LeetCode #3427 — EASY

Sum of Variable Length Subarrays

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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]).

Return the total sum of all elements from the subarray defined for each index in the array.

Example 1:

Input: nums = [2,3,1]

Output: 11

Explanation:

i	Subarray	Sum
0	`nums[0] = [2]`	2
1	`nums[0 ... 1] = [2, 3]`	5
2	`nums[1 ... 2] = [3, 1]`	4
Total Sum		11

The total sum is 11. Hence, 11 is the output.

Example 2:

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

Output: 13

Explanation:

i	Subarray	Sum
0	`nums[0] = [3]`	3
1	`nums[0 ... 1] = [3, 1]`	4
2	`nums[1 ... 2] = [1, 1]`	2
3	`nums[1 ... 3] = [1, 1, 2]`	4
Total Sum		13

The total sum is 13. Hence, 13 is the output.

Constraints:

1 <= n == nums.length <= 100
1 <= nums[i] <= 1000

Step 01

Brute Force Baseline

Problem summary: 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]). Return the total sum of all elements from the subarray defined for each index in the array.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[2,3,1]

Example 2

[3,1,1,2]

Core Insight

What unlocks the optimal approach

The constraints are small, so brute force for each index.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3427: Sum of Variable Length Subarrays
class Solution {
    public int subarraySum(int[] nums) {
        int n = nums.length;
        int[] s = new int[n + 1];
        for (int i = 1; i <= n; ++i) {
            s[i] = s[i - 1] + nums[i - 1];
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans += s[i + 1] - s[Math.max(0, i - nums[i])];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3427: Sum of Variable Length Subarrays
func subarraySum(nums []int) (ans int) {
	s := make([]int, len(nums)+1)
	for i, x := range nums {
		s[i+1] = s[i] + x
	}
	for i, x := range nums {
		ans += s[i+1] - s[max(0, i-x)]
	}
	return
}

# Accepted solution for LeetCode #3427: Sum of Variable Length Subarrays
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))

// Accepted solution for LeetCode #3427: Sum of Variable Length Subarrays
fn subarray_sum(nums: Vec<i32>) -> i32 {
    let mut ret = 0;
    for i in 0..nums.len() {
        let start = std::cmp::max(0, i as i32 - nums[i]) as usize;
        for j in start..=i {
            ret += nums[j];
        }
    }
    ret
}

fn main() {
    let nums = vec![2, 3, 1];
    let ret = subarray_sum(nums);
    println!("ret={ret}");
}

#[test]
fn test() {
    {
        let nums = vec![2, 3, 1];
        let ret = subarray_sum(nums);
        assert_eq!(ret, 11);
    }
    {
        let nums = vec![3, 1, 1, 2];
        let ret = subarray_sum(nums);
        assert_eq!(ret, 13);
    }
}

// Accepted solution for LeetCode #3427: Sum of Variable Length Subarrays
function subarraySum(nums: number[]): number {
    const n = nums.length;
    const s: number[] = Array(n + 1).fill(0);
    for (let i = 0; i < n; ++i) {
        s[i + 1] = s[i] + nums[i];
    }
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        ans += s[i + 1] - s[Math.max(0, i - nums[i])];
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.