What is the time complexity of Minimum Cost to Make Array Equal?

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Minimum Cost to Make Array Equal?

This problem uses Array, Binary Search, Greedy, Sorting, Prefix Sum. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Cost to Make Array Equal?

Insight 1: The target value to which we should equalize the array elements is the weighted median of the `nums` array, with `cost` serving as the weights. This is because the cost function (sum of absolute differences multiplied by corresponding costs) is a convex function, and its minimum lies at the weighted median. Insight 2: Sorting the `nums` array allows us to efficiently calculate the cumulative weighted sum, which is crucial for identifying the weighted median.

Which companies ask LeetCode 2448. Minimum Cost to Make Array Equal in interviews?

This problem is frequently asked at Cisco, HashedIn, J.P. Morgan.

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Minimum Cost to Make Array Equal - LeetCode 2448 Solution

Minimum Cost to Make Array Equal - Complete Solution Guide

Minimum Cost to Make Array Equal is LeetCode problem 2448, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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: - I

Detailed Explanation

The problem asks us to find the minimum cost to make all elements of an array `nums` equal. We are given another array `cost`, where `cost[i]` represents the cost of increasing or decreasing `nums[i]` by 1. The goal is to determine the target value to which we should equalize all the elements of `nums` such that the total cost, calculated based on `cost`, is minimized. The inputs are two arrays: `nums` and `cost`, both of size `n`, containing positive integers. The output is a single integer representing the minimum total cost.

Solution Approach

The solution utilizes the concept of the weighted median to minimize the cost. First, pairs of (number, cost) are created and sorted by the number. Then, the algorithm iterates through these pairs, calculating the cumulative sum of the costs. The optimal target value is the number at which the cumulative cost reaches or exceeds half of the total cost sum. This is based on the property that the minimum cost is achieved when equalizing all numbers to the weighted median. Finally, the solution calculates the total cost of making all numbers equal to the target value.

Step-by-Step Algorithm

Step 1: Create an array of pairs (num, cost) from the input arrays `nums` and `cost`.
Step 2: Sort the array of pairs in ascending order based on the `num` value.
Step 3: Calculate the total sum of all costs.
Step 4: Iterate through the sorted pairs, calculating the cumulative cost.
Step 5: During iteration, check if `cumulative_cost * 2 >= total_cost_sum`. If true, the current `num` is the weighted median (target number). Break the loop.
Step 6: Calculate the total cost by summing the absolute differences between each number in the original `nums` array and the target number, multiplied by the corresponding cost from the `cost` array.
Step 7: Return the calculated total cost.

Key Insights

Insight 1: The target value to which we should equalize the array elements is the weighted median of the `nums` array, with `cost` serving as the weights. This is because the cost function (sum of absolute differences multiplied by corresponding costs) is a convex function, and its minimum lies at the weighted median.
Insight 2: Sorting the `nums` array allows us to efficiently calculate the cumulative weighted sum, which is crucial for identifying the weighted median.
Insight 3: The total cost can exceed the range of `int`, requiring the use of `long` data type in languages like Java and C++ to prevent overflow.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Binary Search, Greedy, Sorting, Prefix Sum.

Companies

Asked at: Cisco, HashedIn, J.P. Morgan.

Minimum Cost to Make Array Equal | LeetCode 2448

Hard

Array Binary Search Greedy Sorting Prefix Sum

Time: O(n log n)Space: O(n)

Solution Code

Available in:

import java.util.Arrays;
import java.util.Comparator;

class Solution {
    public long minCost(int[] nums, int[] cost) {
        // Create pairs of (num, cost) and sort them by num.
        int n = nums.length;
        int[][] pairs = new int[n][2];
        for (int i = 0; i < n; i++) {
            pairs[i][0] = nums[i];
            pairs[i][1] = cost[i];
        }

        Arrays.sort(pairs, Comparator.comparingInt(a -> a[0]));

        // Calculate the total sum of costs which represents the total "weight".
        long totalCostSum = 0;
        for (int c : cost) {
            totalCostSum += c;
        }

        // Find the weighted median. The optimal target value to make all numbers
        // equal to is the weighted median of the `nums` array with `cost` as weights.
        // The cost function f(x) = sum(abs(num[i] - x) * cost[i]) is convex,
        // and its minimum is at the weighted median.
        long cumulativeCost = 0;
        int targetNum = 0;
        for (int[] pair : pairs) {
            int num = pair[0];
            int c = pair[1];
            cumulativeCost += c;
            // When the cumulative cost reaches or exceeds half of the total cost,
            // we have found the weighted median.
            if (cumulativeCost * 2 >= totalCostSum) {
                targetNum = num;
                break;
            }
        }

        // Calculate the minimum cost to make all elements equal to the target number.
        long minTotalCost = 0;
        for (int i = 0; i < n; i++) {
            minTotalCost += (long) Math.abs(nums[i] - targetNum) * cost[i];
        }

        return minTotalCost;
    }
}

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Problem Info

Description

Problem 2448: Minimum Cost to Make Array Equal - Detailed explanation and solution approaches.

