What is the time complexity of Minimum Operations to Make All Array Elements Equal?

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

What data structures are used to solve Minimum Operations to Make All Array Elements Equal?

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

What are the key insights for solving Minimum Operations to Make All Array Elements Equal?

Insight 1: Sorting the `nums` array allows us to efficiently determine the boundary between elements smaller than the query and elements larger than the query using binary search. Insight 2: Using prefix sums enables the calculation of the total cost of transforming elements less than or greater than the query value in O(1) time, once the boundary is found.

Which companies ask LeetCode 2602. Minimum Operations to Make All Array Elements Equal in interviews?

This problem is frequently asked at IBM, J.P. Morgan.

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Minimum Operations to Make All Array Elements Equal - LeetCode 2602 Solution

Minimum Operations to Make All Array Elements Equal - Complete Solution Guide

Minimum Operations to Make All Array Elements Equal is LeetCode problem 2602, a Medium 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 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 arr

Detailed Explanation

The problem requires calculating the minimum number of operations needed to make all elements in an array `nums` equal to each element in a given array of `queries`. An operation involves increasing or decreasing an element by 1. For each query, we need to compute the total cost to transform all elements of `nums` to the query value. The array `nums` is reset to its original state for each query. The input arrays `nums` and `queries` contain positive integers, and the goal is to return an array `answer` where `answer[i]` represents the minimum operations for `queries[i]`.

Solution Approach

The solution sorts the input array `nums` to enable the use of binary search. A prefix sum array is constructed to efficiently calculate the sum of elements up to a given index. For each query, binary search (specifically, `bisect_right` or `upper_bound`) is used to find the index `k` which represents the boundary. The cost to transform elements less than or equal to index `k-1` (inclusive) is calculated using `k * query - prefix_sum[k]`. The cost to transform elements greater than or equal to index `k` is calculated using `(prefix_sum[n] - prefix_sum[k]) - (n - k) * query`. These two costs are added to get the total cost, which is appended to the `answer` array.

Step-by-Step Algorithm

Step 1: Sort the `nums` array in ascending order.
Step 2: Create a prefix sum array `prefix_sum` where `prefix_sum[i]` is the sum of the first `i` elements of the sorted `nums` array.
Step 3: Iterate through each query `q` in the `queries` array.
Step 4: Use binary search (`bisect_right` or `upper_bound`) to find the index `k` in the sorted `nums` array such that all elements before index `k` are less than or equal to `q`.
Step 5: Calculate the cost to transform elements less than `q`: `left_cost = k * q - prefix_sum[k]`.
Step 6: Calculate the cost to transform elements greater than `q`: `right_cost = (prefix_sum[n] - prefix_sum[k]) - (n - k) * q`.
Step 7: Calculate the total cost: `total_cost = left_cost + right_cost`.
Step 8: Append the `total_cost` to the `answer` array.
Step 9: Return the `answer` array.

Key Insights

Insight 1: Sorting the `nums` array allows us to efficiently determine the boundary between elements smaller than the query and elements larger than the query using binary search.
Insight 2: Using prefix sums enables the calculation of the total cost of transforming elements less than or greater than the query value in O(1) time, once the boundary is found.
Insight 3: The cost calculation can be broken down into two parts: the cost to increase elements less than the query value and the cost to decrease elements greater than the query value.

Complexity Analysis

Time Complexity: O(m log n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: IBM, J.P. Morgan.

Minimum Operations to Make All Array Elements Equal | LeetCode 2602

Medium

Array Binary Search Sorting Prefix Sum

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

class Solution {
    public List<Long> minOperations(int[] nums, int[] queries) {
        int n = nums.length;
        Arrays.sort(nums);

        long[] prefixSum = new long[n + 1];
        for (int i = 0; i < n; i++) {
            prefixSum[i + 1] = prefixSum[i] + nums[i];
        }

        List<Long> answer = new ArrayList<>();
        for (int q : queries) {
            int k = binarySearchRight(nums, q);

            long leftCost = (long) k * q - prefixSum[k];
            long rightCost = (prefixSum[n] - prefixSum[k]) - (long) (n - k) * q;

            long totalOps = leftCost + rightCost;
            answer.add(totalOps);
        }

        return answer;
    }

    private int binarySearchRight(int[] nums, int target) {
        int low = 0;
        int high = nums.length;
        while (low < high) {
            int mid = low + (high - low) / 2;
            if (nums[mid] <= target) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        return low;
    }
}

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Problem Info

Description

Problem 2602: Minimum Operations to Make All Array Elements Equal - Detailed explanation and solution approaches.

