%d\n",dp[n][n]);
return 0;
}```

### HackerRank Common Child Solution in Java

```import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) throws IOException
{
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */

int a[][]=new int[x.length+1][];
int dir[][]=new int[x.length+1][];//0 for terminating condtion,1=diagonal,2=left,3=upper
for(int i=0;i<a.length;i++)
{
a[i]=new int[y.length+1];
dir[i]=new int[y.length+1];
//System.out.println(a[i].length);
}
for(int i=1;i<x.length+1;i++)
{
for(int j=1;j<a[0].length;j++)
{
/*   if(i==0||j==0)
{
a[i][j]=0;
dir[i][j]=0;
continue;
}*/
if(x[i-1]==y[j-1])
{
a[i][j]=a[i-1][j-1]+1;
dir[i][j]=1;//diagonal
}
else
{
if(a[i-1][j]>a[i][j-1])//upper is greater
{
a[i][j]=a[i-1][j];
dir[i][j]=3;

}
else//left is greater
{
a[i][j]=a[i][j-1];
dir[i][j]=2;
}
}
}
}

int row=a.length-1;
int col=a[0].length-1;
System.out.println(a[row][col]);
}
}
```

### HackerRank Common Child Solution in Python

```from __future__ import division
from sys import stdin
from collections import defaultdict

def lcs(s1, s2):
prev = defaultdict(int)
for i in range(len(s1)):
cur = defaultdict(int)
for j in range(len(s2)):
cur[j] = prev[j - 1] + 1 if s1[i] == s2[j] else max(cur[j - 1], prev[j])
prev = cur
return prev[len(s2)-1]

def main():
s, t = stdin.next().strip(), stdin.next().strip()
print lcs(s, t)

main()```

### HackerRank Common Child Solution using JavaScript

```function processData(input) {
var parts = input.split("\n"),
firstStr = parts[0],
secondStr = parts[1],
strLen = firstStr.length,
arrPrev = new Array(strLen + 1),
arrCurr = new Array(strLen + 1);

for (var ii = 0; ii <= strLen; ii++) {
arrPrev[ii] = 0;
arrCurr[ii] = 0;
}

//for (var ii = 0; ii <= strLen; ii++) {
//  console.log(arrPrev[ii]);
//}

for (ii = 1; ii <= strLen; ii++) {
for (var jj = 1; jj <= strLen; jj++) {
if (firstStr[ii - 1] == secondStr[jj - 1]) {
arrCurr[jj] = arrPrev[jj - 1] + 1;
}
else {
arrCurr[jj] = Math.max(arrCurr[jj - 1], arrPrev[jj]);
}
}
arrPrev = arrCurr.slice(0);
}

console.log(arrCurr[strLen]);
}

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
_input += input;
});

process.stdin.on("end", function () {
processData(_input);
});```

### HackerRank Common Child Solution in Scala

```object Solution {
def lcs(x: String, y: String): Int = {
val m = x.length
val n = y.length
val c = Array.ofDim[Int](m + 1, n + 1)
for (i <- 1 to m; j <- 1 to n) {
if (x(i - 1) == y(j - 1)) c(i)(j) = c(i - 1)(j - 1) + 1
else c(i)(j) = c(i)(j - 1) max c(i - 1)(j)
}
c(m)(n)
}

def main(args: Array[String]): Unit = {
val a = readLine()
val b = readLine()
println(lcs(a, b))
}
}```

### HackerRank Common Child Solution in Pascal

```var s1,s2:ansistring;
i,j,n:integer;
f:array[0..5000,0..5000] of integer;
begin
for i:=0 to n do f[0,i]:=0;
for j:=0 to n do f[j,0]:=0;
for i:=1 to n do
for j:=1 to n do
if s1[i]=s2[j] then f[i,j]:=f[i-1,j-1]+1 else
if f[i-1,j]>f[i,j-1] then f[i,j]:=f[i-1,j] else f[i,j]:=f[i,j-1];
writeln(f[n,n]);
end.

```

Disclaimer: This problem (Common Child) is generated by HackerRank but the solution is provided by Chase2learn. This tutorial is only for Educational and Learning purposes.

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