Special Positions in a Binary Matrix

Description

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.

Solution(javascript)

/**
 * @param {number[][]} mat
 * @return {number}
 */
const numSpecial = function (mat) {
  const count = (nums = []) => nums.reduce((acc, num) => (acc + (num === 1 ? 1 : 0)), 0)

  const columns = mat.reduce((acc, items) => {
    if (count(items) === 1) {
      const columnIndex = items.findIndex(x => x === 1)
      acc.add(columnIndex)
    }
    return acc
  }, new Set())
  let result = 0
  for (const column of columns.values()) {
    let ones = 0
    for (let row = 0; row < mat.length; row++) {
      if (mat[row][column] === 1) {
        ones += 1
      }
    }
    if (ones === 1) {
      result += 1
    }
  }
  return result
}