Given an N x N matrix mat[][], the task is to find the maximum trace possible for any sub-matrix of the given matrix.
Examples:
Input: mat[][] = {
{10, 2, 5},
{6, 10, 4},
{2, 7, -10}}
Output: 20
{{10, 2},
{6, 10}}
is the sub-matrix with the maximum trace.Input: mat[][] = {
{1, 2, 5},
{6, 3, 4},
{2, 7, 1}}
Output: 13
Approach: An efficient approach is to take the sum along the diagonal from each element of the matrix and update the maximum sum as the trace of any square matrix is the sum of the elements on its main diagonal.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define N 3// Function to return the maximum trace possible// for a sub-matrix of the given matrixint MaxTraceSub(int mat[][N]){ int max_trace = 0; for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { int r = i, s = j, trace = 0; // Calculate the trace for each of // the sub-matrix with top left corner // at cell (r, s) while (r < N && s < N) { trace += mat[r][s]; r++; s++; // Update the maximum trace max_trace = max(trace, max_trace); } } } // Return the maximum trace return max_trace;}// Driver codeint main(){ int mat[N][N] = { { 10, 2, 5 }, { 6, 10, 4 }, { 2, 7, -10 } }; cout << MaxTraceSub(mat); return 0;} |
Java
// Java program for the above approachclass GFG { static int N = 3;// Function to return the maximum trace possible// for a sub-matrix of the given matrixstatic int MaxTraceSub(int mat[][]){ int max_trace = 0; for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { int r = i, s = j, trace = 0; // Calculate the trace for each of // the sub-matrix with top left corner // at cell (r, s) while (r < N && s < N) { trace += mat[r][s]; r++; s++; // Update the maximum trace max_trace = Math.max(trace, max_trace); } } } // Return the maximum trace return max_trace;}// Driver codepublic static void main(String[] args){ int mat[][] = { { 10, 2, 5 }, { 6, 10, 4 }, { 2, 7, -10 } }; System.out.println(MaxTraceSub(mat));}}// This code has been contributed by 29AjayKumar |
Python3
# Python 3 implementation of the approachN = 3# Function to return the maximum trace possible# for a sub-matrix of the given matrixdef MaxTraceSub(mat): max_trace = 0 for i in range(N): for j in range(N): r = i s = j trace = 0 # Calculate the trace for each of # the sub-matrix with top left corner # at cell (r, s) while (r < N and s < N): trace += mat[r][s] r += 1 s += 1 # Update the maximum trace max_trace = max(trace, max_trace) # Return the maximum trace return max_trace# Driver codeif __name__ == '__main__': mat = [[10, 2, 5],[6, 10, 4],[2, 7, -10]] print(MaxTraceSub(mat))# This code is contributed by# Surendra_Gangwar |
C#
// C# program for the above approachusing System;class GFG { static int N = 3;// Function to return the maximum trace possible// for a sub-matrix of the given matrixstatic int MaxTraceSub(int [][]mat){ int max_trace = 0; for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { int r = i, s = j, trace = 0; // Calculate the trace for each of // the sub-matrix with top left corner // at cell (r, s) while (r < N && s < N) { trace += mat[r][s]; r++; s++; // Update the maximum trace max_trace = Math.Max(trace, max_trace); } } } // Return the maximum trace return max_trace;}// Driver codepublic static void Main(){ int[][] mat = new int[][]{new int[]{ 10, 2, 5 }, new int[]{ 6, 10, 4 }, new int[]{ 2, 7, -10 } }; Console.WriteLine(MaxTraceSub(mat));}}// This code has been contributed// by Akanksha Rai |
PHP
<?php// PHP implementation of the approach$N = 3;// Function to return the maximum // trace possible for a sub-matrix // of the given matrixfunction MaxTraceSub($mat){ global $N; $max_trace = 0; for ($i = 0; $i < $N; $i++) { for ($j = 0; $j < $N; $j++) { $r = $i; $s = $j; $trace = 0; // Calculate the trace for // each of the sub-matrix // with top left corner at // cell (r, s) while ($r < $N && $s < $N) { $trace += $mat[$r][$s]; $r++; $s++; // Update the maximum trace $max_trace = max($trace, $max_trace); } } } // Return the maximum trace return $max_trace;}// Driver code$mat = array(array( 10, 2, 5 ), array( 6, 10, 4 ), array( 2, 7, -10 ));print(MaxTraceSub($mat));// This code is contributed by mits?> |
Javascript
<script>// Javascript program for the above approach var N = 3; // Function to return the maximum trace possible // for a sub-matrix of the given matrix function MaxTraceSub(mat) { var max_trace = 0; for (i = 0; i < N; i++) { for (j = 0; j < N; j++) { var r = i, s = j, trace = 0; // Calculate the trace for each of // the sub-matrix with top left corner // at cell (r, s) while (r < N && s < N) { trace += mat[r][s]; r++; s++; // Update the maximum trace max_trace = Math.max(trace, max_trace); } } } // Return the maximum trace return max_trace; } // Driver code var mat = [ [ 10, 2, 5 ], [ 6, 10, 4 ], [ 2, 7, -10 ] ]; document.write(MaxTraceSub(mat));// This code contributed by umadevi9616</script> |
Output:
20
Time Complexity: O(N*N), as we are using nested loops for traversing the matrix.
Auxiliary Space: O(1), as we are not using any extra space.
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