HackerRank The Grid Search problem solution

In this HackerRank The Grid Search problem, you have Given an array of strings of digits, try to find the occurrence of a given pattern of digits. In the grid and pattern arrays, each string represents a row in the grid.

Problem solution in Python programming.

tests = int(input())
n = 0
n2 = 0
for i in range(0, tests):
    inp = input()
    inp = inp.split(' ')
    n = int(inp[0])
    thegrid = []
    for i in range(0, n):
        s = input()
        thegrid.append(s)
    inp2 = input()
    inp2 = inp2.split(' ')
    n2 = int(inp2[0])
    subgrid = []
    for i in range(0, n2):
        s = input()
        subgrid.append(s)
    found = False
    for i in range(0, len(thegrid)):
        for j in range(0, len(subgrid)):
            if(subgrid[j] in thegrid[i + j]):
                if (j == len(subgrid)-1):
                    found = True
                    print ("YES")
                    break
                continue
            else:
                break
            break
    if not found:
        print ("NO")

Problem solution in Java Programming.

import java.io.*;
import java.util.*;

public class Solution {
    
    public static boolean isMatch(String[] grid, int r, int c, String[] pattern) {
        for(int i = r; i < r + pattern.length; i++) {
            if(!grid[i].substring(c, c + pattern[0].length()).equals(pattern[i - r]))
                return false;
        }
        return true;
    }

    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your clas should be named Solution. */
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        for(int k = 0; k < T; k++){
            int R = sc.nextInt();
        int C = sc.nextInt();
        String[] grid = new String[R];
        for(int i = 0; i < R; i++){
            grid[i] = sc.next();
        }
        int r = sc.nextInt();
        int c = sc.nextInt();
        String[] pattern = new String[r];
        for(int i = 0; i < r; i++) {
            pattern[i] = sc.next();
        }
        boolean ret = false;
        for(int i = 0; i <= R - r; i++){
            for(int j = 0; j <= C - c; j++){
                if(grid[i].charAt(j) == pattern[0].charAt(0)){
                    ret = isMatch(grid, i, j, pattern);
                    if(ret)
                        break;
                }
            }
            if(ret)
                break;
        }
        if(ret){
            System.out.println("YES");
        }
        else{
            System.out.println("NO");
        }
        }
    }
}

Problem solution in C++ programming.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
char arr[1001][1001]; 
char find2[1001][1001]; 
int main() {
    int a; cin >> a;
    for (int g=0;g<a; g++)
    {
        int b,c;cin>>b>>c;
        for (int g=0;g<b; g++)
            {
            for (int y=0;y<c; y++)
                {
                cin >> arr[g][y]; 
            }
        }
        int d,e;cin>>d>>e;
        for (int g=0;g<d; g++)
            {
            for (int y=0;y<e; y++)
                {
                cin >> find2[g][y]; 
            }
        }
        if (d>b || e>c)
            {
            cout << "NO" << 'n'; continue; 
        }int realcheck=0; 
        for (int z=0;z<=b-d; z++)
            {
            for (int zz=0; zz<=c-e; zz++)
                {int checker=0; 
                for (int l=z; l<z+d; l++)
                    {
                    for (int ll=zz; ll<zz+e; ll++)
                        {
                        if (find2[l-z][ll-zz]!=arr[l][ll])
                         {
                            checker=1; break;    
                        }
                    }if (checker==1)break; 
                }
                 if (checker==0) {realcheck=1; break;}
            }if (realcheck==1) break; 
        }
        if (realcheck) cout << "YES" << 'n'; 
        else cout << "NO" << 'n';
    }
    return 0;
}

Problem solution in C programming.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int is_match(char** a,int r, int c,char **pat,int pr,int pc,int starty,int startx){
    int i,j;
    if(pr+starty>r||pc+startx>c)
        return 0;
    for(i=0;i<pr;i++){
        for(j=0;j<pc;j++){
            if(a[starty+i][startx+j]!=pat[i][j]){
                return 0;
            }
        }
    }
    return 1;
}
int main() {
    int ntest;
    scanf("%i",&ntest);
    int s;
    int found=0;
    for(s=0;s<ntest;s++){
        int r,c;
        scanf("%i %i",&r,&c);
        char** test;
        test=malloc(sizeof(char*)*r);
        int i;
        for(i=0;i<r;i++){
            test[i]=calloc(sizeof(char),1+c);
            scanf("%s",test[i]);
        }
        int R,C;
        scanf("%i %i",&R,&C);
        
        char **pattern;
        pattern=malloc(sizeof(char*)*R);
        for(i=0;i<R;i++){
            pattern[i]=calloc(sizeof(char),1+C);
            scanf("%s",pattern[i]);
        }
  
        for(i=0;i<r;i++){
            int j;
            for(j=0;j<c;j++){
                if(is_match(test,r,c,pattern,R,C,i,j))
                   {
                       found=1;
                       goto quit;
                   }
            }
        }
            quit:
        if(found)
            printf("YESn");
        else
            printf("NOn");
        found=0;
        for(i=0;i<R||i<r;i++){
            if(i<R)
            free(pattern[i]);
            free(test[i]);
         
        }
        
    }

    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
    return 0;
}

Problem solution in JavaScript programming.

function processData(input) {
    var lines = input.split('n'),
        nCases = parseInt(lines[0]);
    var contains = function(gridStart, gridRows, patternStart, patternRows) {
        var stopIndex = gridStart + gridRows - patternRows + 1,
            innerStopIndex = patternStart + patternRows;
        var cmpIdx;
        for(var i = gridStart; i < stopIndex; i++) {
            cmpIdx = lines[i].indexOf(lines[patternStart]);
            if(cmpIdx > -1) {
                for(var j = i + 1, k = patternStart + 1; k < innerStopIndex && cmpIdx === lines[j].indexOf(lines[k]); ++j,++k);
                if(k === innerStopIndex)
                    return true;
            }
        }
        return false;
    };
    var i = 1, 
        gridRows = parseInt(lines[i].split(' ')[0]),
        gridStart = i + 1,
        j = i + gridRows + 1,
        patternRows = parseInt(lines[j].split(' ')[0]),
        patternStart = j + 1;
    do {
        if(contains(gridStart, gridRows, patternStart, patternRows))
            console.log('YES');
        else
            console.log('NO');
        i = patternStart + patternRows;
        gridRows = parseInt(lines[i].split(' ')[0]);
        gridStart = i + 1;
        j = i + gridRows + 1;
        patternRows = parseInt(lines[j].split(' ')[0]);
        patternStart = j + 1;
        nCases--;
    } while(nCases > 0)
} 

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

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