LeetCode #2567 — MEDIUM

Minimum Score by Changing Two Elements

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given an integer array nums.

The low score of nums is the minimum absolute difference between any two integers.
The high score of nums is the maximum absolute difference between any two integers.
The score of nums is the sum of the high and low scores.

Return the minimum score after changing two elements of nums.

Example 1:

Input: nums = [1,4,7,8,5]

Output: 3

Explanation:

Change nums[0] and nums[1] to be 6 so that nums becomes [6,6,7,8,5].
The low score is the minimum absolute difference: |6 - 6| = 0.
The high score is the maximum absolute difference: |8 - 5| = 3.
The sum of high and low score is 3.

Example 2:

Input: nums = [1,4,3]

Output: 0

Explanation:

Change nums[1] and nums[2] to 1 so that nums becomes [1,1,1].
The sum of maximum absolute difference and minimum absolute difference is 0.

Constraints:

3 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums. The low score of nums is the minimum absolute difference between any two integers. The high score of nums is the maximum absolute difference between any two integers. The score of nums is the sum of the high and low scores. Return the minimum score after changing two elements of nums.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[1,4,7,8,5]

Example 2

[1,4,3]

Step 02

Core Insight

What unlocks the optimal approach

Changing the minimum or maximum values will only minimize the score.
Think about what all possible pairs of minimum and maximum values can be changed to form the minimum score.

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 #2567: Minimum Score by Changing Two Elements
class Solution {
    public int minimizeSum(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        int a = nums[n - 1] - nums[2];
        int b = nums[n - 2] - nums[1];
        int c = nums[n - 3] - nums[0];
        return Math.min(a, Math.min(b, c));
    }
}

// Accepted solution for LeetCode #2567: Minimum Score by Changing Two Elements
func minimizeSum(nums []int) int {
	sort.Ints(nums)
	n := len(nums)
	return min(nums[n-1]-nums[2], min(nums[n-2]-nums[1], nums[n-3]-nums[0]))
}

# Accepted solution for LeetCode #2567: Minimum Score by Changing Two Elements
class Solution:
    def minimizeSum(self, nums: List[int]) -> int:
        nums.sort()
        return min(nums[-1] - nums[2], nums[-2] - nums[1], nums[-3] - nums[0])

// Accepted solution for LeetCode #2567: Minimum Score by Changing Two Elements
impl Solution {
    pub fn minimize_sum(mut nums: Vec<i32>) -> i32 {
        nums.sort();
        let n = nums.len();
        (nums[n - 1] - nums[2])
            .min(nums[n - 2] - nums[1])
            .min(nums[n - 3] - nums[0])
    }
}

// Accepted solution for LeetCode #2567: Minimum Score by Changing Two Elements
function minimizeSum(nums: number[]): number {
    nums.sort((a, b) => a - b);
    const n = nums.length;
    return Math.min(nums[n - 3] - nums[0], nums[n - 2] - nums[1], nums[n - 1] - nums[2]);
}

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 log n)

Space

O(log n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.