LeetCode #3139 — HARD

Minimum Cost to Equalize Array

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

The Problem

Problem Statement

You are given an integer array nums and two integers cost1 and cost2. You are allowed to perform either of the following operations any number of times:

Choose an index i from nums and increase nums[i] by 1 for a cost of cost1.
Choose two different indices i, j, from nums and increase nums[i] and nums[j] by 1 for a cost of cost2.

Return the minimum cost required to make all elements in the array equal.

Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [4,1], cost1 = 5, cost2 = 2

Output: 15

Explanation:

The following operations can be performed to make the values equal:

Increase nums[1] by 1 for a cost of 5. nums becomes [4,2].
Increase nums[1] by 1 for a cost of 5. nums becomes [4,3].
Increase nums[1] by 1 for a cost of 5. nums becomes [4,4].

The total cost is 15.

Example 2:

Input: nums = [2,3,3,3,5], cost1 = 2, cost2 = 1

Output: 6

Explanation:

The following operations can be performed to make the values equal:

Increase nums[0] and nums[1] by 1 for a cost of 1. nums becomes [3,4,3,3,5].
Increase nums[0] and nums[2] by 1 for a cost of 1. nums becomes [4,4,4,3,5].
Increase nums[0] and nums[3] by 1 for a cost of 1. nums becomes [5,4,4,4,5].
Increase nums[1] and nums[2] by 1 for a cost of 1. nums becomes [5,5,5,4,5].
Increase nums[3] by 1 for a cost of 2. nums becomes [5,5,5,5,5].

The total cost is 6.

Example 3:

Input: nums = [3,5,3], cost1 = 1, cost2 = 3

Output: 4

Explanation:

The following operations can be performed to make the values equal:

Increase nums[0] by 1 for a cost of 1. nums becomes [4,5,3].
Increase nums[0] by 1 for a cost of 1. nums becomes [5,5,3].
Increase nums[2] by 1 for a cost of 1. nums becomes [5,5,4].
Increase nums[2] by 1 for a cost of 1. nums becomes [5,5,5].

The total cost is 4.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁶
1 <= cost1 <= 10⁶
1 <= cost2 <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums and two integers cost1 and cost2. You are allowed to perform either of the following operations any number of times: Choose an index i from nums and increase nums[i] by 1 for a cost of cost1. Choose two different indices i, j, from nums and increase nums[i] and nums[j] by 1 for a cost of cost2. Return the minimum cost required to make all elements in the array equal. Since the answer may be very large, return it modulo 109 + 7.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[4,1]
5
2

Example 2

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

Example 3

[3,5,3]
1
3

Step 02

Core Insight

What unlocks the optimal approach

How can you determine the minimum cost if you know the maximum value in the array once all values are made equal?
If <code>cost2 > cost1 * 2</code>, we should just use <code>cost1</code> to change all the values to the maximum one.
Otherwise, it's optimal to choose the smallest two values and use <code>cost2</code> to increase both of them.
Since the maximum value is known, calculate the required increases to equalize all values, instead of naively simulating the operations.
There are not a lot of candidates for the maximum; we can try all of them and choose which uses the minimum number of operations.

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 #3139: Minimum Cost to Equalize Array
class Solution {
  public int minCostToEqualizeArray(int[] nums, int cost1, int cost2) {
    final int MOD = 1_000_000_007;
    final int n = nums.length;
    final int minNum = Arrays.stream(nums).min().getAsInt();
    final int maxNum = Arrays.stream(nums).max().getAsInt();
    final long sum = Arrays.stream(nums).asLongStream().sum();
    long ans = Long.MAX_VALUE;

    if (cost1 * 2 <= cost2 || n < 3) {
      final long totalGap = 1L * maxNum * n - sum;
      return (int) ((cost1 * totalGap) % MOD);
    }

    for (int target = maxNum; target < 2 * maxNum; ++target) {
      final int maxGap = target - minNum;
      final long totalGap = 1L * target * n - sum;
      final long pairs = Math.min(totalGap / 2, totalGap - maxGap);
      ans = Math.min(ans, cost1 * (totalGap - 2 * pairs) + cost2 * pairs);
    }

    return (int) (ans % MOD);
  }
}

// Accepted solution for LeetCode #3139: Minimum Cost to Equalize Array
package main

import "slices"

