LeetCode 327. Count of Range Sum

Description

https://leetcode.com/problems/count-of-range-sum/

Given an integer array nums and two integers lower and upper, return the number of range sums that lie in [lower, upper] inclusive.

Range sum S(i, j) is defined as the sum of the elements in nums between indices i and j inclusive, where i <= j.

Example 1:

Input: nums = [-2,5,-1], lower = -2, upper = 2
Output: 3
Explanation: The three ranges are: [0,0], [2,2], and [0,2] and their respective sums are: -2, -1, 2.

Example 2:

Input: nums = [0], lower = 0, upper = 0
Output: 1

Constraints:

• 1 <= nums.length <= 105
• -231 <= nums[i] <= 231 - 1
• -105 <= lower <= upper <= 105
• The answer is guaranteed to fit in a 32-bit integer.

Explanation

Create a map where the value is the count of prefix sum equals to key. If there is a number in [lower, upper] equals the sum of nums[i, j], then prefix[i – 1] – number must be in the map.

Python Solution

class Solution:
    def countRangeSum(self, nums: List[int], lower: int, upper: int) -> int:
        count_sum = {0:1}
        
        pre_sum = 0
        count = 0
        
        for num in nums:
            pre_sum += num
            
            for d in range(lower, upper + 1):
                if pre_sum - d in count_sum:
                    count += count_sum[pre_sum - d]
            
            count_sum[pre_sum] = count_sum.get(pre_sum, 0) + 1
            
        return count