Problem Description

You are given an integer array nums of size n. For each index i where 0 <= i < n, define a subarray nums[start ... i] where start = max(0, i - nums[i]). Your task is to compute the total sum of all elements from this subarray defined for each index in the array.

Intuition

To solve the problem, observe that for each index i, you can derive a subarray starting from max(0, i - nums[i]) to i. To efficiently sum all elements in these subarrays, you can use prefix sums. A prefix sum array provides cumulative totals, allowing us to rapidly calculate the sum of any subarray by taking advantage of previously computed results.

In the implementation, you first calculate the prefix sum array s, where each element at index j represents the sum of the nums array elements from the start up to j-1. By storing a zero as an initial element, s[i + 1] - s[start] will give the sum of elements from start to i, where start is max(0, i - nums[i]). This allows us to compute the sum across all defined subarrays in a single pass through the nums array.

Learn more about Prefix Sum patterns.

Solution Approach

The solution employs a prefix sum array to efficiently calculate the total sum of elements in the specified subarrays for each index. Here's how it works:

1. Prefix Sum Calculation:

   • Use Python's accumulate from the itertools module to compute the prefix sums of the nums array.
   • Initialize the accumulate with 0 to handle the sum from the beginning of the array conveniently.
   • The resulting array s will have s[i] storing the sum of elements from nums[0] to nums[i-1].

2. Compute Subarray Sums:

   • Use a generator expression within the sum function to iterate over each index i of nums.
   • For each index i, compute the sum of subarray elements using the formula s[i + 1] - s[max(0, i - nums[i])].
   • This formula calculates the sum from the start of the subarray, determined by max(0, i - nums[i]), to the current index i.

3. Efficiency:

   • The prefix sum allows for efficient computation, reducing the problem to linear time complexity, O(n), while using O(n) additional space for storing prefix sums.

Here's the implementation:

class Solution:
    def subarraySum(self, nums: List[int]) -> int:
        s = list(accumulate(nums, initial=0))
        return sum(s[i + 1] - s[max(0, i - x)] for i, x in enumerate(nums))

Example Walkthrough

Let's walk through the problem using a small example to illustrate the solution approach.

Example Input:
nums = [3, 2, 1]

Steps:

1. Compute the Prefix Sums:

   We first compute the prefix sum array s. Initialize s by using accumulate with an initial value of 0:

   • s[0] = 0 (initial value)
   • s[1] = nums[0] = 3
   • s[2] = nums[0] + nums[1] = 3 + 2 = 5
   • s[3] = nums[0] + nums[1] + nums[2] = 3 + 2 + 1 = 6

   So, the prefix sum array s becomes [0, 3, 5, 6].

2. Calculate the Subarray Sums:

   Now, calculate the sum of subarrays defined by each index i:

   • For i = 0:
     • start = max(0, 0 - nums[0]) = 0
     • Sum of subarray nums[0...0] = s[1] - s[0] = 3 - 0 = 3
   • For i = 1:
     • start = max(0, 1 - nums[1]) = 0
     • Sum of subarray nums[0...1] = s[2] - s[0] = 5 - 0 = 5
   • For i = 2:
     • start = max(0, 2 - nums[2]) = 1
     • Sum of subarray nums[1...2] = s[3] - s[1] = 6 - 3 = 3

3. Total Sum of All Subarrays:

   Add up the results from step 2:

   • Total sum = 3 (from i = 0) + 5 (from i = 1) + 3 (from i = 2) = 11

Thus, by summarizing the subarrays for each index i, we achieve an efficient solution with a total sum of 11 for the array nums = [3, 2, 1].

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n^2), where n is the length of the list nums. This comes from the nested operations where for each index i, the expression s[i + 1] - s[max(0, i - x)] is computed, resulting in O(n) computations per element of nums.

The space complexity of the code is O(n), as the code constructs a list s using accumulate that stores the prefix sums, which takes space proportional to the length of nums.

Learn more about how to find time and space complexity quickly.