Description
You are given the array nums consisting of n positive integers. You computed the sum of all non-empty continuous subarrays from the array and then sorted them in non-decreasing order, creating a new array of n * (n + 1) / 2 numbers.
Return the sum of the numbers from index left to index right (indexed from 1), inclusive, in the new array. Since the answer can be a huge number return it modulo 109 + 7.
Example
Input: nums = [1,2,3,4], n = 4, left = 1, right = 5
Output: 13
Explanation: All subarray sums are 1, 3, 6, 10, 2, 5, 9, 3, 7, 4. After sorting them in non-decreasing order we have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 1 to ri = 5 is 1 + 2 + 3 + 3 + 4 = 13.
分析
1. 采用二分
以 n=4, nums = [1,2,3,4] 为例， prefixSum = [1, 3, 6, 10]
step1
猜一个数 M
for i in range(4):
二分查找出 prefixSum 小于等于 M + nums[i] - nums[0] 个数
step2
如小于等于 M 的个数大于 rightnum 或 leftnum， 则缩小 M 的值 。 否则增大 M 值 （常规的 二分更新法则）
如小小于等于 M 的个数等于 rightnum 或者。leftnum。则进入到第三步
step3
total = 0
for i in range(4):
curTotal = 0
for j in range(i, 4):
curTotal += nums[j]
if curTotal > M:
break
total += curTotal
2. 采用最小堆， 代码如下
import heapq
class Solution(object):
def rangeSum(self, nums, n, left, right):
"""
:type nums: List[int]
:type n: int
:type left: int
:type right: int
:rtype: int
"""
q, total_sum = [], 0
for i in range(n):
heapq.heappush(q, [nums[i], i])
for i in range(1, right+1):
v, idx = heapq.heappop(q)
if i >= left:
total_sum += v
if n-1 > idx:
idx += 1
heapq.heappush(q, [v+nums[idx], idx])
return total_sum % 1000000007
总结
Runtime: 560 ms, faster than 48.15% of Python online submissions for Range Sum of Sorted Subarray Sums.
Memory Usage: 29.2 MB, less than 88.89% of Python online submissions for Range Sum of Sorted Subarray Sums.
