LeetCode #862 — HARD

Shortest Subarray with Sum at Least K

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

Given an integer array nums and an integer k, return the length of the shortest non-empty subarray of nums with a sum of at least k. If there is no such subarray, return -1.

A subarray is a contiguous part of an array.

Example 1:

Input: nums = [1], k = 1
Output: 1

Example 2:

Input: nums = [1,2], k = 4
Output: -1

Example 3:

Input: nums = [2,-1,2], k = 3
Output: 3

Constraints:

1 <= nums.length <= 10⁵
-10⁵ <= nums[i] <= 10⁵
1 <= k <= 10⁹

Step 01

Brute Force Baseline

Problem summary: Given an integer array nums and an integer k, return the length of the shortest non-empty subarray of nums with a sum of at least k. If there is no such subarray, return -1. A subarray is a contiguous part of an array.

Baseline thinking

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

Pattern signal: Array · Binary Search · Sliding Window · Monotonic Queue

Example 1

[1]
1

Example 2

[1,2]
4

Example 3

[2,-1,2]
3

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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

Largest constraint values

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 #862: Shortest Subarray with Sum at Least K
class Solution {
    public int shortestSubarray(int[] nums, int k) {
        int n = nums.length;
        long[] s = new long[n + 1];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
        Deque<Integer> q = new ArrayDeque<>();
        int ans = n + 1;
        for (int i = 0; i <= n; ++i) {
            while (!q.isEmpty() && s[i] - s[q.peek()] >= k) {
                ans = Math.min(ans, i - q.poll());
            }
            while (!q.isEmpty() && s[q.peekLast()] >= s[i]) {
                q.pollLast();
            }
            q.offer(i);
        }
        return ans > n ? -1 : ans;
    }
}

// Accepted solution for LeetCode #862: Shortest Subarray with Sum at Least K
func shortestSubarray(nums []int, k int) int {
	n := len(nums)
	s := make([]int, n+1)
	for i, x := range nums {
		s[i+1] = s[i] + x
	}
	q := []int{}
	ans := n + 1
	for i, v := range s {
		for len(q) > 0 && v-s[q[0]] >= k {
			ans = min(ans, i-q[0])
			q = q[1:]
		}
		for len(q) > 0 && s[q[len(q)-1]] >= v {
			q = q[:len(q)-1]
		}
		q = append(q, i)
	}
	if ans > n {
		return -1
	}
	return ans
}

# Accepted solution for LeetCode #862: Shortest Subarray with Sum at Least K
class Solution:
    def shortestSubarray(self, nums: List[int], k: int) -> int:
        s = list(accumulate(nums, initial=0))
        q = deque()
        ans = inf
        for i, v in enumerate(s):
            while q and v - s[q[0]] >= k:
                ans = min(ans, i - q.popleft())
            while q and s[q[-1]] >= v:
                q.pop()
            q.append(i)
        return -1 if ans == inf else ans

// Accepted solution for LeetCode #862: Shortest Subarray with Sum at Least K
struct Solution;

use std::collections::VecDeque;

impl Solution {
    fn shortest_subarray(a: Vec<i32>, k: i32) -> i32 {
        let n = a.len();
        let mut queue: VecDeque<usize> = VecDeque::new();
        let mut prefix = vec![0; n + 1];
        queue.push_back(0);
        let mut res = std::usize::MAX;
        for i in 0..n {
            prefix[i + 1] = prefix[i] + a[i];
            while let Some(&j) = queue.front() {
                if prefix[i + 1] - prefix[j] >= k {
                    res = res.min(i + 1 - j);
                    queue.pop_front();
                } else {
                    break;
                }
            }
            while let Some(&j) = queue.back() {
                if prefix[i + 1] <= prefix[j] {
                    queue.pop_back();
                } else {
                    break;
                }
            }
            queue.push_back(i + 1);
        }
        if res == std::usize::MAX {
            -1
        } else {
            res as i32
        }
    }
}

#[test]
fn test() {
    let a = vec![1];
    let k = 1;
    let res = 1;
    assert_eq!(Solution::shortest_subarray(a, k), res);
    let a = vec![1, 2];
    let k = 4;
    let res = -1;
    assert_eq!(Solution::shortest_subarray(a, k), res);
    let a = vec![2, -1, 2];
    let k = 3;
    let res = 3;
    assert_eq!(Solution::shortest_subarray(a, k), res);
}

// Accepted solution for LeetCode #862: Shortest Subarray with Sum at Least K
function shortestSubarray(nums: number[], k: number): number {
    const [n, MAX] = [nums.length, Number.POSITIVE_INFINITY];
    const s = Array(n + 1).fill(0);
    const q: number[] = [];
    let ans = MAX;

    for (let i = 0; i < n; i++) {
        s[i + 1] = s[i] + nums[i];
    }

    for (let i = 0; i < n + 1; i++) {
        while (q.length && s[i] - s[q[0]] >= k) {
            ans = Math.min(ans, i - q.shift()!);
        }

        while (q.length && s[i] <= s[q.at(-1)!]) {
            q.pop();
        }

        q.push(i);
    }

    return ans === MAX ? -1 : 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(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.