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

• Difficulty Level : Easy
• Last Updated : 22 Apr, 2021

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 = {{-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```

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++

 `// 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

 `// 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

 `# 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#

 `// 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 - rowSum;``        ``int` `min_element = rowSum;``        ` `        ``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.`

## PHP

 ` ``\$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``?>`

## Javascript

 ``

Output:

`9`

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

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