The K Weakest Rows in a Matrix

Description

Given a m * n matrix mat of ones (representing soldiers) and zeros (representing civilians), return the indexes of the k weakest rows in the matrix ordered from the weakest to the strongest.

A row i is weaker than row j, if the number of soldiers in row i is less than the number of soldiers in row j, or they have the same number of soldiers but i is less than j. Soldiers are always stand in the frontier of a row, that is, always ones may appear first and then zeros.

 

Example 1:

Input: mat = 
[[1,1,0,0,0],
 [1,1,1,1,0],
 [1,0,0,0,0],
 [1,1,0,0,0],
 [1,1,1,1,1]], 
k = 3
Output: [2,0,3]
Explanation: 
The number of soldiers for each row is: 
row 0 -> 2 
row 1 -> 4 
row 2 -> 1 
row 3 -> 2 
row 4 -> 5 
Rows ordered from the weakest to the strongest are [2,0,3,1,4]

Example 2:

Input: mat = 
[[1,0,0,0],
 [1,1,1,1],
 [1,0,0,0],
 [1,0,0,0]], 
k = 2
Output: [0,2]
Explanation: 
The number of soldiers for each row is: 
row 0 -> 1 
row 1 -> 4 
row 2 -> 1 
row 3 -> 1 
Rows ordered from the weakest to the strongest are [0,2,3,1]

 

Constraints:

• m == mat.length
• n == mat[i].length
• 2 <= n, m <= 100
• 1 <= k <= m
• matrix[i][j] is either 0 or 1.

Solution(javascript)

/** 为了更好的性能， 可以使用 Heap(klog(n))，不过这里我就排序来做了
 * @param {number[][]} mat
 * @param {number} k
 * @return {number[]}
 */
const kWeakestRows = function (mat, k) {
  const sum = arr => arr.reduce((acc, num) => acc + num)
  mat.forEach((row, index) => {
    row[1] = row[1] + row[0] - index
    row[0] = index
  })
  mat.sort((a, b) => {
    const sumA = sum(a)
    const sumB = sum(b)
    return sumA === sumB ? a[0] - b[0] : sumA - sumB
  })
  return mat.slice(0, k).map(a => a[0])
}