What is the time complexity of Make K-Subarray Sums Equal?

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

What data structures are used to solve Make K-Subarray Sums Equal?

This problem uses Array, Math, Greedy, Sorting, Number Theory. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Make K-Subarray Sums Equal?

Insight 1: The key insight is that if the sum of each k-length subarray is equal, then `arr[i] == arr[i+k] == arr[i+2k]...`. This means that elements at indices `i + j*k` are equivalent for all `j`. Insight 2: Elements belonging to the same group are separated by `gcd(n, k)` indices. Therefore, we iterate through the indices `i` from `0` to `gcd(n, k) - 1`, and elements `arr[i], arr[i + gcd(n,k)], arr[i + 2*gcd(n,k)], ...` should be made equal.

Which companies ask LeetCode 2607. Make K-Subarray Sums Equal in interviews?

This problem is frequently asked at Morgan Stanley.

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Make K-Subarray Sums Equal - LeetCode 2607 Solution

Make K-Subarray Sums Equal - Complete Solution Guide

Make K-Subarray Sums Equal is LeetCode problem 2607, 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 a 0-indexed integer array arr and an integer k . The array arr is circular. In other words, the first element of the array is the next element of the last element, and the last element of the array is the previous element of the first element. You can do the following operation any number of times: Pick any element from arr and increase or decrease it by 1 . Return the minimum number of operations such that the sum of each subarray of length k is equal . A subarray is a contiguous

Detailed Explanation

The problem asks us to find the minimum number of operations needed to make the sum of all k-length subarrays of a circular array `arr` equal. An operation involves incrementing or decrementing an element of the array by 1. The array is circular, meaning the last element is considered adjacent to the first element. To achieve equal k-length subarray sums, all elements within a cycle determined by `gcd(n,k)` should become equal. The core task is to find the best target value (median) for each cycle and count the moves necessary to achieve it.

Solution Approach

The provided solution efficiently calculates the minimum number of operations by identifying groups of elements that need to be equalized and then computing the cost of transforming each group to its median value. The solution starts by finding the greatest common divisor (GCD) of the array's length `n` and `k`. The GCD represents the number of independent groups of elements that need to be considered. For each group, the elements are collected, sorted, and the median is identified. The sum of the absolute differences between each element in the group and the median gives the minimum number of operations required for that group. The total operations across all groups are then summed to give the final result.

Step-by-Step Algorithm

Step 1: Calculate the greatest common divisor (GCD) of `n` (length of `arr`) and `k` using the Euclidean algorithm or `math.gcd()` function.
Step 2: Initialize `total_ops` to 0. This variable will store the total number of operations required.
Step 3: Iterate from `i = 0` to `gcd(n, k) - 1`. This loop iterates through each independent group.
Step 4: Within the loop, create a list called `group` and populate it with elements from `arr` at indices `i, i + gcd(n, k), i + 2*gcd(n, k), ...` until you reach the end of the array.
Step 5: Sort the `group` list in ascending order.
Step 6: Calculate the median of the `group` list. For a sorted list, the median is simply the element at index `len(group) // 2`.
Step 7: Iterate through each element `x` in the `group` list and add `abs(x - median)` to `total_ops`. This calculates the number of operations required to make each element equal to the median.
Step 8: After the outer loop completes, return `total_ops`.

Key Insights

Insight 1: The key insight is that if the sum of each k-length subarray is equal, then `arr[i] == arr[i+k] == arr[i+2k]...`. This means that elements at indices `i + j*k` are equivalent for all `j`.
Insight 2: Elements belonging to the same group are separated by `gcd(n, k)` indices. Therefore, we iterate through the indices `i` from `0` to `gcd(n, k) - 1`, and elements `arr[i], arr[i + gcd(n,k)], arr[i + 2*gcd(n,k)], ...` should be made equal.
Insight 3: The optimal value to make the elements in a group equal to is the median of the group. This minimizes the sum of absolute differences, which corresponds to the number of increment/decrement operations needed.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Math, Greedy, Sorting, Number Theory.

Companies

Asked at: Morgan Stanley.

