LeetCode #3430 — HARD

Maximum and Minimum Sums of at Most Size K Subarrays

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

Solve on LeetCode
The Problem

Problem Statement

You are given an integer array nums and a positive integer k. Return the sum of the maximum and minimum elements of all subarrays with at most k elements.

Example 1:

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

Output: 20

Explanation:

The subarrays of nums with at most 2 elements are:

Subarray Minimum Maximum Sum
[1] 1 1 2
[2] 2 2 4
[3] 3 3 6
[1, 2] 1 2 3
[2, 3] 2 3 5
Final Total     20

The output would be 20.

Example 2:

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

Output: -6

Explanation:

The subarrays of nums with at most 2 elements are:

Subarray Minimum Maximum Sum
[1] 1 1 2
[-3] -3 -3 -6
[1] 1 1 2
[1, -3] -3 1 -2
[-3, 1] -3 1 -2
Final Total     -6

The output would be -6.

Constraints:

  • 1 <= nums.length <= 80000
  • 1 <= k <= nums.length
  • -106 <= nums[i] <= 106
Patterns Used
📚Monotonic Stack→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums and a positive integer k. Return the sum of the maximum and minimum elements of all subarrays with at most k elements.

Baseline thinking

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

Pattern signal: Array · Math · Stack

Example 1

[1,2,3]
2

Example 2

[1,-3,1]
2

Related Problems

  • Next Greater Element II (next-greater-element-ii)
Step 02

Core Insight

What unlocks the optimal approach

  • Use a monotonic stack.
  • How can we calculate the number of subarrays where an element is the largest?
  • Enforce the condition on size too.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #3430: Maximum and Minimum Sums of at Most Size K Subarrays
class Solution {
  // Similar to 2104. Sum of Subarray Ranges
  public long minMaxSubarraySum(int[] nums, int k) {
    Pair<int[], int[]> gt = getPrevNext(nums, (a, b) -> a < b);
    Pair<int[], int[]> lt = getPrevNext(nums, (a, b) -> a > b);
    int[] prevGt = gt.getKey();
    int[] nextGt = gt.getValue();
    int[] prevLt = lt.getKey();
    int[] nextLt = lt.getValue();
    return subarraySum(nums, k, prevGt, nextGt) + subarraySum(nums, k, prevLt, nextLt);
  }

  // Returns the sum of all subarrays with a size <= k, The `prev` and `next`
  // arrays are used to store the indices of the nearest numbers that are
  // smaller or larger than the current number, respectively.
  private long subarraySum(int[] nums, int k, int[] prev, int[] next) {
    long res = 0;
    for (int i = 0; i < nums.length; ++i) {
      final int l = Math.min(i - prev[i], k);
      final int r = Math.min(next[i] - i, k);
      final int extra = Math.max(0, l + r - 1 - k);
      res += (long) nums[i] * (l * r - extra * (extra + 1) / 2);
    }
    return res;
  }

  // Returns `prev` and `next`, that store the indices of the nearest numbers
  // that are smaller or larger than the current number depending on `op`.
  private Pair<int[], int[]> getPrevNext(int[] nums, BiFunction<Integer, Integer, Boolean> op) {
    final int n = nums.length;
    int[] prev = new int[n];
    int[] next = new int[n];
    Arrays.fill(prev, -1);
    Arrays.fill(next, n);
    Deque<Integer> stack = new ArrayDeque<>();
    for (int i = 0; i < n; ++i) {
      while (!stack.isEmpty() && op.apply(nums[stack.peek()], nums[i]))
        next[stack.pop()] = i;
      if (!stack.isEmpty())
        prev[i] = stack.peek();
      stack.push(i);
    }
    return new Pair<>(prev, next);
  }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(n)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK
O(n) time
O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).
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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.

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