Skip to content
Related Articles

Related Articles

Maximum trace possible for any sub-matrix of the given matrix

Improve Article
Save Article
  • Difficulty Level : Basic
  • Last Updated : 26 May, 2022
Improve Article
Save Article

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 matrix
int 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 code
int 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 approach
class GFG
{
     
static int N = 3;
 
// Function to return the maximum trace possible
// for a sub-matrix of the given matrix
static 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 code
public 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 approach
N = 3
 
# Function to return the maximum trace possible
# for a sub-matrix of the given matrix
def 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 code
if __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 approach
using System;
 
class GFG
{
     
static int N = 3;
 
// Function to return the maximum trace possible
// for a sub-matrix of the given matrix
static 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 code
public 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 matrix
function 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.


My Personal Notes arrow_drop_up
Related Articles

Start Your Coding Journey Now!