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O(N) solution, generalization of rearrange string.
Pete Stenger
Oct 27, 2024
from heapq import * from collections import deque, Counter class Solution: def reorganizeString(self, str, k): counts = Counter(str) heap = [ (-value, char) for char, value in counts.items() ] heapify(heap) if -heap[0][0] > (len(str) + k - 1)//k: return '' res = list('_' * len(str)) i = 0 offset = 0 while heap: count, char = heappop(heap) while count < 0: res[i] = char i += k if i >= len(res): offset += 1 i = offset count += 1 return ''.join(res)
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