Make K-Subarray Sums Equal | LeetCode 2607

Medium

Array Math Greedy Sorting Number Theory

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

Solution Code

Available in:

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

class Solution {
    public long makeSubKSumEqual(int[] arr, int k) {
        int n = arr.length;
        int g = gcd(n, k);
        
        long total_ops = 0;
        
        for (int i = 0; i < g; i++) {
            List<Integer> groupList = new ArrayList<>();
            for (int j = i; j < n; j += g) {
                groupList.add(arr[j]);
            }
            
            int[] group = new int[groupList.size()];
            for (int j = 0; j < groupList.size(); j++) {
                group[j] = groupList.get(j);
            }
            
            Arrays.sort(group);
            int median = group[group.length / 2];
            
            for (int x : group) {
                total_ops += Math.abs(x - median);
            }
        }
        
        return total_ops;
    }
    
    private int gcd(int a, int b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }
}

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Problem Info

Description

Problem 2607: Make K-Subarray Sums Equal - Detailed explanation and solution approaches.

Topics

Array Math Greedy Sorting Number Theory

Companies

Morgan Stanley

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2607

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

8 steps

1
Step 1: Calculate the greatest common divisor (GCD) of `n` (length of `arr`) and `k` using the Euclidean algorithm or `math.gcd()` function.
2
Step 2: Initialize `total_ops` to 0. This variable will store the total number of operations required.
3
Step 3: Iterate from `i = 0` to `gcd(n, k) - 1`. This loop iterates through each independent group.
4
Step 4: Within the loop, create a list called `group` and populate it with elements from `arr` at indices `i, i + gcd(n, k), i + 2*gcd(n, k), ...` until you reach the end of the array.
5
Step 5: Sort the `group` list in ascending order.
6
Step 6: Calculate the median of the `group` list. For a sorted list, the median is simply the element at index `len(group) // 2`.
7
Step 7: Iterate through each element `x` in the `group` list and add `abs(x - median)` to `total_ops`. This calculates the number of operations required to make each element equal to the median.
8
Step 8: After the outer loop completes, return `total_ops`.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is derived from iterating through `g` groups and sorting each group. In the worst case, when `g` is 1, the complexity is O(n log n) due to the sorting. The space complexity comes from storing the elements of each group, which in the worst case is O(n). Overall space is divided amongst each group

📖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

Dynamic Programming

Although unlikely to be efficient, a DP approach could be conceived where dp[i][j] stores min operations to make first i elements have k subarray sums = j. However given constraints this could be computationally expensive and memory intensive.

Time: O(n*sum(arr))Space: O(n*sum(arr))

Trade-offs: Not recommended, due to high time and space complexity and inability to deal with large constraints

✨

Key Insights

1Insight 1: The key insight is that if the sum of each k-length subarray is equal, then `arr[i] == arr[i+k] == arr[i+2k]...`. This means that elements at indices `i + j*k` are equivalent for all `j`.
2Insight 2: Elements belonging to the same group are separated by `gcd(n, k)` indices. Therefore, we iterate through the indices `i` from `0` to `gcd(n, k) - 1`, and elements `arr[i], arr[i + gcd(n,k)], arr[i + 2*gcd(n,k)], ...` should be made equal.
3Insight 3: The optimal value to make the elements in a group equal to is the median of the group. This minimizes the sum of absolute differences, which corresponds to the number of increment/decrement operations needed.

📚

Key Terms Explained

For Students

Array

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

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: `k = 1`. In this case, all elements should be equal, and the optimal target is the median of the entire array.
Edge case 2: `k = n`. In this case, all elements should be equal, and the optimal target is the median of the entire array.
Edge case 3: `n = 1`. In this case, there is only one element, and no operations are needed.
Edge case 4: Array contains identical numbers. In this case, the operations count should return 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to take the absolute difference when calculating the operations count. The number of operations is the absolute difference between the element and the median.

✗Common mistake 2: Not considering the circular nature of the array. The solution should properly handle the indices when creating the groups.

✗Common mistake 3: Failing to recognize that the median minimizes the sum of absolute differences. Choosing the mean or another value will not minimize the operations count.

✅

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: Data balancing: Imagine sensor readings in a circular arrangement. The algorithm can be used to minimize the adjustments needed to make the readings more uniform, for example in sensor calibration.

🏦Real-world scenario 2: Inventory Management: Consider inventory levels in a circular supply chain. The algorithm helps determine the minimum adjustments (transfers) needed to equalize inventory levels at different points in the chain.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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 27, 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 14

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

@CodeNinjaJul 7

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

@TechLeadJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Make K-Subarray Sums 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 math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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