LeetCode #1674 — MEDIUM

Minimum Moves to Make Array Complementary

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

The Problem

Problem Statement

You are given an integer array nums of even length n and an integer limit. In one move, you can replace any integer from nums with another integer between 1 and limit, inclusive.

The array nums is complementary if for all indices i (0-indexed), nums[i] + nums[n - 1 - i] equals the same number. For example, the array [1,2,3,4] is complementary because for all indices i, nums[i] + nums[n - 1 - i] = 5.

Return the minimum number of moves required to make nums complementary.

Example 1:

Input: nums = [1,2,4,3], limit = 4
Output: 1
Explanation: In 1 move, you can change nums to [1,2,2,3] (underlined elements are changed).
nums[0] + nums[3] = 1 + 3 = 4.
nums[1] + nums[2] = 2 + 2 = 4.
nums[2] + nums[1] = 2 + 2 = 4.
nums[3] + nums[0] = 3 + 1 = 4.
Therefore, nums[i] + nums[n-1-i] = 4 for every i, so nums is complementary.

Example 2:

Input: nums = [1,2,2,1], limit = 2
Output: 2
Explanation: In 2 moves, you can change nums to [2,2,2,2]. You cannot change any number to 3 since 3 > limit.

Example 3:

Input: nums = [1,2,1,2], limit = 2
Output: 0
Explanation: nums is already complementary.

Constraints:

n == nums.length
2 <= n <= 10⁵
1 <= nums[i] <= limit <= 10⁵
n is even.

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums of even length n and an integer limit. In one move, you can replace any integer from nums with another integer between 1 and limit, inclusive. The array nums is complementary if for all indices i (0-indexed), nums[i] + nums[n - 1 - i] equals the same number. For example, the array [1,2,3,4] is complementary because for all indices i, nums[i] + nums[n - 1 - i] = 5. Return the minimum number of moves required to make nums complementary.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

[1,2,4,3]
4

Example 2

[1,2,2,1]
2

Example 3

[1,2,1,2]
2

Core Insight

What unlocks the optimal approach

Given a target sum x, each pair of nums[i] and nums[n-1-i] would either need 0, 1, or 2 modifications.
Can you find the optimal target sum x value such that the sum of modifications is minimized?
Create a difference array to efficiently sum all the modifications.

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 #1674: Minimum Moves to Make Array Complementary
class Solution {
    public int minMoves(int[] nums, int limit) {
        int[] d = new int[2 * limit + 2];
        int n = nums.length;
        for (int i = 0; i < n / 2; ++i) {
            int x = Math.min(nums[i], nums[n - i - 1]);
            int y = Math.max(nums[i], nums[n - i - 1]);
            d[2] += 2;
            d[x + 1] -= 2;
            d[x + 1] += 1;
            d[x + y] -= 1;
            d[x + y + 1] += 1;
            d[y + limit + 1] -= 1;
            d[y + limit + 1] += 2;
        }
        int ans = n;
        for (int i = 2, s = 0; i < d.length; ++i) {
            s += d[i];
            ans = Math.min(ans, s);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1674: Minimum Moves to Make Array Complementary
func minMoves(nums []int, limit int) int {
	n := len(nums)
	d := make([]int, 2*limit+2)
	for i := 0; i < n/2; i++ {
		x, y := nums[i], nums[n-1-i]
		if x > y {
			x, y = y, x
		}
		d[2] += 2
		d[x+1] -= 2
		d[x+1] += 1
		d[x+y] -= 1
		d[x+y+1] += 1
		d[y+limit+1] -= 1
		d[y+limit+1] += 2
	}
	ans, s := n, 0
	for _, x := range d[2:] {
		s += x
		ans = min(ans, s)
	}
	return ans
}

# Accepted solution for LeetCode #1674: Minimum Moves to Make Array Complementary
class Solution:
    def minMoves(self, nums: List[int], limit: int) -> int:
        d = [0] * (2 * limit + 2)
        n = len(nums)
        for i in range(n // 2):
            x, y = nums[i], nums[-i - 1]
            if x > y:
                x, y = y, x
            d[2] += 2
            d[x + 1] -= 2
            d[x + 1] += 1
            d[x + y] -= 1
            d[x + y + 1] += 1
            d[y + limit + 1] -= 1
            d[y + limit + 1] += 2
        return min(accumulate(d[2:]))

// Accepted solution for LeetCode #1674: Minimum Moves to Make Array Complementary
struct Solution;

impl Solution {
    fn min_moves(nums: Vec<i32>, limit: i32) -> i32 {
        let n = nums.len();
        let limit = limit as usize;
        let mut delta = vec![0; limit * 2 + 2];
        for i in 0..n / 2 {
            let a = nums[i] as usize;
            let b = nums[n - 1 - i] as usize;
            delta[2] += 2;
            delta[a.min(b) + 1] -= 1;
            delta[a + b] -= 1;
            delta[a + b + 1] += 1;
            delta[a.max(b) + limit + 1] += 1;
        }

        let mut res = std::i32::MAX;
        let mut cur = 0;
        for sum in 2..=limit * 2 {
            cur += delta[sum];
            res = res.min(cur);
        }
        res
    }
}

#[test]
fn test() {
    let nums = vec![1, 2, 4, 3];
    let limit = 4;
    let res = 1;
    assert_eq!(Solution::min_moves(nums, limit), res);
    let nums = vec![1, 2, 2, 1];
    let limit = 2;
    let res = 2;
    assert_eq!(Solution::min_moves(nums, limit), res);
    let nums = vec![1, 2, 1, 2];
    let limit = 2;
    let res = 0;
    assert_eq!(Solution::min_moves(nums, limit), res);
}

// Accepted solution for LeetCode #1674: Minimum Moves to Make Array Complementary
function minMoves(nums: number[], limit: number): number {
    const n = nums.length;
    const d: number[] = Array(limit * 2 + 2).fill(0);
    for (let i = 0; i < n >> 1; ++i) {
        const x = Math.min(nums[i], nums[n - 1 - i]);
        const y = Math.max(nums[i], nums[n - 1 - i]);
        d[2] += 2;
        d[x + 1] -= 2;
        d[x + 1] += 1;
        d[x + y] -= 1;
        d[x + y + 1] += 1;
        d[y + limit + 1] -= 1;
        d[y + limit + 1] += 2;
    }
    let ans = n;
    let s = 0;
    for (let i = 2; i < d.length; ++i) {
        s += d[i];
        ans = Math.min(ans, s);
    }
    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(n)

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.

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.