# Maximum difference of sum of elements in two rows in a matrix

Given a matrix of m*n order, the task is to find the maximum difference between two rows Rj and Ri such that i < j, i.e., we need to find maximum value of sum(Rj) – sum(Ri) such that row i is above row j.

Examples:

Input : mat[5][4] = {{-1, 2, 3, 4},
{5, 3, -2, 1},
{6, 7, 2, -3},
{2, 9, 1, 4},
{2, 1, -2, 0}}
Output: 9
// difference of R3 - R1 is maximum

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

A simple solution for this problem is to one by one select each row, compute sum of elements in it and take difference from sum of next rows in forward direction. Finally return the maximum difference. Time complexity for this approach will be O(n*m2).

An efficient solution solution for this problem is to first calculate the sum of all elements of each row and store them in an auxiliary array rowSum[] and then calculate maximum difference of two elements max(rowSum[j] – rowSum[i]) such that rowSum[i] < rowSum[j] in linear time. See this article. In this method, we need to keep track of 2 things:
1) Maximum difference found so far (max_diff).
2) Minimum number visited so far (min_element).

 // C++ program to find maximum difference of sum of // elements of two rows #include #define MAX 100 using namespace std;    // Function to find maximum difference of sum of // elements of two rows such that second row appears // before first row. int maxRowDiff(int mat[][MAX], int m, int n) {     // auxiliary array to store sum of all elements     // of each row     int rowSum[m];        // calculate sum of each row and store it in     // rowSum array     for (int i=0; i max_diff)             max_diff = rowSum[i] - min_element;            // if new element is less than previous minimum         // element then update it so that         // we may get maximum difference in remaining array         if (rowSum[i] < min_element)             min_element = rowSum[i];     }        return max_diff; }    // Driver program to run the case int main() {     int m = 5, n = 4;     int mat[][MAX] = {{-1, 2, 3, 4},                      {5, 3, -2, 1},                      {6, 7, 2, -3},                      {2, 9, 1, 4},                      {2, 1, -2, 0}};        cout << maxRowDiff(mat, m, n);     return 0; }

 // Java program to find maximum difference // of sum of elements of two rows class GFG { static final int MAX = 100;    // Function to find maximum difference of sum  // of elements of two rows such that second  // row appears before first row. static int maxRowDiff(int mat[][], int m, int n) {        // auxiliary array to store sum      // of all elements of each row     int rowSum[] = new int[m];        // calculate sum of each row and      // store it in rowSum array     for (int i = 0; i < m; i++) {     int sum = 0;     for (int j = 0; j < n; j++)         sum += mat[i][j];     rowSum[i] = sum;     }        // calculating maximum difference of two elements     // such that rowSum[i] max_diff)         max_diff = rowSum[i] - min_element;        // if new element is less than previous      // minimum element then update it so that     // we may get maximum difference in remaining array     if (rowSum[i] < min_element)         min_element = rowSum[i];     }        return max_diff; }    // Driver code public static void main(String[] args) {     int m = 5, n = 4;     int mat[][] = {{-1, 2,  3, 4 },                      {5,  3, -2, 1 },                     {6,  7,  2, -3},                    {2,  9,  1, 4 },                    {2,  1, -2, 0}};        System.out.print(maxRowDiff(mat, m, n)); } } // This code is contributed by Anant Agarwal.

 # Python3 program to find maximum difference  # of sum of elements of two rows    # Function to find maximum difference of  # sum of elements of two rows such that  # second row appears before first row. def maxRowDiff(mat, m, n):            # auxiliary array to store sum of      # all elements of each row     rowSum = [0] * m            # calculate sum of each row and      # store it in rowSum array     for i in range(0, m):         sum = 0         for j in range(0, n):             sum += mat[i][j]          rowSum[i] = sum            # calculating maximum difference of      # two elements such that      # rowSum[i] max_diff):             max_diff = rowSum[i] - min_element                    # if new element is less than previous         # minimum element then update it so          # that we may get maximum difference          # in remaining array         if (rowSum[i] < min_element):             min_element = rowSum[i]      return max_diff     # Driver program to run the case m = 5 n = 4 mat = [[-1, 2, 3, 4],        [5, 3, -2, 1],        [6, 7, 2, -3],        [2, 9, 1, 4],        [2, 1, -2, 0]]    print( maxRowDiff(mat, m, n))    # This code is contributed by Swetank Modi

 // C# program to find maximum difference // of sum of elements of two rows using System;    class GFG {            // Function to find maximum difference      // of sum of elements of two rows such     // that second row appears before     // first row.     static int maxRowDiff(int [,] mat,                                int m, int n)     {                // auxiliary array to store sum          // of all elements of each row         int [] rowSum = new int[m];                // calculate sum of each row and          // store it in rowSum array         for (int i = 0; i < m; i++)         {             int sum = 0;                            for (int j = 0; j < n; j++)                 sum += mat[i,j];                                rowSum[i] = sum;         }                // calculating maximum difference         // of two elements such that          // rowSum[i] < rowsum[j]         int max_diff = rowSum[1] - rowSum[0];         int min_element = rowSum[0];                    for (int i = 1; i < m; i++)         {                    // if current difference is              // greater than previous then              // update it             if (rowSum[i] - min_element                                    > max_diff)                 max_diff = rowSum[i]                               - min_element;                        // if new element is less than             // previous minimum element then             // update it so that we may get              // maximum difference in              // remaining array             if (rowSum[i] < min_element)                 min_element = rowSum[i];         }                return max_diff;     }            // Driver code     public static void Main()     {         int m = 5, n = 4;         int [,] mat = { {-1, 2, 3, 4 },                         {5, 3, -2, 1 },                         {6, 7, 2, -3},                         {2, 9, 1, 4 },                         {2, 1, -2, 0} };                Console.Write(maxRowDiff(mat, m, n));     } }    // This code is contributed by KRV.

 \$max_diff)             \$max_diff = \$rowSum[\$i] - \$min_element;            // if new element is less          // than previous minimum         // element then update it          // so that we may get maximum         // difference in remaining array         if (\$rowSum[\$i] < \$min_element)             \$min_element = \$rowSum[\$i];     }        return \$max_diff; }    // Driver Code \$m = 5; \$n = 4; \$mat = array(array(-1, 2, 3, 4),              array(5, 3, -2, 1),              array(6, 7, 2, -3),              array(2, 9, 1, 4),              array(2, 1, -2, 0));    echo maxRowDiff(\$mat, \$m, \$n);    // This code is contributed by ajit ?>

Output:
9

Time complexity : O(m*n)
Auxiliary space : O(m)

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Improved By : KRV, jit_t

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