# LeetCode: Special Positions in a Binary Matrix Solution

*Count number of 1s*

## Approach

Count number of 1s in each row, column

A special position is the position that

- has the value of 1
- number of 1s in its row and column are both 1

## Implementation

1var numSpecial = function (mat) {2 const [m, n] = [mat.length, mat[0].length]3 const [numberOf1sInRow, numberOf1sInCol] = [4 new Uint8Array(m),5 new Uint8Array(n),6 ]7 let res = 089 for (let row = 0; row < m; row++) {10 for (let col = 0; col < n; col++) {11 if (mat[row][col]) {12 numberOf1sInRow[row]++13 numberOf1sInCol[col]++14 }15 }16 }1718 for (let row = 0; row < m; row++) {19 for (let col = 0; col < n; col++) {20 res +=21 mat[row][col] &&22 numberOf1sInRow[row] === 1 &&23 numberOf1sInCol[col] === 124 // boolean coerce to 1 if true, 0 if false25 }26 }2728 return res29}

## References

## Similar problems

N/A

## Comments

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## Tags

leetcode

array

matrix

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