Hello Programmers, In this post, you will Know how to solve HackerRank Common Child Solution. This problem is a part of the HackerRank Algorithms Series.
One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.
HackerRank Common Child Solution
Task
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
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;
}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;
}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. */
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
char x[]=br.readLine().toCharArray();
char y[]=br.readLine().toCharArray();
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]);
}
}
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()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);
});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))
}
}Common Child Solution in Pascal
var s1,s2:ansistring;
i,j,n:integer;
f:array[0..5000,0..5000] of integer;
begin
readln(s1);readln(s2); n:=length(s1);
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 BrokenProgrammers. This tutorial is only for Educational and Learning purposes.