2482. Difference Between Ones and Zeros in Row and Column
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You are given a 0-indexed m x n binary matrix grid.
A 0-indexed m x n difference matrix diff is created with the following procedure:
- Let the number of ones in the
ithrow beonesRowi. - Let the number of ones in the
jthcolumn beonesColj. - Let the number of zeros in the
ithrow bezerosRowi. - Let the number of zeros in the
jthcolumn bezerosColj. diff[i][j] = onesRowi + onesColj - zerosRowi - zerosColj
Return the difference matrix diff.
Example 1:
Input: grid = [[0,1,1],[1,0,1],[0,0,1]] Output: [[0,0,4],[0,0,4],[-2,-2,2]] Explanation: - diff[0][0] =onesRow0 + onesCol0 - zerosRow0 - zerosCol0= 2 + 1 - 1 - 2 = 0 - diff[0][1] =onesRow0 + onesCol1 - zerosRow0 - zerosCol1= 2 + 1 - 1 - 2 = 0 - diff[0][2] =onesRow0 + onesCol2 - zerosRow0 - zerosCol2= 2 + 3 - 1 - 0 = 4 - diff[1][0] =onesRow1 + onesCol0 - zerosRow1 - zerosCol0= 2 + 1 - 1 - 2 = 0 - diff[1][1] =onesRow1 + onesCol1 - zerosRow1 - zerosCol1= 2 + 1 - 1 - 2 = 0 - diff[1][2] =onesRow1 + onesCol2 - zerosRow1 - zerosCol2= 2 + 3 - 1 - 0 = 4 - diff[2][0] =onesRow2 + onesCol0 - zerosRow2 - zerosCol0= 1 + 1 - 2 - 2 = -2 - diff[2][1] =onesRow2 + onesCol1 - zerosRow2 - zerosCol1= 1 + 1 - 2 - 2 = -2 - diff[2][2] =onesRow2 + onesCol2 - zerosRow2 - zerosCol2= 1 + 3 - 2 - 0 = 2
Example 2:
Input: grid = [[1,1,1],[1,1,1]] Output: [[5,5,5],[5,5,5]] Explanation: - diff[0][0] = onesRow0 + onesCol0 - zerosRow0 - zerosCol0 = 3 + 2 - 0 - 0 = 5 - diff[0][1] = onesRow0 + onesCol1 - zerosRow0 - zerosCol1 = 3 + 2 - 0 - 0 = 5 - diff[0][2] = onesRow0 + onesCol2 - zerosRow0 - zerosCol2 = 3 + 2 - 0 - 0 = 5 - diff[1][0] = onesRow1 + onesCol0 - zerosRow1 - zerosCol0 = 3 + 2 - 0 - 0 = 5 - diff[1][1] = onesRow1 + onesCol1 - zerosRow1 - zerosCol1 = 3 + 2 - 0 - 0 = 5 - diff[1][2] = onesRow1 + onesCol2 - zerosRow1 - zerosCol2 = 3 + 2 - 0 - 0 = 5
Constraints:
m == grid.lengthn == grid[i].length1 <= m, n <= 1051 <= m * n <= 105grid[i][j]is either0or1.
Hints
Hint 1
You need to reuse information about a row or a column many times. Try storing it to avoid computing it multiple times.
Hint 2
Use an array to store the number of 1’s in each row and another array to store the number of 1’s in each column. Once you know the number of 1’s in each row or column, you can also easily calculate the number of 0’s.