Maximize matrix as per given condition

You are given a N*N matrix, where each element of matrix lies in the range 0 to M. You can apply the below operation on matrix any number of times:

• Choose any two consecutive elements
• Increment one of them by 1 and decrease other by 1

Note: The elements should remain within the range 0 to M after applying 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 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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Algorithm :
Below is the step by step algorithm to do this:

1. First of all, calculate the sum of all elements of given matrix as SUM.
2. 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)…..
3. Fill up each cell of matrix with M and update SUM = SUM- M for each element till SUM < M. Now, Fill the SUM value at next place in order if it is greater than zero and all other remaining place as zero.
4. Finally you can calculate RESULT as per above mentioned formula.

Example :
Input Matrix: Solution Matrix after applying above algorithm : Below is the implementation of above idea :

C++

 // CPP to maximize matrix result #include 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; i0; j--)     {         for (int i=0; i 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

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

 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. ?>

Output:

425

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Improved By : vt_m, Rajput-Ji, 29AjayKumar

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