What is the time complexity of Minimum Deletions to Make Array Divisible?

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

What data structures are used to solve Minimum Deletions to Make Array Divisible?

This problem uses Array, Math, Sorting, Heap (Priority Queue), Number Theory. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Deletions to Make Array Divisible?

Insight 1: The smallest element in `nums` after deletions must divide every element in `numsDivide`. This is equivalent to the smallest element dividing the Greatest Common Divisor (GCD) of all elements in `numsDivide`. Insight 2: Sorting `nums` allows us to efficiently find the smallest element that satisfies the divisibility condition. We can then easily determine the number of elements smaller than it, which represent the minimum deletions required.

Which companies ask LeetCode 2344. Minimum Deletions to Make Array Divisible in interviews?

This problem is frequently asked at LinkedIn.

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Minimum Deletions to Make Array Divisible - LeetCode 2344 Solution

Minimum Deletions to Make Array Divisible - Complete Solution Guide

Minimum Deletions to Make Array Divisible is LeetCode problem 2344, 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 positive integer arrays nums and numsDivide . You can delete any number of elements from nums . Return the minimum number of deletions such that the smallest element in nums divides all the elements of numsDivide . If this is not possible, return -1 . Note that an integer x divides y if y % x == 0 . Example 1: Input: nums = [2,3,2,4,3], numsDivide = [9,6,9,3,15] Output: 2 Explanation: The smallest element in [2,3,2,4,3] is 2, which does not divide all the elements of numsDivide

Detailed Explanation

The problem asks us to find the minimum number of elements we need to delete from an array `nums` such that the smallest element remaining in `nums` divides all elements in another array `numsDivide`. If it's not possible to achieve this divisibility, we return -1. An integer 'x' divides 'y' if 'y % x == 0'. The goal is to minimize the number of deletions.

Solution Approach

The solution approach involves finding the GCD of all elements in `numsDivide`, sorting the `nums` array, and then iterating through the sorted `nums` array to find the smallest element that divides the GCD. The index of this element in the sorted array represents the minimum number of deletions required, as all elements before it must be deleted to make it the smallest.

Step-by-Step Algorithm

Step 1: Calculate the Greatest Common Divisor (GCD) of all the numbers in `numsDivide`. The GCD represents the largest number that divides all the numbers in `numsDivide` without any remainder. This step is crucial because any number that divides all elements in `numsDivide` must also divide their GCD.
Step 2: Sort the `nums` array in ascending order. Sorting allows us to iterate through the elements in increasing order and find the smallest element that could potentially divide the GCD.
Step 3: Iterate through the sorted `nums` array. For each element, check if it divides the GCD (i.e., GCD % element == 0). If it does, return the index of that element, as it represents the minimum number of deletions needed to make it the smallest element.
Step 4: If the loop completes without finding any element in `nums` that divides the GCD, it means no solution exists. Return -1 in this case.

Key Insights

Insight 1: The smallest element in `nums` after deletions must divide every element in `numsDivide`. This is equivalent to the smallest element dividing the Greatest Common Divisor (GCD) of all elements in `numsDivide`.
Insight 2: Sorting `nums` allows us to efficiently find the smallest element that satisfies the divisibility condition. We can then easily determine the number of elements smaller than it, which represent the minimum deletions required.
Insight 3: If no element in `nums` divides the GCD of `numsDivide`, there is no solution and we return -1.

Complexity Analysis

Time Complexity: O(n log n + m)

Space Complexity: O(1)

Topics

This problem involves: Array, Math, Sorting, Heap (Priority Queue), Number Theory.

Companies

Asked at: LinkedIn.

Minimum Deletions to Make Array Divisible | LeetCode 2344

Hard

Array Math Sorting Heap (Priority Queue)Number Theory

Time: O(n log n + m)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int minOperations(int[] nums, int[] numsDivide) {
        /**
         * Calculates the minimum number of deletions from `nums` such that the smallest
         * remaining element divides all elements in `numsDivide`.
         */

        // Step 1: Find the greatest common divisor (GCD) of all elements in numsDivide.
        // An integer `x` divides all elements of `numsDivide` if and only if
        // `x` divides the GCD of `numsDivide`.
        int g = numsDivide[0];
        for (int i = 1; i < numsDivide.length; i++) {
            g = gcd(g, numsDivide[i]);
        }

        // Step 2: Sort `nums` to easily find the smallest potential divisor.
        // We want to find the smallest number in `nums` that divides `g`.
        // This will minimize the number of required deletions.
        Arrays.sort(nums);

        // Step 3: Iterate through the sorted `nums` to find the first element
        // that divides the GCD.
        for (int i = 0; i < nums.length; i++) {
            if (g % nums[i] == 0) {
                // This `num` is the smallest element in the original `nums` that satisfies
                // the divisibility condition. To make it the smallest element in the
                // modified `nums`, we must delete all elements smaller than it.
                // In the sorted array, these are the elements at indices 0 to i-1.
                // The count of these elements is exactly `i`.
                return i;
            }
        }

        // Step 4: If no element in `nums` divides the GCD, it's impossible.
        return -1;
    }

    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 2344: Minimum Deletions to Make Array Divisible - Detailed explanation and solution approaches.

