Back to course home
0% completed
Vote For New Content
Simplified Solution for #3
Pete Stenger
Oct 28, 2024
from heapq import heappop, heappush class Solution: def countSmallerEq(self, matrix, val): n, m = len(matrix), len(matrix[0]) # nxm row, col = 0, m - 1 smaller = 0 while row < n and col >= 0: if matrix[row][col] <= val: smaller += (col + 1) row += 1 elif matrix[row][col] > val: col -= 1 return smaller def findKthSmallest(self, matrix, k): n, m = len(matrix), len(matrix[0]) lo = matrix[0][0] hi = matrix[n-1][m-1] while lo <= hi: mid = (lo + hi) // 2 n_smaller = self.countSmallerEq(matrix, mid) if n_smaller >= k: hi = mid - 1 else: lo = mid + 1 return lo
0
0
Comments
Comments
On this page