LeetCode #2541 — MEDIUM

Minimum Operations to Make Array Equal II

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

The Problem

Problem Statement

You are given two integer arrays nums1 and nums2 of equal length n and an integer k. You can perform the following operation on nums1:

Choose two indexes i and j and increment nums1[i] by k and decrement nums1[j] by k. In other words, nums1[i] = nums1[i] + k and nums1[j] = nums1[j] - k.

nums1 is said to be equal to nums2 if for all indices i such that 0 <= i < n, nums1[i] == nums2[i].

Return the minimum number of operations required to make nums1 equal to nums2. If it is impossible to make them equal, return -1.

Example 1:

Input: nums1 = [4,3,1,4], nums2 = [1,3,7,1], k = 3
Output: 2
Explanation: In 2 operations, we can transform nums1 to nums2.
1^st operation: i = 2, j = 0. After applying the operation, nums1 = [1,3,4,4].
2^nd operation: i = 2, j = 3. After applying the operation, nums1 = [1,3,7,1].
One can prove that it is impossible to make arrays equal in fewer operations.

Example 2:

Input: nums1 = [3,8,5,2], nums2 = [2,4,1,6], k = 1
Output: -1
Explanation: It can be proved that it is impossible to make the two arrays equal.

Constraints:

n == nums1.length == nums2.length
2 <= n <= 10⁵
0 <= nums1[i], nums2[j] <= 10⁹
0 <= k <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given two integer arrays nums1 and nums2 of equal length n and an integer k. You can perform the following operation on nums1: Choose two indexes i and j and increment nums1[i] by k and decrement nums1[j] by k. In other words, nums1[i] = nums1[i] + k and nums1[j] = nums1[j] - k. nums1 is said to be equal to nums2 if for all indices i such that 0 <= i < n, nums1[i] == nums2[i]. Return the minimum number of operations required to make nums1 equal to nums2. If it is impossible to make them equal, return -1.

Baseline thinking

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

Pattern signal: Array · Math · Greedy

Example 1

[4,3,1,4]
[1,3,7,1]
3

Example 2

[3,8,5,2]
[2,4,1,6]
1

Core Insight

What unlocks the optimal approach

What are the cases for which we cannot make nums1 == nums2?
For minimum moves, if nums1[i] < nums2[i], then we should never decrement nums1[i]. If nums1[i] > nums2[i], then we should never increment nums1[i].

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 #2541: Minimum Operations to Make Array Equal II
class Solution {
    public long minOperations(int[] nums1, int[] nums2, int k) {
        long ans = 0, x = 0;
        for (int i = 0; i < nums1.length; ++i) {
            int a = nums1[i], b = nums2[i];
            if (k == 0) {
                if (a != b) {
                    return -1;
                }
                continue;
            }
            if ((a - b) % k != 0) {
                return -1;
            }
            int y = (a - b) / k;
            ans += Math.abs(y);
            x += y;
        }
        return x == 0 ? ans / 2 : -1;
    }
}

// Accepted solution for LeetCode #2541: Minimum Operations to Make Array Equal II
func minOperations(nums1 []int, nums2 []int, k int) int64 {
	ans, x := 0, 0
	for i, a := range nums1 {
		b := nums2[i]
		if k == 0 {
			if a != b {
				return -1
			}
			continue
		}
		if (a-b)%k != 0 {
			return -1
		}
		y := (a - b) / k
		ans += abs(y)
		x += y
	}
	if x != 0 {
		return -1
	}
	return int64(ans / 2)
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #2541: Minimum Operations to Make Array Equal II
class Solution:
    def minOperations(self, nums1: List[int], nums2: List[int], k: int) -> int:
        ans = x = 0
        for a, b in zip(nums1, nums2):
            if k == 0:
                if a != b:
                    return -1
                continue
            if (a - b) % k:
                return -1
            y = (a - b) // k
            ans += abs(y)
            x += y
        return -1 if x else ans // 2

// Accepted solution for LeetCode #2541: Minimum Operations to Make Array Equal II
impl Solution {
    pub fn min_operations(nums1: Vec<i32>, nums2: Vec<i32>, k: i32) -> i64 {
        let k = k as i64;
        let n = nums1.len();
        if k == 0 {
            return if nums1.iter().enumerate().all(|(i, &v)| v == nums2[i]) {
                0
            } else {
                -1
            };
        }
        let mut sum1 = 0;
        let mut sum2 = 0;
        for i in 0..n {
            let diff = (nums1[i] - nums2[i]) as i64;
            sum1 += diff;
            if diff % k != 0 {
                return -1;
            }
            sum2 += diff.abs();
        }
        if sum1 != 0 {
            return -1;
        }
        sum2 / (k * 2)
    }
}

// Accepted solution for LeetCode #2541: Minimum Operations to Make Array Equal II
function minOperations(nums1: number[], nums2: number[], k: number): number {
    const n = nums1.length;
    if (k === 0) {
        return nums1.every((v, i) => v === nums2[i]) ? 0 : -1;
    }
    let sum1 = 0;
    let sum2 = 0;
    for (let i = 0; i < n; i++) {
        const diff = nums1[i] - nums2[i];
        sum1 += diff;
        if (diff % k !== 0) {
            return -1;
        }
        sum2 += Math.abs(diff);
    }
    if (sum1 !== 0) {
        return -1;
    }
    return sum2 / (k * 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)

Space

O(1)

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.

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.

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.