Topics

Array Math Sorting Heap (Priority Queue)Number Theory

Companies

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2344

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: Calculate the Greatest Common Divisor (GCD) of all the numbers in `numsDivide`. The GCD represents the largest number that divides all the numbers in `numsDivide` without any remainder. This step is crucial because any number that divides all elements in `numsDivide` must also divide their GCD.
2
Step 2: Sort the `nums` array in ascending order. Sorting allows us to iterate through the elements in increasing order and find the smallest element that could potentially divide the GCD.
3
Step 3: Iterate through the sorted `nums` array. For each element, check if it divides the GCD (i.e., GCD % element == 0). If it does, return the index of that element, as it represents the minimum number of deletions needed to make it the smallest element.
4
Step 4: If the loop completes without finding any element in `nums` that divides the GCD, it means no solution exists. Return -1 in this case.

💡

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 + m)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

Sorting the `nums` array using an efficient algorithm such as mergesort or quicksort takes O(n log n) time. Calculating the GCD iteratively takes O(m) time, where 'm' is the size of the `numsDivide` array. The iteration through the sorted `nums` array takes O(n) time, but is dominated by the sorting complexity. The space complexity is O(1) as only constant extra space is used.

📖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

Using a Min-Heap

Instead of sorting, we can use a min-heap to find the smallest element in `nums` that divides the GCD. We iterate through `nums` and maintain a min-heap. Then we iteratively pop the smallest elements until we find one which divides the GCD.

Time: O(n log n + m), where n is the length of `nums` and m is the length of `numsDivide`. Building the heap takes O(n) and popping elements can take O(log n) in the worst case. Calculating GCD takes O(m).Space: O(n), as the heap stores at most n elements.

Trade-offs: Using a heap provides a similar time complexity to sorting but consumes more space. It's less efficient in practice for this particular problem due to the extra overhead of heap operations compared to the optimized sorting implementations.

✨

Key Insights

1Insight 1: The smallest element in `nums` after deletions must divide every element in `numsDivide`. This is equivalent to the smallest element dividing the Greatest Common Divisor (GCD) of all elements in `numsDivide`.
2Insight 2: Sorting `nums` allows us to efficiently find the smallest element that satisfies the divisibility condition. We can then easily determine the number of elements smaller than it, which represent the minimum deletions required.
3Insight 3: If no element in `nums` divides the GCD of `numsDivide`, there is no solution and we return -1.

📚

Key Terms Explained

For Students

Array

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

Sorting

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

Edge Cases

Edge case 1: `nums` is empty. The code will handle this case since the loop won't execute, and -1 will be returned, which is technically correct as no element can divide numsDivide.
Edge case 2: `numsDivide` is empty. The GCD will be initialized to the first element, but the loop will not execute, and sorting will still happen on nums. The result might not be meaningful in the context of the problem description. However, the code still handles this gracefully without crashing.
Edge case 3: One or more elements in `numsDivide` are zero. The GCD might be incorrectly computed. However, the constraints specify that the elements are positive integers, so this case won't happen.
Edge case 4: The smallest element that divides the GCD is also the smallest element in original `nums`. The algorithm correctly identifies this by returning 0, since no deletions are needed.
Edge case 5: No element in nums divides gcd(numsDivide). The algorithm correctly returns -1

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not calculating the GCD of `numsDivide` correctly. Using only one element of `numsDivide` or implementing the GCD algorithm incorrectly will lead to incorrect results.

✗Common mistake 2: Not sorting `nums` before iterating. Without sorting, you might find an element that divides the GCD, but it might not be the smallest element in the array after deletions, resulting in a suboptimal number of deletions.

✗Common mistake 3: Incorrectly calculating the number of deletions. Forgetting that the index `i` of the element in the sorted `nums` array that divides the GCD directly corresponds to the number of elements that must be deleted.

✗Common mistake 4: Not considering the case where no element divides the GCD. Failing to return -1 when no solution exists leads to incorrect answers.

✅

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: Optimizing database queries. Imagine you have a database table and need to filter rows based on a condition involving divisibility. This algorithm can help determine the minimum number of rows to remove (e.g., by marking them as inactive) to ensure the divisibility condition holds for the remaining rows.

🏦Real-world scenario 2: Resource allocation. Suppose you need to allocate resources to different tasks. Each task has a resource requirement. This problem can be adapted to determine the minimum number of tasks to remove so that the remaining tasks can be fulfilled by a certain resource with a given capacity, where divisibility represents whether a resource can fulfil the task's requirements

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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: June 5, 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

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

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 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 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 Deletions to Make Array Divisible"?

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 math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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