327-Count-Of-Range-Sum
Sat 17 May 2025
https://leetcode.com/problems/count-of-range-sum
import pyutil as pyu
pyu.get_local_pyinfo()
print(pyu.ps2("python-dotenv"))
from typing import List
class Solution:
def countRangeSum(self, nums: List[int], lower: int, upper: int) -> int:
n = len(nums)
self.ans = 0
prefix = [0] + list(itertools.accumulate(nums))
self._mergeSort(prefix, 0, n, lower, upper)
return self.ans
def _mergeSort(self, prefix: List[int], l: int, r: int, lower: int, upper: int) -> None:
if l >= r:
return
m = (l + r) // 2
self._mergeSort(prefix, l, m, lower, upper)
self._mergeSort(prefix, m + 1, r, lower, upper)
self._merge(prefix, l, m, r, lower, upper)
def _merge(self, prefix: List[int], l: int, m: int, r: int, lower: int, upper: int) -> None:
lo = m + 1 # 1st index s.t. prefix[lo] - prefix[i] >= lower
hi = m + 1 # 1st index s.t. prefix[hi] - prefix[i] > upper
# For each index i in range [l, m], add hi - lo to ans
for i in range(l, m + 1):
while lo <= r and prefix[lo] - prefix[i] < lower:
lo += 1
while hi <= r and prefix[hi] - prefix[i] <= upper:
hi += 1
self.ans += hi - lo
sorted = [0] * (r - l + 1)
k = 0 # sorted's index
i = l # left's index
j = m + 1 # right's index
while i <= m and j <= r:
if prefix[i] < prefix[j]:
sorted[k] = prefix[i]
k += 1
i += 1
else:
sorted[k] = prefix[j]
k += 1
j += 1
# Put possible remaining left part to the sorted array
while i <= m:
sorted[k] = prefix[i]
k += 1
i += 1
# Put possible remaining right part to the sorted array
while j <= r:
sorted[k] = prefix[j]
k += 1
j += 1
prefix[l:l + len(sorted)] = sorted
new Solution().countRangeSum()
Score: 5
Category: leetcode