LeetCode #3510 — HARD

Minimum Pair Removal to Sort Array II

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

Solve on LeetCode
The Problem

Problem Statement

Given an array nums, you can perform the following operation any number of times:

  • Select the adjacent pair with the minimum sum in nums. If multiple such pairs exist, choose the leftmost one.
  • Replace the pair with their sum.

Return the minimum number of operations needed to make the array non-decreasing.

An array is said to be non-decreasing if each element is greater than or equal to its previous element (if it exists).

Example 1:

Input: nums = [5,2,3,1]

Output: 2

Explanation:

  • The pair (3,1) has the minimum sum of 4. After replacement, nums = [5,2,4].
  • The pair (2,4) has the minimum sum of 6. After replacement, nums = [5,6].

The array nums became non-decreasing in two operations.

Example 2:

Input: nums = [1,2,2]

Output: 0

Explanation:

The array nums is already sorted.

Constraints:

  • 1 <= nums.length <= 105
  • -109 <= nums[i] <= 109
Patterns Used
🔄In-Place Reversal→📐Segment Tree→

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: Given an array nums, you can perform the following operation any number of times: Select the adjacent pair with the minimum sum in nums. If multiple such pairs exist, choose the leftmost one. Replace the pair with their sum. Return the minimum number of operations needed to make the array non-decreasing. An array is said to be non-decreasing if each element is greater than or equal to its previous element (if it exists).

Baseline thinking

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

Pattern signal: Array · Hash Map · Linked List · Segment Tree

Example 1

[5,2,3,1]

Example 2

[1,2,2]
Step 02

Core Insight

What unlocks the optimal approach

  • We can perform the simulation using data structures.
  • Maintain an array index and value using a map since we need to find the next and previous ones.
  • Maintain the indices to be removed using a hash set.
  • Maintain the neighbor sums with the smaller indices (set or priority queue).
  • Keep the 3 structures in sync during the removals.
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 #3510: Minimum Pair Removal to Sort Array II
class Solution {
    record Pair(long s, int i) implements Comparable<Pair> {
        @Override
        public int compareTo(Pair other) {
            int compareS = Long.compare(this.s, other.s);
            return compareS != 0 ? compareS : Integer.compare(this.i, other.i);
        }
    }

    public int minimumPairRemoval(int[] nums) {
        int n = nums.length;
        int inv = 0;
        TreeSet<Pair> sl = new TreeSet<>();
        for (int i = 0; i < n - 1; ++i) {
            if (nums[i] > nums[i + 1]) {
                ++inv;
            }
            sl.add(new Pair(nums[i] + nums[i + 1], i));
        }
        TreeSet<Integer> idx = new TreeSet<>();
        long[] arr = new long[n];
        for (int i = 0; i < n; ++i) {
            idx.add(i);
            arr[i] = nums[i];
        }

        int ans = 0;
        while (inv > 0) {
            ++ans;
            var p = sl.pollFirst();
            long s = p.s;
            int i = p.i;
            int j = idx.higher(i);
            if (arr[i] > arr[j]) {
                --inv;
            }
            Integer h = idx.lower(i);
            if (h != null) {
                if (arr[h] > arr[i]) {
                    --inv;
                }
                sl.remove(new Pair(arr[h] + arr[i], h));
                if (arr[h] > s) {
                    ++inv;
                }
                sl.add(new Pair(arr[h] + s, h));
            }
            Integer k = idx.higher(j);
            if (k != null) {
                if (arr[j] > arr[k]) {
                    --inv;
                }
                sl.remove(new Pair(arr[j] + arr[k], j));
                if (s > arr[k]) {
                    ++inv;
                }
                sl.add(new Pair(s + arr[k], i));
            }
            arr[i] = s;
            idx.remove(j);
        }
        return ans;
    }
}
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 log n)
Space
O(n)

Approach Breakdown

COPY TO ARRAY
O(n) time
O(n) space

Copy all n nodes into an array (O(n) time and space), then use array indexing for random access. Operations like reversal or middle-finding become trivial with indices, but the O(n) extra space defeats the purpose of using a linked list.

IN-PLACE POINTERS
O(n) time
O(1) space

Most linked list operations traverse the list once (O(n)) and re-wire pointers in-place (O(1) extra space). The brute force often copies nodes to an array to enable random access, costing O(n) space. In-place pointer manipulation eliminates that.

Shortcut: Traverse once + re-wire pointers → O(n) time, O(1) space. Dummy head nodes simplify edge cases.
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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Losing head/tail while rewiring

Wrong move: Pointer updates overwrite references before they are saved.

Usually fails on: List becomes disconnected mid-operation.

Fix: Store next pointers first and use a dummy head for safer joins.

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