you are given an N*N matrix, where each element of the matrix lies in the range 0 to M. You can apply the below operation on the matrix any number of times:
- Choose any two consecutive elements
- Increment one of them by 1 and decrease the other by 1
Note: The elements should remain within the range of 0 to M after applying the above operations.
The task is to find the maximum value of the expression shown below that can be obtained after performing the above operation on the matrix if required:
res += (i+j)*A[i][j] for 0 <= i, j <= N
Examples:
Input : A[][] = {1, 2,
5, 1}
M = 5
Output : RESULT = 27
Matrix : 0 0
4 5
Input : A[][] = {3, 4,
5, 4}
M = 6
Output : RESULT = 43
Matrix : 0 4
6 6
Algorithm :
Below is the step-by-step algorithm to do this:
- First of all, calculate the sum of all elements of the given matrix as SUM.
- Start from last element that i.e. A(n, n) and move backward towards A(0,0) anti-diagonally as A(n, n), A(n, n-1), A(n-1, n), A(n, n-2), A(n-1, n-1), A(n-2, n)…..
- Fill up each cell of a matrix with M and update SUM = SUM- M for each element till SUM < M. Now, Fill the SUM value at the next place in order if it is greater than zero and all other remaining places as zero.
- Finally, you can calculate RESULT as per the above-mentioned formula.
Example :
Input Matrix:
Solution Matrix after applying the above algorithm :
Below is the implementation of the above idea :
C++
// CPP to maximize matrix result #include<bits/stdc++.h> using namespace std; #define n 4 // utility function for maximize matrix result int maxMatrix(int A[][n], int M) { int sum = 0, res = 0; for ( int i=0; i<n ; i++) for ( int j=0; j<n; j++) sum += A[i][j]; // diagonals below longest diagonal // starting from last element of matrix for (int j=n-1; j>0; j--) { for (int i=0; i<n-j; i++) { if (sum > M) { A[n-1-i][j+i] = M; sum -= M; } else { A[n-1-i][j+i] = sum; sum -= sum; } } } // diagonals above longest diagonal for (int i=n-1; i>=0; i--) { for (int j=0; j<=i; j++) { if (sum > M) { A[i-j][j] = M; sum -= M; } else { A[i-j][j] = sum; sum -= sum; } } } // calculating result for (int i=0; i<n; i++) { for (int j=0; j<n;j++) res += (i+j+2) * A[i][j]; } return res; } // driver program int main() { int A[n][n] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 1, 2, 3, 4, 5, 6}; int m = 9; cout << maxMatrix(A, m); return 0; } |
Java
// Java to maximize matrix result class GFG { static final int n = 4; // utility function for maximize matrix result static int maxMatrix(int A[][], int M) { int sum = 0, res = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { sum += A[i][j]; } } // diagonals below longest diagonal // starting from last element of matrix for (int j = n - 1; j > 0; j--) { for (int i = 0; i < n - j; i++) { if (sum > M) { A[n - 1 - i][j + i] = M; sum -= M; } else { A[n - 1 - i][j + i] = sum; sum -= sum; } } } // diagonals above longest diagonal for (int i = n - 1; i >= 0; i--) { for (int j = 0; j <= i; j++) { if (sum > M) { A[i - j][j] = M; sum -= M; } else { A[i - j][j] = sum; sum -= sum; } } } // calculating result for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { res += (i + j + 2) * A[i][j]; } } return res; } // driver program static public void main(String[] args) { int A[][] = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 1, 1, 2}, {3, 4, 5, 6}}; int m = 9; System.out.println(maxMatrix(A, m)); } } // This code is contributed by Rajput-Ji |
Python3
