LeetCode #2602 — MEDIUM

Minimum Operations to Make All Array Elements Equal

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

The Problem

Problem Statement

You are given an array nums consisting of positive integers.

You are also given an integer array queries of size m. For the i^th query, you want to make all of the elements of nums equal to queries[i]. You can perform the following operation on the array any number of times:

Increase or decrease an element of the array by 1.

Return an array answer of size m where answer[i] is the minimum number of operations to make all elements of nums equal to queries[i].

Note that after each query the array is reset to its original state.

Example 1:

Input: nums = [3,1,6,8], queries = [1,5]
Output: [14,10]
Explanation: For the first query we can do the following operations:
- Decrease nums[0] 2 times, so that nums = [1,1,6,8].
- Decrease nums[2] 5 times, so that nums = [1,1,1,8].
- Decrease nums[3] 7 times, so that nums = [1,1,1,1].
So the total number of operations for the first query is 2 + 5 + 7 = 14.
For the second query we can do the following operations:
- Increase nums[0] 2 times, so that nums = [5,1,6,8].
- Increase nums[1] 4 times, so that nums = [5,5,6,8].
- Decrease nums[2] 1 time, so that nums = [5,5,5,8].
- Decrease nums[3] 3 times, so that nums = [5,5,5,5].
So the total number of operations for the second query is 2 + 4 + 1 + 3 = 10.

Example 2:

Input: nums = [2,9,6,3], queries = [10]
Output: [20]
Explanation: We can increase each value in the array to 10. The total number of operations will be 8 + 1 + 4 + 7 = 20.

Constraints:

n == nums.length
m == queries.length
1 <= n, m <= 10⁵
1 <= nums[i], queries[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an array nums consisting of positive integers. You are also given an integer array queries of size m. For the ith query, you want to make all of the elements of nums equal to queries[i]. You can perform the following operation on the array any number of times: Increase or decrease an element of the array by 1. Return an array answer of size m where answer[i] is the minimum number of operations to make all elements of nums equal to queries[i]. Note that after each query the array is reset to its original state.

Baseline thinking

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

Pattern signal: Array · Binary Search

Example 1

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

Example 2

[2,9,6,3]
[10]

Core Insight

What unlocks the optimal approach

For each query, you should decrease all elements greater than queries[i] and increase all elements less than queries[i].
The answer is the sum of absolute differences between queries[i] and every element of the array. How do you calculate that optimally?

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 #2602: Minimum Operations to Make All Array Elements Equal
class Solution {
    public List<Long> minOperations(int[] nums, int[] queries) {
        Arrays.sort(nums);
        int n = nums.length;
        long[] s = new long[n + 1];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
        List<Long> ans = new ArrayList<>();
        for (int x : queries) {
            int i = search(nums, x + 1);
            long t = s[n] - s[i] - 1L * (n - i) * x;
            i = search(nums, x);
            t += 1L * x * i - s[i];
            ans.add(t);
        }
        return ans;
    }

    private int search(int[] nums, int x) {
        int l = 0, r = nums.length;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

// Accepted solution for LeetCode #2602: Minimum Operations to Make All Array Elements Equal
func minOperations(nums []int, queries []int) (ans []int64) {
	sort.Ints(nums)
	n := len(nums)
	s := make([]int, n+1)
	for i, x := range nums {
		s[i+1] = s[i] + x
	}
	for _, x := range queries {
		i := sort.SearchInts(nums, x+1)
		t := s[n] - s[i] - (n-i)*x
		i = sort.SearchInts(nums, x)
		t += x*i - s[i]
		ans = append(ans, int64(t))
	}
	return
}

# Accepted solution for LeetCode #2602: Minimum Operations to Make All Array Elements Equal
class Solution:
    def minOperations(self, nums: List[int], queries: List[int]) -> List[int]:
        nums.sort()
        s = list(accumulate(nums, initial=0))
        ans = []
        for x in queries:
            i = bisect_left(nums, x + 1)
            t = s[-1] - s[i] - (len(nums) - i) * x
            i = bisect_left(nums, x)
            t += x * i - s[i]
            ans.append(t)
        return ans

// Accepted solution for LeetCode #2602: Minimum Operations to Make All Array Elements Equal
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2602: Minimum Operations to Make All Array Elements Equal
// class Solution {
//     public List<Long> minOperations(int[] nums, int[] queries) {
//         Arrays.sort(nums);
//         int n = nums.length;
//         long[] s = new long[n + 1];
//         for (int i = 0; i < n; ++i) {
//             s[i + 1] = s[i] + nums[i];
//         }
//         List<Long> ans = new ArrayList<>();
//         for (int x : queries) {
//             int i = search(nums, x + 1);
//             long t = s[n] - s[i] - 1L * (n - i) * x;
//             i = search(nums, x);
//             t += 1L * x * i - s[i];
//             ans.add(t);
//         }
//         return ans;
//     }
// 
//     private int search(int[] nums, int x) {
//         int l = 0, r = nums.length;
//         while (l < r) {
//             int mid = (l + r) >> 1;
//             if (nums[mid] >= x) {
//                 r = mid;
//             } else {
//                 l = mid + 1;
//             }
//         }
//         return l;
//     }
// }

// Accepted solution for LeetCode #2602: Minimum Operations to Make All Array Elements Equal
function minOperations(nums: number[], queries: number[]): number[] {
    nums.sort((a, b) => a - b);
    const n = nums.length;
    const s: number[] = new Array(n + 1).fill(0);
    for (let i = 0; i < n; ++i) {
        s[i + 1] = s[i] + nums[i];
    }
    const search = (x: number): number => {
        let l = 0;
        let r = n;
        while (l < r) {
            const mid = (l + r) >> 1;
            if (nums[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    const ans: number[] = [];
    for (const x of queries) {
        const i = search(x + 1);
        let t = s[n] - s[i] - (n - i) * x;
        const j = search(x);
        t += x * j - s[j];
        ans.push(t);
    }
    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(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.