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

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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]
                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

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C#

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// 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

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PHP

Output:

20


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