Topics

Array Binary Search Greedy Sorting Prefix Sum

Companies

CiscoHashedInJ.P. Morgan

Stats

Difficulty:Hard

Languages:4

Companies:3

Problem #:2448

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Create an array of pairs (num, cost) from the input arrays `nums` and `cost`.
2
Step 2: Sort the array of pairs in ascending order based on the `num` value.
3
Step 3: Calculate the total sum of all costs.
4
Step 4: Iterate through the sorted pairs, calculating the cumulative cost.
5
Step 5: During iteration, check if `cumulative_cost * 2 >= total_cost_sum`. If true, the current `num` is the weighted median (target number). Break the loop.
6
Step 6: Calculate the total cost by summing the absolute differences between each number in the original `nums` array and the target number, multiplied by the corresponding cost from the `cost` array.
7
Step 7: Return the calculated total cost.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The dominant operation is the sorting of the pairs array, which uses an efficient sorting algorithm (e.g., quicksort or mergesort) that typically has a time complexity of O(n log n). The creation of the pairs array takes O(n) time. The iterations to compute the weighted median and the final cost also take O(n) time. The space complexity is O(n) to store the pairs.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Binary Search

We can perform a binary search on the range of possible target values (from the minimum to the maximum value in `nums`). For each potential target, we calculate the total cost. The goal is to find the target that minimizes the cost. This works because the cost function is convex.

Time: O(n log m), where n is the length of the array and m is the range of values in nums.Space: O(1)

Trade-offs: Binary search can be more efficient than sorting if the range of values in `nums` is relatively small. However, implementing the cost calculation within the binary search loop adds complexity. Sorting might be simpler to implement in practice.

✨

Key Insights

1Insight 1: The target value to which we should equalize the array elements is the weighted median of the `nums` array, with `cost` serving as the weights. This is because the cost function (sum of absolute differences multiplied by corresponding costs) is a convex function, and its minimum lies at the weighted median.
2Insight 2: Sorting the `nums` array allows us to efficiently calculate the cumulative weighted sum, which is crucial for identifying the weighted median.
3Insight 3: The total cost can exceed the range of `int`, requiring the use of `long` data type in languages like Java and C++ to prevent overflow.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: When all numbers in `nums` are already equal. The algorithm will correctly identify the target number as any of the existing numbers, and the cost will be 0. The loop will iterate and the first num will be chosen as the target.
Edge case 2: When `nums` and `cost` arrays are very large, potential integer overflow can occur when calculating the total cost. The solutions using languages like Java, C++, and C employ `long` data types to handle these large values.
Edge case 3: When the `cost` array contains a very large value at a specific index. The weighted median might coincide with the corresponding value in `nums`, significantly influencing the total cost calculation. The solution correctly handles this case as it focuses on identifying the precise weighted median.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using the simple median instead of the weighted median. The costs associated with each number need to be considered; simply finding the middle element after sorting `nums` will not necessarily yield the minimum total cost.

✗Common mistake 2: Integer overflow. The total cost can be very large, exceeding the capacity of an `int`. Using `long` data types is crucial for preventing incorrect calculations.

✗Common mistake 3: Incorrectly calculating the cumulative cost. It's important to add the cost associated with each `num` value in the sorted pairs array to the cumulative cost to accurately determine the weighted median.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation with varying costs. Imagine distributing resources (e.g., bandwidth, server capacity) to different users where the cost of allocating to each user differs. The algorithm helps find the optimal allocation point that minimizes the overall cost.

🏦Real-world scenario 2: Setting prices for products with varying customer sensitivity. Consider a scenario where adjusting the price of a product affects demand, and each customer segment has a different sensitivity (cost) to price changes. The algorithm helps determine the optimal price point that minimizes the total cost of lost sales or customer dissatisfaction due to price adjustments.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 17, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Cost to Make Array Equal"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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