Given an n x n matrix and an integer x, find the position of x in the matrix if it is present. Otherwise, print “Element not found”. Every row and column of the matrix is sorted in increasing order. The designed algorithm should have linear time complexity
Examples:
Input: mat[4][4] = { {10, 20, 30, 40}, x = 29
{15, 25, 35, 45},
{27, 29, 37, 48},
{32, 33, 39, 50}}
Output: Found at (2, 1)
Explanation: Element at (2,1) is 29Input : mat[4][4] = { {10, 20, 30, 40}, x = 100
{15, 25, 35, 45},
{27, 29, 37, 48},
{32, 33, 39, 50}};
Output: Element not found
Explanation: Element 100 does not exist in the matrix
Naive approach: To solve the problem follow the below idea:
The simple idea is to traverse the array and search elements one by one
Follow the given steps to solve the problem:
- Run a nested loop, outer loop for the row, and inner loop for the column
- Check every element with x and if the element is found then print “element found”
- If the element is not found, then print “element not found”
Below is the implementation of the above approach:
C++
// C++ program to search an element in row-wise// and column-wise sorted matrix#include <bits/stdc++.h>using namespace std;/* Searches the element x in mat[][]. If theelement is found, then prints its positionand returns true, otherwise prints "not found"and returns false */int search(int mat[4][4], int n, int x){ if (n == 0) return -1; // traverse through the matrix for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) // if the element is found if (mat[i][j] == x) { cout << "Element found at (" << i << ", " << j << ")\n"; return 1; } } cout << "n Element not found"; return 0;}// Driver codeint main(){ int mat[4][4] = { { 10, 20, 30, 40 }, { 15, 25, 35, 45 }, { 27, 29, 37, 48 }, { 32, 33, 39, 50 } }; // Function call search(mat, 4, 29); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
C
// C program to search an element in row-wise// and column-wise sorted matrix#include <stdio.h>/* Searches the element x in mat[][]. If theelement is found, then prints its positionand returns true, otherwise prints "not found"and returns false */int search(int mat[4][4], int n, int x){ if (n == 0) return -1; // traverse through the matrix for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) // if the element is found if (mat[i][j] == x) { printf("Element found at (%d, %d)\n", i, j); return 1; } } printf("Element not found"); return 0;}// Driver codeint main(){ int mat[4][4] = { { 10, 20, 30, 40 }, { 15, 25, 35, 45 }, { 27, 29, 37, 48 }, { 32, 33, 39, 50 } }; // Function call search(mat, 4, 29); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Java program to search an element in row-wise// and column-wise sorted matrixclass GFG { static int search(int[][] mat, int n, int x) { if (n == 0) return -1; // traverse through the matrix for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) // if the element is found if (mat[i][j] == x) { System.out.print("Element found at (" + i + ", " + j + ")\n"); return 1; } } System.out.print(" Element not found"); return 0; } // Driver code public static void main(String[] args) { int mat[][] = { { 10, 20, 30, 40 }, { 15, 25, 35, 45 }, { 27, 29, 37, 48 }, { 32, 33, 39, 50 } }; // Function call search(mat, 4, 29); }}// This code is contributed by Aditya Kumar (adityakumar129) |
Python3
# Python program to search an element in row-wise# and column-wise sorted matrix# Searches the element x in mat[][]. If the# element is found, then prints its position# and returns true, otherwise prints "not found"# and returns falsedef search(mat, n, x): if(n == 0): return -1 # Traverse through the matrix for i in range(n): for j in range(n): # If the element is found if(mat[i][j] == x): print("Element found at (", i, ",", j, ")") return 1 print(" Element not found") return 0# Driver codeif __name__ == "__main__": mat = [[10, 20, 30, 40], [15, 25, 35, 45], [27, 29, 37, 48], [32, 33, 39, 50]] # Function call search(mat, 4, 29)# This code is contributed by rag2127 |
C#
// C# program to search an element in row-wise// and column-wise sorted matrixusing System;class GFG { /* Searches the element x in mat[][]. If the element is found, then prints its position and returns true, otherwise prints "not found" and returns false */ static int search(int[, ] mat, int n, int x) { if (n == 0) return -1; // Traverse through the matrix for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) // If the element is found if (mat[i, j] == x) { Console.Write("Element found at (" + i + ", " + j + ")\n"); return 1; } } Console.Write(" Element not found"); return 0; } // Driver code static public void Main() { int[, ] mat = { { 10, 20, 30, 40 }, { 15, 25, 35, 45 }, { 27, 29, 37, 48 }, { 32, 33, 39, 50 } }; // Function call search(mat, 4, 29); }}// This code is contributed by avanitrachhadiya2155 |
