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O(nlogk) time and O(n) space using min heap.
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1337. The K Weakest Rows in a Matrix/1337. The K Weakest Rows in a Matrix.py
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""" | ||
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. | ||
""" | ||
class Solution: | ||
def kWeakestRows(self, mat: List[List[int]], k: int) -> List[int]: | ||
min_heap,result = [],[] | ||
heapq.heapify(min_heap) | ||
for i,row in enumerate(mat): | ||
heapq.heappush(min_heap,(sum(row),i,row)) | ||
if k <= len(min_heap): | ||
while k: | ||
temp = heapq.heappop(min_heap) | ||
result.append(temp[1]) | ||
k -= 1 | ||
else: | ||
while min_heap: | ||
temp = heapq.heappop(min_heap) | ||
result.append(temp[1]) | ||
return result |