// https://space.bilibili.com/206214
func minCostToEqualizeArray(nums []int, c1, c2 int) int {
	const mod = 1_000_000_007
	n := len(nums)
	m := slices.Min(nums)
	M := slices.Max(nums)
	base := n * M
	for _, x := range nums {
		base -= x
	}
	if n <= 2 || c1*2 <= c2 {
		return base * c1 % mod
	}

	f := func(x int) int {
		s := base + (x-M)*n
		d := x - m
		return max(s/2*c2+s%2*c1, (s-d)*c2+(d*2-s)*c1)
	}

	i := (n*M - m*2 - base + n - 3) / (n - 2)
	if i <= M {
		return min(f(M), f(M+1)) % mod
	}
	return min(f(M), f(i-1), f(i), f(i+1)) % mod
}

# Accepted solution for LeetCode #3139: Minimum Cost to Equalize Array
class Solution:
  def minCostToEqualizeArray(
      self,
      nums: list[int],
      cost1: int,
      cost2: int,
  ) -> int:
    MOD = 1_000_000_007
    n = len(nums)
    minNum = min(nums)
    maxNum = max(nums)
    summ = sum(nums)

    if cost1 * 2 <= cost2 or n < 3:
      totalGap = maxNum * n - summ
      return (cost1 * totalGap) % MOD

    def getMinCost(target: int) -> int:
      """Returns the minimum cost to make all numbers equal to `target`."""
      maxGap = target - minNum
      totalGap = target * n - summ
      # Pair one shallowest number with one non-shallowest number, so the worst
      # case is that we have `totalGap - maxGap` non-shallowest numbers to pair.
      pairs = min(totalGap // 2, totalGap - maxGap)
      return cost1 * (totalGap - 2 * pairs) + cost2 * pairs

    return min(getMinCost(target)
               for target in range(maxNum, 2 * maxNum)) % MOD

// Accepted solution for LeetCode #3139: Minimum Cost to Equalize Array
// 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 #3139: Minimum Cost to Equalize Array
// class Solution {
//   public int minCostToEqualizeArray(int[] nums, int cost1, int cost2) {
//     final int MOD = 1_000_000_007;
//     final int n = nums.length;
//     final int minNum = Arrays.stream(nums).min().getAsInt();
//     final int maxNum = Arrays.stream(nums).max().getAsInt();
//     final long sum = Arrays.stream(nums).asLongStream().sum();
//     long ans = Long.MAX_VALUE;
// 
//     if (cost1 * 2 <= cost2 || n < 3) {
//       final long totalGap = 1L * maxNum * n - sum;
//       return (int) ((cost1 * totalGap) % MOD);
//     }
// 
//     for (int target = maxNum; target < 2 * maxNum; ++target) {
//       final int maxGap = target - minNum;
//       final long totalGap = 1L * target * n - sum;
//       final long pairs = Math.min(totalGap / 2, totalGap - maxGap);
//       ans = Math.min(ans, cost1 * (totalGap - 2 * pairs) + cost2 * pairs);
//     }
// 
//     return (int) (ans % MOD);
//   }
// }

// Accepted solution for LeetCode #3139: Minimum Cost to Equalize Array
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3139: Minimum Cost to Equalize Array
// class Solution {
//   public int minCostToEqualizeArray(int[] nums, int cost1, int cost2) {
//     final int MOD = 1_000_000_007;
//     final int n = nums.length;
//     final int minNum = Arrays.stream(nums).min().getAsInt();
//     final int maxNum = Arrays.stream(nums).max().getAsInt();
//     final long sum = Arrays.stream(nums).asLongStream().sum();
//     long ans = Long.MAX_VALUE;
// 
//     if (cost1 * 2 <= cost2 || n < 3) {
//       final long totalGap = 1L * maxNum * n - sum;
//       return (int) ((cost1 * totalGap) % MOD);
//     }
// 
//     for (int target = maxNum; target < 2 * maxNum; ++target) {
//       final int maxGap = target - minNum;
//       final long totalGap = 1L * target * n - sum;
//       final long pairs = Math.min(totalGap / 2, totalGap - maxGap);
//       ans = Math.min(ans, cost1 * (totalGap - 2 * pairs) + cost2 * pairs);
//     }
// 
//     return (int) (ans % MOD);
//   }
// }

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(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.

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.