Javascript
<script> // Java Script program to search an element in row-wise// and column-wise sorted matrixfunction search(mat,n,x) { if (n == 0) return -1; // traverse through the matrix for (let i = 0; i < n; i++) { for (let j = 0; j < n; j++) // if the element is found if (mat[i][j] == x) { document.write("Element found at (" + i + ", " + j + ")<br>"); return 1; } } document.write(" Element not found"); return 0;} let mat = [[ 10, 20, 30, 40 ], [15, 25, 35, 45] , [ 27, 29, 37, 48 ], [ 32, 33, 39, 50 ]]; search(mat, 4, 29); //contributed by 171fa07058</script> |
Output
Element found at (2, 1)
Time Complexity: O(N2)
Auxiliary Space: O(1), since no extra space has been taken
Note: A better solution is to use Divide and Conquer to find the element which has a time complexity of O(n1.58). Please refer here for details
Search in a row-wise and column-wise sorted matrix in linear time complexity:
The simple idea is to remove a row or column in each comparison until an element is found. Start searching from the top-right corner of the matrix. There are three possible cases:-
- The given number is greater than the current number: This will ensure that all the elements in the current row are smaller than the given number as the pointer is already at the right-most elements and the row is sorted. Thus, the entire row gets eliminated and continues the search for the next row. Here, elimination means that a row needs not to be searched.
- The given number is smaller than the current number: This will ensure that all the elements in the current column are greater than the given number. Thus, the entire column gets eliminated and continues the search for the previous column, i.e. the column on the immediate left.
- The given number is equal to the current number: This will end the search.
Follow the given steps to solve the problem:
- Let the given element be x, create two variable i = 0, j = n-1 as index of row and column.
- Run a loop until i < n.
- Check if the current element is greater than x then decrease the count of j. Exclude the current column.
- Check if the current element is less than x then increase the count of i. Exclude the current row.
- If the element is equal, then print the position and end.
- Print the Element is not found
Thanks to devendraiiit for suggesting the approach below
Below is the implementation of the above approach:
C++
// C++ program to search an element in row-wise// and column-wise sorted matrix#include <bits/stdc++.h>using namespace std;/* Searches the element x in mat[][]. If theelement is found, then prints its positionand returns true, otherwise prints "not found"and returns false */int search(int mat[4][4], int n, int x){ if (n == 0) return -1; int smallest = mat[0][0], largest = mat[n - 1][n - 1]; if (x < smallest || x > largest) return -1; // set indexes for top right element int i = 0, j = n - 1; while (i < n && j >= 0) { if (mat[i][j] == x) { cout << "Element found at " << i << ", " << j; return 1; } if (mat[i][j] > x) j--; // Check if mat[i][j] < x else i++; } cout << "n Element not found"; return 0;}// Driver codeint main(){ int mat[4][4] = { { 10, 20, 30, 40 }, { 15, 25, 35, 45 }, { 27, 29, 37, 48 }, { 32, 33, 39, 50 } }; // Function call search(mat, 4, 29); return 0;}// This code is contributed// by Akanksha Rai(Abby_akku) |
C
// C program to search an element in row-wise// and column-wise sorted matrix#include <stdio.h>/* Searches the element x in mat[][]. If theelement is found, then prints its positionand returns true, otherwise prints "not found"and returns false */int search(int mat[4][4], int n, int x){ if (n == 0) return -1; int smallest = mat[0][0], largest = mat[n - 1][n - 1]; if (x < smallest || x > largest) return -1; // set indexes for top right element int i = 0, j = n - 1; while (i < n && j >= 0) { if (mat[i][j] == x) { printf("Element found at %d, %d", i, j); return 1; } if (mat[i][j] > x) j--; else // if mat[i][j] < x i++; } printf("n Element not found"); return 0; // if ( i==n || j== -1 )}// Driver codeint main(){ int mat[4][4] = { { 10, 20, 30, 40 }, { 15, 25, 35, 45 }, { 27, 29, 37, 48 }, { 32, 33, 39, 50 }, }; // Function call search(mat, 4, 29); return 0;} |