# Python to maximize matrix result n = 4 # utility function for maximize # matrix result def maxMatrix(A, M): sum, res = 0, 0 for i in range(n): for j in range(n): sum += A[i][j] # diagonals below longest diagonal # starting from last element of matrix for j in range(n - 1, 0, -1): for i in range(n - j): if (sum > M): A[n - 1 - i][j + i] = M sum -= M else: A[n - 1 - i][j + i] = sum sum -= sum # diagonals above longest diagonal for i in range(n - 1, -1, -1): for j in range(i + 1): if (sum > M): A[i - j][j] = M sum -= M else: A[i - j][j] = sum sum -= sum # calculating result for i in range(n): for j in range(n): res += (i + j + 2) * A[i][j] return res # Driver Code if __name__ == '__main__': A = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 1, 1, 2], [3, 4, 5, 6]] m = 9 print(maxMatrix(A, m)) # This code is contributed by 29AjayKumar |
C#
// C# to maximize matrix result using System; public class GFG { static readonly int n = 4; // utility function for maximize matrix result static int maxMatrix(int [,]A, int M) { int sum = 0, res = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { sum += A[i,j]; } } // diagonals below longest diagonal // starting from last element of matrix for (int j = n - 1; j > 0; j--) { for (int i = 0; i < n - j; i++) { if (sum > M) { A[n - 1 - i,j + i] = M; sum -= M; } else { A[n - 1 - i,j + i] = sum; sum -= sum; } } } // diagonals above longest diagonal for (int i = n - 1; i >= 0; i--) { for (int j = 0; j <= i; j++) { if (sum > M) { A[i - j,j] = M; sum -= M; } else { A[i - j,j] = sum; sum -= sum; } } } // calculating result for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { res += (i + j + 2) * A[i,j]; } } return res; } // driver program static public void Main() { int [,]A= {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 1, 1, 2}, {3, 4, 5, 6}}; int m = 9; Console.Write(maxMatrix(A, m)); } } // This code is contributed by Rajput-Ji |
PHP
<?php // PHP to maximize matrix result $n = 4; // function for maximize // matrix result function maxMatrix($A, $M) { global $n; $sum = 0; $res = 0; for ($i = 0; $i < $n ; $i++) for ($j = 0; $j < $n; $j++) $sum += $A[$i][$j]; // diagonals below longest diagonal // starting from last element of matrix for ($j = $n - 1; $j > 0; $j--) { for ($i = 0; $i < $n - $j; $i++) { if ($sum > $M) { $A[$n - 1 - $i][$j + $i] = $M; $sum -= $M; } else { $A[$n - 1 - $i][$j + i] = $sum; $sum -= $sum; } } } // diagonals above longest diagonal for ($i = $n - 1; $i >= 0; $i--) { for ($j = 0; $j <= $i; $j++) { if ($sum > $M) { $A[$i - $j][$j] = $M; $sum -= $M; } else { $A[$i - $j][$j] = $sum; $sum -= $sum; } } } // calculating result for ($i = 0; $i < $n; $i++) { for ($j = 0; $j < $n; $j++) $res += ($i + $j + 2) * $A[$i][$j]; } return $res; } // Driver Code $A = array(array(1, 2, 3, 4), array(5, 6, 7, 8), array(9, 1, 1, 2), array(3, 4, 5, 6)); $m = 9; echo maxMatrix($A, $m); // This code is contributed by anuj_67. ?> |
Javascript
<script> // Javascript to maximize matrix result let n = 4; // utility function for maximize matrix result function maxMatrix(A, M) { let sum = 0, res = 0; for (let i = 0; i < n; i++) { for (let j = 0; j < n; j++) { sum += A[i][j]; } } // diagonals below longest diagonal // starting from last element of matrix for (let j = n - 1; j > 0; j--) { for (let i = 0; i < n - j; i++) { if (sum > M) { A[n - 1 - i][j + i] = M; sum -= M; } else { A[n - 1 - i][j + i] = sum; sum -= sum; } } } // diagonals above longest diagonal for (let i = n - 1; i >= 0; i--) { for (let j = 0; j <= i; j++) { if (sum > M) { A[i - j][j] = M; sum -= M; } else { A[i - j][j] = sum; sum -= sum; } } } // calculating result for (let i = 0; i < n; i++) { for (let j = 0; j < n; j++) { res += (i + j + 2) * A[i][j]; } } return res; } let A = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 1, 1, 2], [3, 4, 5, 6]]; let m = 9; document.write(maxMatrix(A, m)); // This code is contributed by divyesh072019. </script> |
Output
425
Time Complexity: O(n2)
Auxiliary Space: O(1)
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