Given a square matrix of size . Find minimum number of operation are required such that sum of elements on each row and column becomes equals. In one operation, increment any value of cell of matrix by 1. In first line print minimum operation required and in next ‘n’ lines print ‘n’ integers representing the final matrix after operation.
Input: 1 2 3 4 Output: 4 4 3 3 4 Explanation 1. Increment value of cell(0, 0) by 3 2. Increment value of cell(0, 1) by 1 Hence total 4 operation are required Input: 9 1 2 3 4 2 3 3 2 1 Output: 6 2 4 3 4 2 3 3 3 3
The approach is simple, let’s assume that maxSum is the maximum sum among all rows and columns. We just need to increment some cells such that the sum of any row or column becomes ‘maxSum’.
Let’s say Xi is the total number of operation needed to make the sum on row ‘i’ equals to maxSum and Yj is the total number of operation needed to make the sum on column ‘j’ equals to maxSum. Since Xi = Yj so we need to work at any one of them according to the condition.
In order to minimise Xi, we need to choose the maximum from rowSumi and colSumj as it will surely lead to minimum operation. After that, increment ‘i’ or ‘j’ according to the condition satisfied after increment.
Below is the implementation of the above approach.
Output 4 4 3 3 4
Time complexity: O(n2)
Auxiliary space: O(n)
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