Java
// JAVA Code for Search in a row wise and// column wise sorted matrixclass GFG { /* Searches the element x in mat[][]. If the element is found, then prints its position and returns true, otherwise prints "not found" and returns false */ private static void search(int[][] mat, int n, int x) { // set indexes for top right int i = 0, j = n - 1; // element while (i < n && j >= 0) { if (mat[i][j] == x) { System.out.print("Element found at " + i + " " + j); return; } if (mat[i][j] > x) j--; else // if mat[i][j] < x i++; } System.out.print("n Element not found"); return; // if ( i==n || j== -1 ) } // Driver code public static void main(String[] args) { int mat[][] = { { 10, 20, 30, 40 }, { 15, 25, 35, 45 }, { 27, 29, 37, 48 }, { 32, 33, 39, 50 } }; // Function call search(mat, 4, 29); }}// This code is contributed by Arnav Kr. Mandal. |
Python3
# Python3 program to search an element# in row-wise and column-wise sorted matrix# Searches the element x in mat[][]. If the# element is found, then prints its position# and returns true, otherwise prints "not found"# and returns falsedef search(mat, n, x): i = 0 # set indexes for top right element j = n - 1 while (i < n and j >= 0): if (mat[i][j] == x): print("Element found at ", i, ", ", j) return 1 if (mat[i][j] > x): j -= 1 # if mat[i][j] < x else: i += 1 print("Element not found") return 0 # if (i == n || j == -1 )# Driver Codeif __name__ == "__main__": mat = [[10, 20, 30, 40], [15, 25, 35, 45], [27, 29, 37, 48], [32, 33, 39, 50]] # Function call search(mat, 4, 29)# This code is contributed by Anant Agarwal. |
C#
// C# Code for Search in a row wise and// column wise sorted matrixusing System;class GFG { /* Searches the element x in mat[][]. If the element is found, then prints its position and returns true, otherwise prints "not found" and returns false */ private static void search(int[, ] mat, int n, int x) { // set indexes for top right // element int i = 0, j = n - 1; while (i < n && j >= 0) { if (mat[i, j] == x) { Console.Write("Element found at " + i + ", " + j); return; } if (mat[i, j] > x) j--; else // if mat[i][j] < x i++; } Console.Write("n Element not found"); return; // if ( i==n || j== -1 ) } // Driver code public static void Main() { int[, ] mat = { { 10, 20, 30, 40 }, { 15, 25, 35, 45 }, { 27, 29, 37, 48 }, { 32, 33, 39, 50 } }; // Function call search(mat, 4, 29); }}// This code is contributed by Sam007 |
Javascript
<script>// JAVA SCRIPT Code for Search in a row wise and// column wise sorted matrix/* Searches the element x in mat[][]. If theelement is found, then prints its positionand returns true, otherwise prints "not found"and returns false */function search(mat,n,x){ // set indexes for top right let i = 0, j = n - 1; // element while (i < n && j >= 0) { if (mat[i][j] == x) { document.write("Element found at " + i + " " + j); return; } if (mat[i][j] > x) j--; else // if mat[i][j] < x i++; } document.write("n Element not found"); return; // if ( i==n || j== -1 )}// driver program to test above function let mat = [[10, 20, 30, 40 ], [ 15, 25, 35, 45 ], [ 27, 29, 37, 48 ], [ 32, 33, 39, 50 ]]; search(mat, 4, 29); // This code is contributed by bobby</script> |
PHP
<?php // PHP program to search an // element in row-wise and // column-wise sorted matrix/* Searches the element $x in mat[][]. If the element is found, then prints its position and returns true, otherwise prints"not found" and returns false */function search(&$mat, $n, $x){ $i = 0; $j = $n - 1; // set indexes for // top right element while ($i < $n && $j >= 0) { if ($mat[$i][$j] == $x) { echo "Element found at " . $i. ", " . $j; return 1; } if ($mat[$i][$j] > $x) $j--; else // if $mat[$i][$j] < $x $i++; } echo "n Element not found"; return 0; // if ( $i==$n || $j== -1 )}// Driver Code$mat = array(array(10, 20, 30, 40), array(15, 25, 35, 45), array(27, 29, 37, 48), array(32, 33, 39, 50));// Function callsearch($mat, 4, 29);// This code is contributed// by ChitraNayal?> |
Output
Element found at 2, 1
Time Complexity: O(N), Only one traversal is needed, i.e, i from 0 to n and j from n-1 to 0 with at most 2*N steps. The above approach will also work for the M x N matrix (not only for N x N). Complexity would be O(M + N)
Auxiliary Space: O(1), No extra space is required
Related Article: Search element in a sorted matrix
Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.
