What is the time complexity of Minimize Length of Array Using Operations?

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

What data structures are used to solve Minimize Length of Array Using Operations?

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

What are the key insights for solving Minimize Length of Array Using Operations?

Insight 1: The minimum value in the array plays a crucial role. If any element is not divisible by the minimum value, the array can always be reduced to a single element. Insight 2: If all elements are divisible by the minimum value, the operation `nums[i] % nums[j]` where `nums[j]` is the minimum element will always result in 0 if `nums[i]` is not the minimum element and `nums[i]` modulo the minimum element, if `nums[i]` is not the minimum element, or 0 if `nums[i]` is also equal to the minimum element.

Which companies ask LeetCode 3012. Minimize Length of Array Using Operations in interviews?

This problem is frequently asked at BNY Mellon, HashedIn.

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Minimize Length of Array Using Operations - LeetCode 3012 Solution

Minimize Length of Array Using Operations - Complete Solution Guide

Minimize Length of Array Using Operations is LeetCode problem 3012, 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 nums containing positive integers. Your task is to minimize the length of nums by performing the following operations any number of times (including zero): Select two distinct indices i and j from nums , such that nums[i] > 0 and nums[j] > 0 . Insert the result of nums[i] % nums[j] at the end of nums . Delete the elements at indices i and j from nums . Return an integer denoting the minimum length of nums after performing the operation any number of times.

Detailed Explanation

The problem asks us to minimize the length of an array of positive integers by repeatedly performing an operation: selecting two distinct elements, calculating their modulo, appending the result to the array, and then removing the original two elements. The goal is to find the shortest possible length of the array after applying this operation any number of times.

Solution Approach

The solution involves first finding the minimum value in the array. Then, it checks if all other elements are divisible by this minimum value. If not, it means the array can be reduced to a single element. If all elements are divisible by the minimum value, the solution counts the number of occurrences of the minimum value in the array and returns `(count + 1) // 2`. This is because each pair of minimum values can be combined to effectively remove two elements from the array at a time.

Step-by-Step Algorithm

Step 1: Find the minimum value in the array.
Step 2: Iterate through the array and check if each element is divisible by the minimum value. If any element is not divisible, return 1 immediately.
Step 3: If all elements are divisible by the minimum value, count the occurrences of the minimum value in the array.
Step 4: Return `(count + 1) // 2`, where `count` is the number of occurrences of the minimum value.

Key Insights

Insight 1: The minimum value in the array plays a crucial role. If any element is not divisible by the minimum value, the array can always be reduced to a single element.
Insight 2: If all elements are divisible by the minimum value, the operation `nums[i] % nums[j]` where `nums[j]` is the minimum element will always result in 0 if `nums[i]` is not the minimum element and `nums[i]` modulo the minimum element, if `nums[i]` is not the minimum element, or 0 if `nums[i]` is also equal to the minimum element.
Insight 3: Once all numbers are divisible by the smallest element. the only things that matter is the count of the smallest value. Any pair of smallest values will eventually form a 0 or another minimum element. The key is to find how many pairs you can form using the count of the minimum element

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

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

Companies

Asked at: BNY Mellon, HashedIn.

Minimize Length of Array Using Operations | LeetCode 3012

Medium

Array Math Greedy Number Theory

Time: O(n)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int minimumArrayLength(int[] nums) {
        int min_val = Arrays.stream(nums).min().orElse(Integer.MAX_VALUE);
        
        for (int num : nums) {
            if (num % min_val != 0) {
                return 1;
            }
        }
        
        int count_min = 0;
        for (int num : nums) {
            if (num == min_val) {
                count_min++;
            }
        }
        return (count_min + 1) / 2;
    }
}

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Problem Info

Description

Problem 3012: Minimize Length of Array Using Operations - Detailed explanation and solution approaches.

Topics

Array Math Greedy Number Theory

Companies

BNY MellonHashedIn

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:3012

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

4 steps

1
Step 1: Find the minimum value in the array.
2
Step 2: Iterate through the array and check if each element is divisible by the minimum value. If any element is not divisible, return 1 immediately.
3
Step 3: If all elements are divisible by the minimum value, count the occurrences of the minimum value in the array.
4
Step 4: Return `(count + 1) // 2`, where `count` is the number of occurrences of the minimum value.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because the algorithm iterates through the array a fixed number of times. The space complexity is constant because only a few variables are used to store the minimum value and its count.

📖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

Sorting-based approach

Sort the array and then apply the modulo operation repeatedly. This approach would be significantly less efficient

Time: O(n log n) due to sorting.Space: O(1) if sorting in place, or O(n) if sorting requires extra space.

Trade-offs: This approach is less efficient than the linear-time solution and is not recommended.

✨

Key Insights

1Insight 1: The minimum value in the array plays a crucial role. If any element is not divisible by the minimum value, the array can always be reduced to a single element.
2Insight 2: If all elements are divisible by the minimum value, the operation `nums[i] % nums[j]` where `nums[j]` is the minimum element will always result in 0 if `nums[i]` is not the minimum element and `nums[i]` modulo the minimum element, if `nums[i]` is not the minimum element, or 0 if `nums[i]` is also equal to the minimum element.
3Insight 3: Once all numbers are divisible by the smallest element. the only things that matter is the count of the smallest value. Any pair of smallest values will eventually form a 0 or another minimum element. The key is to find how many pairs you can form using the count of the minimum element

📚

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.

Edge Cases

Edge case 1: Array contains only one element. The code correctly calculates the minimum value, finds that all elements are divisible, counts the minimum value (which is 1), and returns (1+1)//2 = 1.
Edge case 2: Array contains multiple elements, but none of them are divisible by the minimum. The code finds the minimum, encounters an element that isn't divisible by it, and immediately returns 1.
Edge case 3: Array with all the same numbers. All elements will be divisible by the minimum, then the code finds count of the smallest number that would be the length of the original array. It correctly computes the result as `(count + 1) // 2`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where some elements are not divisible by the minimum value. Failing to return 1 in this scenario leads to incorrect results.

✗Common mistake 2: Incorrectly calculating the number of minimum values needed. Using `count // 2` will result in an off by one error when the count is odd because at the very end we can have one single number left

✅

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. Minimizing the number of resources needed by combining smaller units. Similar logic can be applied to merge similar elements to reduce overall count.

🏦Real-world scenario 2: Data compression. Identifying patterns in data to reduce its size, where the modulo operation can represent a similarity or relationship between data points.

💡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: October 11, 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

❓

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.

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 "Minimize Length of Array Using Operations"?

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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