What is the time complexity of Number of Different Subsequences GCDs?

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

What data structures are used to solve Number of Different Subsequences GCDs?

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

What are the key insights for solving Number of Different Subsequences GCDs?

Insight 1: The GCD of any subsequence must be a divisor of at least one element in the subsequence, and therefore a divisor of some number in the original array `nums`. This means we only need to consider numbers up to the maximum value in `nums` as potential GCDs. Insight 2: For a given potential GCD `g`, we can efficiently check if `g` is a valid GCD of some subsequence by iterating through the multiples of `g` present in `nums`. If `g` is a valid GCD, it must be equal to the GCD of a subsequence consisting of some multiples of `g` from `nums`.

Which companies ask LeetCode 1819. Number of Different Subsequences GCDs in interviews?

This problem is frequently asked at Akuna Capital.

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Number of Different Subsequences GCDs - LeetCode 1819 Solution

Number of Different Subsequences GCDs - Complete Solution Guide

Number of Different Subsequences GCDs is LeetCode problem 1819, 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 an array nums that consists of positive integers. The GCD of a sequence of numbers is defined as the greatest integer that divides all the numbers in the sequence evenly. For example, the GCD of the sequence [4,6,16] is 2 . A subsequence of an array is a sequence that can be formed by removing some elements (possibly none) of the array. For example, [2,5,10] is a subsequence of [1,2,1, 2 ,4,1, 5 , 10 ] . Return the number of different GCDs among all non-empty subsequences of nums .

Detailed Explanation

The problem asks us to find the number of distinct greatest common divisors (GCDs) that can be obtained from all non-empty subsequences of a given array of positive integers, `nums`. A subsequence is formed by selecting elements from the original array without changing their order, but possibly skipping some. The GCD of a sequence is the largest positive integer that divides all numbers in the sequence evenly. The goal is to iterate through all possible subsequences, calculate their GCDs, and count the number of *unique* GCDs.

Solution Approach

The solution iterates through all possible GCDs from 1 to the maximum value in the input array. For each potential GCD `g`, it checks if `g` is a valid GCD of some subsequence. This is done by iterating through the multiples of `g` in the range [g, max_num], and checking if each multiple exists in the input `nums`. If a multiple exists, it's included in calculating the GCD of the potential subsequence. If, after iterating through all multiples, the GCD of the considered multiples (divided by `g`) is equal to 1, this confirms that `g` can be a valid subsequence GCD, meaning that the GCD of the multiples is equal to `g` and 'g' exists as a subsequence GCD. The solution keeps track of a boolean array `has_num` to check if each number exists in the input array in O(1) time.

Step-by-Step Algorithm

Step 1: Find the maximum number (`max_num`) in the input array `nums`. This is needed to determine the range of potential GCDs to check.
Step 2: Create a boolean array `has_num` of size `max_num + 1` to quickly check if a number exists in the input array `nums`. Initialize it to `false` for all elements.
Step 3: Iterate through the input array `nums` and set `has_num[num]` to `true` for each number in `nums`.
Step 4: Initialize a counter `count` to 0. This counter will store the number of distinct subsequence GCDs.
Step 5: Iterate through all possible GCDs `g` from 1 to `max_num`.
Step 6: For each `g`, initialize `subsequence_k_gcd` to 0. This variable will store the GCD of the multiples of g, divided by g, that are present in `nums`.
Step 7: Iterate through the multiples of `g` (starting from `g` up to `max_num`, incrementing by `g` in each step).
Step 8: Inside the inner loop, check if the current multiple exists in `nums` using the `has_num` array. If `has_num[multiple]` is `true`, it means the multiple is present.
Step 9: If the multiple is present, update `subsequence_k_gcd` by taking the GCD of the current `subsequence_k_gcd` and `multiple / g`. Note how we are calculating the gcd of the multiples divided by g.
Step 10: Optimization: If at any point the `subsequence_k_gcd` becomes 1, it means the GCD of the subsequence is exactly `g`, and `g` is a valid subsequence GCD, thus break out of the inner loop. This avoids further unnecessary GCD calculations.
Step 11: After the inner loop finishes, check if `subsequence_k_gcd` is equal to 1. If it is, increment the `count` because `g` is a valid subsequence GCD.
Step 12: After the outer loop finishes, return the `count`, which represents the total number of distinct subsequence GCDs.

Key Insights

Insight 1: The GCD of any subsequence must be a divisor of at least one element in the subsequence, and therefore a divisor of some number in the original array `nums`. This means we only need to consider numbers up to the maximum value in `nums` as potential GCDs.
Insight 2: For a given potential GCD `g`, we can efficiently check if `g` is a valid GCD of some subsequence by iterating through the multiples of `g` present in `nums`. If `g` is a valid GCD, it must be equal to the GCD of a subsequence consisting of some multiples of `g` from `nums`.
Insight 3: An optimization can be made: for each potential GCD 'g', once the subsequence GCD derived from its multiples becomes 'g', we know g is a valid GCD, so we can stop computing. This significantly reduces unnecessary calculations

Complexity Analysis

Time Complexity: O(n + mlogm)

Space Complexity: O(m)

Topics

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

Companies

Asked at: Akuna Capital.

