The Grid Search - HackerRank Solution

In this post, we will solve The Grid Search HackerRank Solution. This problem (The Grid Search) is a part of HackerRak Problem Solving series.

Table of Contents

Task

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. For example, consider the following grid:

The pattern array is:

876543  
111111  
111111

The pattern begins at the second row and the third column of the grid and continues in the following two rows. The pattern is said to be present in the grid. The return value should be YES or NO, depending on whether the pattern is found. In this case, return YES.

Function Description

Complete the gridSearch function in the editor below. It should return YES if the pattern exists in the grid, or NO otherwise.

gridSearch has the following parameter(s):

string G[R]: the grid to search
string P[r]: the pattern to search for

Input Format

The first line contains an integer t, the number of test cases.

Each of the t test cases is represented as follows:
The first line contains two space-separated integers R and C, the number of rows in the search grid G and the length of each row string.
This is followed by R lines, each with a string of C digits that represent the grid G.
The following line contains two space-separated integers, r and c, the number of rows in the pattern grid P and the length of each pattern row string.
This is followed by r lines, each with a string of c digits that represent the pattern grid P.

Returns

string: either YES or NO

Constraints

1 <= t <= 5
1 <= R, r, C, c <= 1000
1 <= r <= R
1 <= c <= C

Sample Input

2
10 10
7283455864
6731158619
8988242643
3830589324
2229505813
5633845374
6473530293
7053106601
0834282956
4607924137
3 4
9505
3845
3530
15 15
400453592126560
114213133098692
474386082879648
522356951189169
887109450487496
252802633388782
502771484966748
075975207693780
511799789562806
404007454272504
549043809916080
962410809534811
445893523733475
768705303214174
650629270887160
2 2
99
99

Sample Output

YES
NO

Explanation

The first test in the input file is:

The pattern is present in the larger grid as marked in bold below.

The second test in the input file is:

15 15
400453592126560
114213133098692
474386082879648
522356951189169
887109450487496
252802633388782
502771484966748
075975207693780
511799789562806
404007454272504
549043809916080
962410809534811
445893523733475
768705303214174
650629270887160
2 2
99
99

The search pattern is:

99
99

This pattern is not found in the larger grid.

Solution – The Grid Search – HackerRank Solution

C++

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main(){
    int t;
    cin >> t;
    for(int a0 = 0; a0 < t; a0++){
        int R;
        int C;
        cin >> R >> C;
        vector<string> G(R);
        for(int G_i = 0;G_i < R;G_i++){
           cin >> G[G_i];
        }
        int r;
        int c;
        cin >> r >> c;
        vector<string> P(r);
        for(int P_i = 0;P_i < r;P_i++){
           cin >> P[P_i];
        }
        // Optimization - continue if P > G
        if(r > R || c > C) {
            cout<<"NO\n";
            continue;
        }
        bool patternExists = false;
        // Main algorithm
        for(int i = 0; i < R; i++) {
            bool foundMatch = false;
            for(int j = 0; j < C; j++) {
                if(G[i][j] == P[0][0] && (i+r <= R) && (j+c <= C)) {
                    // Potential match
                    bool match = true;
                    for(int x = 0; x < r; x++) {
                        bool rowMatch = true;
                        for(int y = 0; y < c; y++) {
                            if(P[x][y] != G[x+i][y+j]) {
                                rowMatch = false;
                                match = false;
                                break;
                            }
                        }
                        if(!rowMatch) {
                            break;
                        }
                    }
                    if(match) {
                        foundMatch = true;
                        break;
                    }
                }
            }
            if(foundMatch) {
                patternExists = true;
                break;
            }
        }
        if(patternExists) {
            cout<<"YES\n";
        } else {
            cout<<"NO\n";
        }
    }
    return 0;
}

Python

import sys
import re

def occurrences(string, sub):
    res = []
    ind = 0
    while ind < len(string) - len(sub) + 1:
        found = string.find(sub, ind)
        #print("ind = {} found = {}".format(ind, found))
        if found != -1:
            res.append(found)
            ind = found + 1
        else:
            break
    return res

def gridSearch(G, P):
    for ind_g in range(len(G) - len(P) + 1):
        cur = -1
        all_occurrences = []
        for ind_p, s_pat in enumerate(P):
            all_occurrences.append(occurrences(G[ind_g + ind_p], s_pat))
            #ret = G[ind_g + ind_p].find(s_pat)
            #print("ind_g = {} ind_p = {} check {} in {}".format(ind_g, ind_p, s_pat, G[ind_g + ind_p]))
            
            #if ret == -1 or (ind_p > 0 and cur != ret):
            #    break
            #elif ind_p == len(P) - 1:
            #    return 'YES'
            #else:
            #    cur = ret
        
        #print(all_occurrences)
        ourset = set(all_occurrences[0])
        for lst in all_occurrences:
            ourset &= set(lst)
            
        #print("ourset = {}".format(ourset))
        if len(ourset) >= 1:
            return 'YES'
            
    
    return 'NO'
        

if __name__ == "__main__":
    t = int(input().strip())
    for a0 in range(t):
        R, C = input().strip().split(' ')
        R, C = [int(R), int(C)]
        G = []
        G_i = 0
        for G_i in range(R):
            G_t = str(input().strip())
            G.append(G_t)
        r, c = input().strip().split(' ')
        r, c = [int(r), int(c)]
        P = []
        P_i = 0
        for P_i in range(r):
            P_t = str(input().strip())
            P.append(P_t)
        result = gridSearch(G, P)
        print(result)

Java

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");
        }
        }
    }
}

Note: This problem (The Grid Search) is generated by HackerRank but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.

The Grid Search – HackerRank Solution

Task

Input Format

Constraints

Solution – The Grid Search – HackerRank Solution

C++

Python

Java

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