Given an N×N matrix mat[][] and an integer K, the task is to search K in mat[][] and if found return its indices.
Examples:
Input: mat[][] = {{1, 2, 3}, {7, 6, 8}, {9, 2, 5}}, K = 6
Output: {1, 1}
Explanation: mat[1][1] = 6Input: mat[][] = {{3, 4, 5, 0}, {2, 9, 8, 7}}, K = 10
Output: Not Found
Approach:
Traverse the matrix using nested loops and if the target is found then, return its indices
Follow the steps to solve this problem:
- Take the matrix (2-D array) and target as input.
- Apply two nested for-loop with i and j variables respectively.
- If mat[i][j] = K return the indices pair {i, j}
- If the target is not found print “Not Found”.
Below is the implementation of this approach
C++
// C++ code to search in a Matrix#include <bits/stdc++.h>using namespace std;// Function to search a number in a matrixpair<int, int> FindPosition(vector<vector<int> > M, int K){ for (int i = 0; i < M.size(); i++) { for (int j = 0; j < M[i].size(); j++) { if (M[i][j] == K) return { i, j }; } } return { -1, -1 };}// Driver codeint main(){ // Taking the 2D Array vector<vector<int> > M{ { 1, 2, 3 }, { 7, 6, 8 }, { 9, 2, 5 } }; int K = 6; // Function call pair<int, int> p = FindPosition(M, K); if (p.first == -1) cout << "Not Found" << endl; else cout << p.first << " " << p.second; return 0;} |
Java
// Java code to search in a Matrixclass Pair { int first; int second; Pair(int _first, int _second) { first = _first; second = _second; }}class GFG { // Function to search a number in a matrix static Pair FindPosition(int[][] M, int K) { for (int i = 0; i < M.length; i++) { for (int j = 0; j < M[i].length; j++) { if (M[i][j] == K) return new Pair(i, j); } } return new Pair(-1, -1); } // Driver code public static void main(String args[]) { // Taking the 2D Array int[][] M = { { 1, 2, 3 }, { 7, 6, 8 }, { 9, 2, 5 } }; int K = 6; // Function call Pair p = FindPosition(M, K); if (p.first == -1) System.out.println("Not Found"); else System.out.println(p.first + " " + p.second); }}// This code is contributed by _saurabh_jaiswal. |
Python3
# Function to search a number in a matrixdef FindPosition(M,K): ans = [] for i in range(0,len(M)): for j in range(0,len(M[0])): if (M[i][j] is K): ans.append(i) ans.append(j) return ans ans.append(-1) ans.append(-1) return ans M = [[ 1, 2, 3 ],[ 7, 6, 8 ],[ 9, 2, 5]]K = 6# Function callp = FindPosition(M,K) if (p[0] is -1): print("Not Found")else: print(p[0],p[1]) # This code is contributed by akashish__ |
C#
// Include namespace systemusing System;public class GFG{ // Function to search a number in a matrix public static int[] FindPosition(int[,] M, int K) { for (int i = 0; i < M.GetLength(0); i++) { for (int j = 0; j < M.GetLength(1); j++) { if (M[i,j] == K) { return new int[]{i, j}; } } } return new int[]{-1, -1}; } // Driver code public static void Main(String[] args) { // Taking the 2D Array int[,] M = {{1, 2, 3}, {7, 6, 8}, {9, 2, 5}}; var K = 6; // Function call int[] p = GFG.FindPosition(M, K); if (p[0] == -1) { Console.WriteLine("Not Found"); } else { Console.WriteLine(p[0].ToString() + " " + p[1].ToString()); } }}// This code is contributed by aadityapburujwale |
Javascript
<script>// Javascript code to search in a Matrix// Function to search a number in a matrixfunction FindPosition( M, K){ for (let i = 0; i < M.length; i++) { for (let j = 0; j < M[i].length; j++) { if (M[i][j] == K){ let ans = new Array( i, j ); return ans; } } } let ans=new Array( -1, -1 ); return ans; }// Driver code // Taking the 2D Array let M=[ [ 1, 2, 3 ], [ 7, 6, 8 ], [ 9, 2, 5 ] ]; let K = 6; // Function call let p=new Array(); p = FindPosition(M, K); if (p[0] == -1){ document.write("Not Found" ); document.write("<br>"); } else document.write(p[0]+" " + p[1]); // This code is contributed by satwik4409. </script> |
Output
1 1
Time Complexity: O(N2)
Auxiliary Space: O(1)
