LeetCode #2448 — HARD

Minimum Cost to Make Array Equal

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

Solve on LeetCode

The Problem

Problem Statement

You are given two 0-indexed arrays nums and cost consisting each of n positive integers.

You can do the following operation any number of times:

Increase or decrease any element of the array nums by 1.

The cost of doing one operation on the i^th element is cost[i].

Return the minimum total cost such that all the elements of the array nums become equal.

Example 1:

Input: nums = [1,3,5,2], cost = [2,3,1,14]
Output: 8
Explanation: We can make all the elements equal to 2 in the following way:
- Increase the 0^th element one time. The cost is 2.
- Decrease the 1^st element one time. The cost is 3.
- Decrease the 2^nd element three times. The cost is 1 + 1 + 1 = 3.
The total cost is 2 + 3 + 3 = 8.
It can be shown that we cannot make the array equal with a smaller cost.

Example 2:

Input: nums = [2,2,2,2,2], cost = [4,2,8,1,3]
Output: 0
Explanation: All the elements are already equal, so no operations are needed.

Constraints:

n == nums.length == cost.length
1 <= n <= 10⁵
1 <= nums[i], cost[i] <= 10⁶
Test cases are generated in a way that the output doesn't exceed 2⁵³-1

Step 01

Brute Force Baseline

Problem summary: You are given two 0-indexed arrays nums and cost consisting each of n positive integers. You can do the following operation any number of times: Increase or decrease any element of the array nums by 1. The cost of doing one operation on the ith element is cost[i]. Return the minimum total cost such that all the elements of the array nums become equal.

Baseline thinking

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

Pattern signal: Array · Binary Search · Greedy

Example 1

[1,3,5,2]
[2,3,1,14]

Example 2

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

Core Insight

What unlocks the optimal approach

Changing the elements into one of the numbers already existing in the array nums is optimal.
Try finding the cost of changing the array into each element, and return the minimum value.

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

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 #2448: Minimum Cost to Make Array Equal
class Solution {
    public long minCost(int[] nums, int[] cost) {
        int n = nums.length;
        int[][] arr = new int[n][2];
        for (int i = 0; i < n; ++i) {
            arr[i] = new int[] {nums[i], cost[i]};
        }
        Arrays.sort(arr, (a, b) -> a[0] - b[0]);
        long[] f = new long[n + 1];
        long[] g = new long[n + 1];
        for (int i = 1; i <= n; ++i) {
            long a = arr[i - 1][0], b = arr[i - 1][1];
            f[i] = f[i - 1] + a * b;
            g[i] = g[i - 1] + b;
        }
        long ans = Long.MAX_VALUE;
        for (int i = 1; i <= n; ++i) {
            long a = arr[i - 1][0];
            long l = a * g[i - 1] - f[i - 1];
            long r = f[n] - f[i] - a * (g[n] - g[i]);
            ans = Math.min(ans, l + r);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2448: Minimum Cost to Make Array Equal
func minCost(nums []int, cost []int) int64 {
	n := len(nums)
	type pair struct{ a, b int }
	arr := make([]pair, n)
	for i, a := range nums {
		b := cost[i]
		arr[i] = pair{a, b}
	}
	sort.Slice(arr, func(i, j int) bool { return arr[i].a < arr[j].a })
	f := make([]int, n+1)
	g := make([]int, n+1)
	for i := 1; i <= n; i++ {
		a, b := arr[i-1].a, arr[i-1].b
		f[i] = f[i-1] + a*b
		g[i] = g[i-1] + b
	}
	var ans int64 = 1e18
	for i := 1; i <= n; i++ {
		a := arr[i-1].a
		l := a*g[i-1] - f[i-1]
		r := f[n] - f[i] - a*(g[n]-g[i])
		ans = min(ans, int64(l+r))
	}
	return ans
}

# Accepted solution for LeetCode #2448: Minimum Cost to Make Array Equal
class Solution:
    def minCost(self, nums: List[int], cost: List[int]) -> int:
        arr = sorted(zip(nums, cost))
        n = len(arr)
        f = [0] * (n + 1)
        g = [0] * (n + 1)
        for i in range(1, n + 1):
            a, b = arr[i - 1]
            f[i] = f[i - 1] + a * b
            g[i] = g[i - 1] + b
        ans = inf
        for i in range(1, n + 1):
            a = arr[i - 1][0]
            l = a * g[i - 1] - f[i - 1]
            r = f[n] - f[i] - a * (g[n] - g[i])
            ans = min(ans, l + r)
        return ans

// Accepted solution for LeetCode #2448: Minimum Cost to Make Array Equal
impl Solution {
    #[allow(dead_code)]
    pub fn min_cost(nums: Vec<i32>, cost: Vec<i32>) -> i64 {
        let mut zip_vec: Vec<_> = nums.into_iter().zip(cost.into_iter()).collect();

        // Sort the zip vector based on nums
        zip_vec.sort_by(|lhs, rhs| lhs.0.cmp(&rhs.0));

        let (nums, cost): (Vec<i32>, Vec<i32>) = zip_vec.into_iter().unzip();

        let mut sum: i64 = 0;
        for &c in &cost {
            sum += c as i64;
        }
        let middle_cost = (sum + 1) / 2;
        let mut cur_sum: i64 = 0;
        let mut i = 0;
        let n = nums.len();

        while i < n {
            if (cost[i] as i64) + cur_sum >= middle_cost {
                break;
            }
            cur_sum += cost[i] as i64;
            i += 1;
        }

        Self::compute_manhattan_dis(&nums, &cost, nums[i])
    }

    #[allow(dead_code)]
    fn compute_manhattan_dis(v: &Vec<i32>, c: &Vec<i32>, e: i32) -> i64 {
        let mut ret = 0;
        let n = v.len();

        for i in 0..n {
            if v[i] == e {
                continue;
            }
            ret += ((v[i] - e).abs() as i64) * (c[i] as i64);
        }

        ret
    }
}

// Accepted solution for LeetCode #2448: Minimum Cost to Make Array Equal
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2448: Minimum Cost to Make Array Equal
// class Solution {
//     public long minCost(int[] nums, int[] cost) {
//         int n = nums.length;
//         int[][] arr = new int[n][2];
//         for (int i = 0; i < n; ++i) {
//             arr[i] = new int[] {nums[i], cost[i]};
//         }
//         Arrays.sort(arr, (a, b) -> a[0] - b[0]);
//         long[] f = new long[n + 1];
//         long[] g = new long[n + 1];
//         for (int i = 1; i <= n; ++i) {
//             long a = arr[i - 1][0], b = arr[i - 1][1];
//             f[i] = f[i - 1] + a * b;
//             g[i] = g[i - 1] + b;
//         }
//         long ans = Long.MAX_VALUE;
//         for (int i = 1; i <= n; ++i) {
//             long a = arr[i - 1][0];
//             long l = a * g[i - 1] - f[i - 1];
//             long r = f[n] - f[i] - a * (g[n] - g[i]);
//             ans = Math.min(ans, l + r);
//         }
//         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.

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.