Topics

Array Binary Search Sorting Prefix Sum

Companies

IBMJ.P. Morgan

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2602

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

9 steps

1
Step 1: Sort the `nums` array in ascending order.
2
Step 2: Create a prefix sum array `prefix_sum` where `prefix_sum[i]` is the sum of the first `i` elements of the sorted `nums` array.
3
Step 3: Iterate through each query `q` in the `queries` array.
4
Step 4: Use binary search (`bisect_right` or `upper_bound`) to find the index `k` in the sorted `nums` array such that all elements before index `k` are less than or equal to `q`.
5
Step 5: Calculate the cost to transform elements less than `q`: `left_cost = k * q - prefix_sum[k]`.
6
Step 6: Calculate the cost to transform elements greater than `q`: `right_cost = (prefix_sum[n] - prefix_sum[k]) - (n - k) * q`.
7
Step 7: Calculate the total cost: `total_cost = left_cost + right_cost`.
8
Step 8: Append the `total_cost` to the `answer` array.
9
Step 9: Return the `answer` array.

💡

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(m log n)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The dominant operations are sorting the array and performing binary search for each query. Sorting `nums` is O(n log n), and the loop iterates `m` times, each with a binary search operation costing O(log n). The prefix sum array is O(n) for storage. Therefore, the total time complexity is O(n log n + m log n) and the space complexity is O(n).

📖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

Brute-force approach

For each query, iterate through the `nums` array and calculate the absolute difference between each element and the query value. Sum up these differences to get the total cost.

Time: O(n * m), where n is the length of `nums` and m is the length of `queries`.Space: O(1)

Trade-offs: This approach is simpler to implement but has a much higher time complexity compared to the optimized approach using sorting and binary search. It's suitable for small input sizes but becomes inefficient for larger inputs. It avoids any auxiliary space.

✨

Key Insights

1Insight 1: Sorting the `nums` array allows us to efficiently determine the boundary between elements smaller than the query and elements larger than the query using binary search.
2Insight 2: Using prefix sums enables the calculation of the total cost of transforming elements less than or greater than the query value in O(1) time, once the boundary is found.
3Insight 3: The cost calculation can be broken down into two parts: the cost to increase elements less than the query value and the cost to decrease elements greater than the query value.

📚

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!

Sorting

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

Edge Cases

Edge case 1: `nums` is empty. The solution handles this implicitly since the loops won't execute, and the prefix sum will be an array of size 1 with a value of 0.
Edge case 2: All elements in `nums` are the same. The binary search will still function correctly and find the appropriate index. The cost calculation remains valid.
Edge case 3: All queries are the same. The code will still correctly compute the operation count for each query individually.
Edge case 4: A query is smaller than all numbers in `nums` or larger than all numbers in `nums`. Binary search will return 0 or `n` respectively, and the cost calculation will still be valid in these extreme scenarios.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the `nums` array. Without sorting, binary search cannot be used effectively, leading to an inefficient solution.

✗Common mistake 2: Incorrectly calculating the prefix sum or using the prefix sum array incorrectly in the cost calculation. Errors in prefix sum calculation will propagate to the final answer.

✗Common mistake 3: Integer overflow. When calculating the cost, multiplying `k` by `q` or `(n - k)` by `q` can result in an integer overflow if the values are large. Using `long` or `long long` data types is necessary to prevent this.

✗Common mistake 4: Using bisect_left instead of bisect_right (or similar functions in other languages). This will cause an incorrect index to be used for calculating prefix sum leading to incorrect cost.

✅

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. Imagine allocating resources (e.g., memory, bandwidth) to different tasks. This algorithm can help determine the minimum cost (e.g., time, money) to adjust resource levels to a target value for each task.

🏦Real-world scenario 2: Data normalization. In data preprocessing, features are often normalized to a specific range. This algorithm can be used to calculate the total cost (effort) to normalize values to a target value, where the cost is proportional to the difference between the original and normalized values.

🎮Real-world scenario 3: Inventory management. Suppose you have several warehouses and need to determine the cost of making the inventory levels in all warehouses equal to a target value. This problem can be modelled using this algorithm.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: August 22, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

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

@CodeNinjaJul 6

❓

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.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 Operations to Make All Array Elements Equal"?

For this medium-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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