Number of Different Subsequences GCDs | LeetCode 1819

Hard

Array Math Counting Number Theory

Time: O(n + mlogm)Space: O(m)

Solution Code

Available in:

class Solution {
    public int countDifferentSubsequenceGCDs(int[] nums) {
        int max_num = 0;
        for (int num : nums) {
            if (num > max_num) {
                max_num = num;
            }
        }

        boolean[] has_num = new boolean[max_num + 1];
        for (int num : nums) {
            has_num[num] = true;
        }

        int count = 0;
        for (int g = 1; g <= max_num; g++) {

            int subsequence_k_gcd = 0;
            // Iterate through multiples of g
            for (int multiple = g; multiple <= max_num; multiple += g) {
                if (has_num[multiple]) {
                    int k = multiple / g;
                    if (subsequence_k_gcd == 0) {
                        subsequence_k_gcd = k;
                    } else {
                        subsequence_k_gcd = gcd(subsequence_k_gcd, k);
                    }

                    if (subsequence_k_gcd == 1) {
                        // Optimization: if gcd of k-values is 1, g is a valid subsequence gcd.
                        break;
                    }
                }
            }

            if (subsequence_k_gcd == 1) {
                count++;
            }

        }

        return count;
    }
    
    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 1819: Number of Different Subsequences GCDs - Detailed explanation and solution approaches.

Topics

Array Math Counting Number Theory

Companies

Akuna Capital

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1819

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

12 steps

1
Step 1: Find the maximum number (`max_num`) in the input array `nums`. This is needed to determine the range of potential GCDs to check.
2
Step 2: Create a boolean array `has_num` of size `max_num + 1` to quickly check if a number exists in the input array `nums`. Initialize it to `false` for all elements.
3
Step 3: Iterate through the input array `nums` and set `has_num[num]` to `true` for each number in `nums`.
4
Step 4: Initialize a counter `count` to 0. This counter will store the number of distinct subsequence GCDs.
5
Step 5: Iterate through all possible GCDs `g` from 1 to `max_num`.
6
Step 6: For each `g`, initialize `subsequence_k_gcd` to 0. This variable will store the GCD of the multiples of g, divided by g, that are present in `nums`.
7
Step 7: Iterate through the multiples of `g` (starting from `g` up to `max_num`, incrementing by `g` in each step).
8
Step 8: Inside the inner loop, check if the current multiple exists in `nums` using the `has_num` array. If `has_num[multiple]` is `true`, it means the multiple is present.
9
Step 9: If the multiple is present, update `subsequence_k_gcd` by taking the GCD of the current `subsequence_k_gcd` and `multiple / g`. Note how we are calculating the gcd of the multiples divided by g.
10
Step 10: Optimization: If at any point the `subsequence_k_gcd` becomes 1, it means the GCD of the subsequence is exactly `g`, and `g` is a valid subsequence GCD, thus break out of the inner loop. This avoids further unnecessary GCD calculations.
11
Step 11: After the inner loop finishes, check if `subsequence_k_gcd` is equal to 1. If it is, increment the `count` because `g` is a valid subsequence GCD.
12
Step 12: After the outer loop finishes, return the `count`, which represents the total number of distinct subsequence GCDs.

💡

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 + mlogm)

💾

Space Complexity

O(m)

📝

Why?

The dominant factor in time complexity comes from the nested loop iterating through potential GCDs and their multiples. The space complexity is dictated by the `has_num` array which needs to be as large as the maximum number in the input array to quickly determine the presence of a number.

📖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

Generate all possible subsequences, calculate their GCDs, and then count the number of distinct GCDs. This involves generating all subsets of the input array and calculating the GCD for each subset.

Time: O(2^n * n), where n is the number of elements in the `nums`. The O(2^n) comes from creating all subsequences and n for the gcd of subsequence.Space: O(n) in the worst case, to store the subsequences.

Trade-offs: This approach is very inefficient due to the exponential time complexity and only suitable for very small input sizes. The provided solution is much more efficient for larger inputs.

✨

Key Insights

1Insight 1: The GCD of any subsequence must be a divisor of at least one element in the subsequence, and therefore a divisor of some number in the original array `nums`. This means we only need to consider numbers up to the maximum value in `nums` as potential GCDs.
2Insight 2: For a given potential GCD `g`, we can efficiently check if `g` is a valid GCD of some subsequence by iterating through the multiples of `g` present in `nums`. If `g` is a valid GCD, it must be equal to the GCD of a subsequence consisting of some multiples of `g` from `nums`.
3Insight 3: An optimization can be made: for each potential GCD 'g', once the subsequence GCD derived from its multiples becomes 'g', we know g is a valid GCD, so we can stop computing. This significantly reduces unnecessary calculations

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: `nums` contains only one element. The GCD will be that single element. The code handles this correctly as the outer loop will iterate once with g = element, and the inner loop will find the element and the GCD is confirmed.
Edge case 2: `nums` contains numbers with GCD equal to 1 (e.g., consecutive integers). The code correctly identifies 1 as a valid subsequence GCD. During iteration 'g' will equal 1. The inner loop will iterate over the multiples of 1 and as soon as a number exists the subsequence GCD calculation will confirm 1 and 'g' will be properly counted.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to consider all potential GCDs. It's crucial to iterate up to the maximum value in the input array since any number from the input list can be a valid GCD for a subsequence.

✗Common mistake 2: Incorrect GCD calculation. It's essential to use an efficient GCD algorithm and correctly apply it when checking for the validity of a potential GCD. Additionally, the solution calculates gcd(k1, k2, ...) efficiently by iteratively calculating gcd(gcd(k1,k2), k3) and then further comparing with gcd values of 1.

✅

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 compression. GCD-based techniques can be used to identify common factors in data, which can then be exploited to compress the data more efficiently.

🏦Real-world scenario 2: Cryptography. GCDs are used in various cryptographic algorithms, such as the RSA algorithm, for key generation and encryption/decryption purposes.

💡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 20, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

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.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Number of Different Subsequences GCDs"?

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.

🤖 Need help with this problem?