Using Linear Search:
Approach:
In this approach, we traverse the matrix row by row and check each element until we find the target element.
Define a function search_element that takes a matrix mat and a target element x as input.
Traverse the matrix mat row by row using a nested loop.
For each element in the matrix, check if it is equal to the target element x.
If the target element is found, return its position as a string in the format “Found at (i, j)”, where i and j are the row and column indices of the element, respectively.
If the target element is not found, return the string “Element not found”.
C++
#include <iostream>#include <vector>using namespace std;string search_element(vector<vector<int>> mat, int x) { for (int i = 0; i < mat.size(); i++) { for (int j = 0; j < mat[0].size(); j++) { if (mat[i][j] == x) { return "Found at (" + to_string(i) + ", " + to_string(j) + ")"; } } } return "Element not found";}int main() { vector<vector<int>> mat = {{10, 20, 30, 40}, {15, 25, 35, 45}, {27, 29, 37, 48}, {32, 33, 39, 50}}; int x = 29; cout << search_element(mat, x) << endl; // Output: Found at (2, 1) x = 100; cout << search_element(mat, x) << endl; // Output: Element not found return 0;} |
Java
// Java Code for the above approachimport java.util.ArrayList;import java.util.List;class GFG { public static String search_element(List<List<Integer>> mat, int x) { for (int i = 0; i < mat.size(); i++) { for (int j = 0; j < mat.get(0).size(); j++) { if (mat.get(i).get(j) == x) { return "Found at (" + i + ", " + j + ")"; } } } return "Element not found"; } public static void main(String[] args) { List<List<Integer>> mat = new ArrayList<>(); mat.add(List.of(10, 20, 30, 40)); mat.add(List.of(15, 25, 35, 45)); mat.add(List.of(27, 29, 37, 48)); mat.add(List.of(32, 33, 39, 50)); int x = 29; System.out.println(search_element(mat, x)); // Output: Found at (2, 1) x = 100; System.out.println(search_element(mat, x)); // Output: Element not found }}// This code is contributed by Kanchan Agarwal |
Python3
def search_element(mat, x): for i in range(len(mat)): for j in range(len(mat[0])): if mat[i][j] == x: return f"Found at ({i}, {j})" return "Element not found" mat = [ [10, 20, 30, 40], [15, 25, 35, 45], [27, 29, 37, 48], [32, 33, 39, 50]]x = 29print(search_element(mat, x)) # Output: Found at (2, 1)x = 100print(search_element(mat, x)) # Output: Element not found |
C#
using System;using System.Collections.Generic;class GFG{ static string SearchElement(List<List<int>> mat, int x) { for (int i = 0; i < mat.Count; i++) { for (int j = 0; j < mat[0].Count; j++) { if (mat[i][j] == x) { return "Found at (" + i + ", " + j + ")"; } } } return "Element not found"; } static void Main(string[] args) { List<List<int>> mat = new List<List<int>> { new List<int> { 10, 20, 30, 40 }, new List<int> { 15, 25, 35, 45 }, new List<int> { 27, 29, 37, 48 }, new List<int> { 32, 33, 39, 50 } }; int x = 29; Console.WriteLine(SearchElement(mat, x)); // Output: Found at (2, 1) x = 100; Console.WriteLine(SearchElement(mat, x)); // Output: Element not found }} |
Javascript
function search_element(mat, x) { for (let i = 0; i < mat.length; i++) { for (let j = 0; j < mat[0].length; j++) { if (mat[i][j] === x) { return "Found at (" + i + ", " + j + ")"; } } } return "Element not found";}// Driver Program to test above functionconst mat = [ [10, 20, 30, 40], [15, 25, 35, 45], [27, 29, 37, 48], [32, 33, 39, 50]];let x = 29;console.log(search_element(mat, x)); // Output: Found at (2, 1)x = 100;console.log(search_element(mat, x)); // Output: Element not found// THIS CODE IS CONTRIBUTED BY KANCHAN AGARWAL |
Output
Found at (2, 1) Element not found
Time Complexity: O(n^2)
Space Complexity: O(1)
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