Skip to content
Related Articles

Related Articles

Improve Article

Maximize sum by traversing diagonally from each cell of a given Matrix

  • Last Updated : 29 May, 2021

Given a 2D square matrix arr[][] of dimensions N x N, the task is to find the maximum path sum by moving diagonally from any cell and each cell must be visited only once i.e., from the cell (i, j), a player can move to the cell (i + 1, j + 1).

 

Examples:

Input: arr[][] = {{1, 2, 3}, {3, 5, 10}, {1 3 5}} 
Output: 12
Explanation:
Sum of cells (1, 1), (2, 2) and (3, 3) is 11.  
The sum of cells (1, 2), (2, 3) and (1, 3) is 3. 
The sum of cells (2, 1) and (3, 2) is 6.
The sum of cell (3, 1) is 1.
The maximum possible sum is 12.

Input: arr[][] = {{1, 1, 1}, {1 1 1}, {1 1 1}} 
Output: 3



Approach: To solve this problem, the idea is to traverse the matrix diagonally for first row and column elements and sum up their diagonal elements within the range of the matrix. 
Follow the steps below to solve the problem:

  1. Initialize a variable, say max with 0.
  2. Choose each cell (i, j) from the first row and from the first column.
  3. Now, from each cell, find the diagonal sum starting from that cell by incrementing i and j by 1, say sum.
  4. Then, update max as max(max, sum).
  5. After traversing, print max as the required answer.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum sum
int MaximumSum(vector<vector<int> >& arr, int n)
{
 
    int ans = 0;
 
    // Loop to traverse through the
    // upper triangular matrix and
    // update the maximum sum to ans
    for (int i = 0; i < n; i++) {
        int x = 0, y = i, sum = 0;
        for (int j = i; j < n; j++) {
            sum += arr[x++][y++];
        }
        if (sum > ans)
            ans = sum;
    }
 
    // Traverse through the
    // lower triangular matrix
    for (int i = 1; i < n; i++) {
 
        int x = i, y = 0, sum = 0;
 
        for (int j = i; j < n; j++) {
 
            sum += arr[x++][y++];
        }
        if (sum > ans)
            ans = sum;
    }
 
    return ans;
}
 
// Driver Code
int main()
{
 
    // Given matrix
    vector<vector<int> > arr;
    arr = { { 1, 2, 3 },
            { 3, 5, 10 },
            { 1, 3, 5 } };
 
    // Given dimension
    int n = arr.size();
 
    cout << MaximumSum(arr, n);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
class GFG{
 
// Function to find the maximum sum
static int MaximumSum(int [][]arr, int n)
{
 
    int ans = 0;
 
    // Loop to traverse through the
    // upper triangular matrix and
    // update the maximum sum to ans
    for (int i = 0; i < n; i++)
    {
        int x = 0, y = i, sum = 0;
        for (int j = i; j < n; j++)
        {
            sum += arr[x++][y++];
        }
        if (sum > ans)
            ans = sum;
    }
 
    // Traverse through the
    // lower triangular matrix
    for (int i = 1; i < n; i++)
    {
        int x = i, y = 0, sum = 0;
        for (int j = i; j < n; j++)
        {
            sum += arr[x++][y++];
        }
        if (sum > ans)
            ans = sum;
    }
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Given matrix
    int [][]arr = { { 1, 2, 3 },
            { 3, 5, 10 },
            { 1, 3, 5 } };
 
    // Given dimension
    int n = arr.length;
    System.out.print(MaximumSum(arr, n));
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 program for the above approach
 
# Function to find the maximum sum
def MaximumSum(arr, n):
    ans = 0;
 
    # Loop to traverse through the
    # upper triangular matrix and
    # update the maximum sum to ans
    for i in range(n):
        x, y, sum = 0, i, 0
        for j in range(i, n):
            sum, x, y =sum + arr[x][y], x + 1, y + 1
        if (sum > ans):
            ans = sum
 
    # Traverse through the
    # lower triangular matrix
    for i in range(1, n):
 
        x, y, sum = i, 0, 0
 
        for j in range(i, n):
 
            sum, x, y =sum + arr[x][y], x + 1, y + 1
        if (sum > ans):
            ans = sum
    return ans
 
# Driver Code
if __name__ == '__main__':
 
    # Given matrix
    arr = [ [ 1, 2, 3],
            [ 3, 5, 10],
            [ 1, 3, 5 ]]
 
    # Given dimension
    n = len(arr)
    print (MaximumSum(arr, n))
 
    # This code is contributed by mohit kumar 29

C#




// C# program for the above approach
using System;
class GFG{
 
  // Function to find the maximum sum
  static int MaximumSum(int [,]arr, int n)
  {
    int ans = 0;
 
    // Loop to traverse through the
    // upper triangular matrix and
    // update the maximum sum to ans
    for (int i = 0; i < n; i++)
    {
      int x = 0, y = i, sum = 0;
      for (int j = i; j < n; j++)
      {
        sum += arr[x++, y++];
      }
      if (sum > ans)
        ans = sum;
    }
 
    // Traverse through the
    // lower triangular matrix
    for (int i = 1; i < n; i++)
    {
      int x = i, y = 0, sum = 0;
      for (int j = i; j < n; j++)
      {
        sum += arr[x++, y++];
      }
      if (sum > ans)
        ans = sum;
    }
    return ans;
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
 
    // Given matrix
    int [,]arr = { { 1, 2, 3 },
                  { 3, 5, 10 },
                  { 1, 3, 5 } };
 
    // Given dimension
    int n = arr.GetLength(0);
    Console.Write(MaximumSum(arr, n));
  }
}
 
// This code is contributed by shikhasingrajput

Javascript




<script>
 
// Javascript program of the above approach
 
// Function to find the maximum sum
function MaximumSum(arr, n)
{
  
    let ans = 0;
  
    // Loop to traverse through the
    // upper triangular matrix and
    // update the maximum sum to ans
    for (let i = 0; i < n; i++)
    {
        let x = 0, y = i, sum = 0;
        for (let j = i; j < n; j++)
        {
            sum += arr[x++][y++];
        }
        if (sum > ans)
            ans = sum;
    }
  
    // Traverse through the
    // lower triangular matrix
    for (let i = 1; i < n; i++)
    {
        let x = i, y = 0, sum = 0;
        for (let j = i; j < n; j++)
        {
            sum += arr[x++][y++];
        }
        if (sum > ans)
            ans = sum;
    }
    return ans;
}
 
    // Driver Code
     
    // Given matrix
    let arr = [[ 1, 2, 3 ],
            [ 3, 5, 10 ],
            [ 1, 3, 5 ]];
  
    // Given dimension
    let n = arr.length;
    document.write(MaximumSum(arr, n));
  
</script>

  

Output: 
12

 

Time Complexity: O(N2)
Auxiliary Space: O(N2)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :