O(nlogk) time and O(n) space using min heap.

Author: 149ps

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