LeetCode [1582] Special Positions in a Binary Matrix

 1582. Special Positions in a Binary Matrix

Easy

Given a rows x cols matrix mat, where mat[i][j] is either 0 or 1, return the number of special positions in mat.

A position (i,j) is called special if mat[i][j] == 1 and all other elements in row i and column j are 0 (rows and columns are 0-indexed).

 

Example 1:

Input: mat = [[1,0,0],
              [0,0,1],
              [1,0,0]]
Output: 1
Explanation: (1,2) is a special position because mat[1][2] == 1 and all other elements in row 1 and column 2 are 0.

Example 2:

Input: mat = [[1,0,0],
              [0,1,0],
              [0,0,1]]
Output: 3
Explanation: (0,0), (1,1) and (2,2) are special positions. 

Example 3:

Input: mat = [[0,0,0,1],
              [1,0,0,0],
              [0,1,1,0],
              [0,0,0,0]]
Output: 2

Example 4:

Input: mat = [[0,0,0,0,0],
              [1,0,0,0,0],
              [0,1,0,0,0],
              [0,0,1,0,0],
              [0,0,0,1,1]]
Output: 3

 

Constraints:

• rows == mat.length
• cols == mat[i].length
• 1 <= rows, cols <= 100
• mat[i][j] is 0 or 1.

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class Solution { public int numSpecial(int[][] mat) { int m = mat.length, n = mat[0].length; int[] r = new int[m]; int[] c = new int[n]; int cnt = 0; for(int i=0; i<m; ++i){ for(int j=0; j<n; ++j){ if(mat[i][j]==1){ r[i]++; c[j]++; } } } for(int i=0; i<m; ++i){ for(int j=0; j<n; ++j){ if(mat[i][j]==1 && r[i]==1 && c[j]==1){ cnt++; } } } return cnt; } }