What is the time complexity of Minimum Non-Zero Product of the Array Elements?

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

What are the key insights for solving Minimum Non-Zero Product of the Array Elements?

Insight 1: The key is realizing that you can redistribute the bits of the numbers in the array `nums`. This means you can effectively decide which numbers have which bits set. Insight 2: To minimize the product, you want as many numbers as possible to be as small as possible. Since the product must be non-zero, we aim to make as many numbers as 1 as possible.

Which companies ask LeetCode 1969. Minimum Non-Zero Product of the Array Elements in interviews?

This problem is frequently asked at PayPal.

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Minimum Non-Zero Product of the Array Elements - LeetCode 1969 Solution

Q: What data structures are used to solve Minimum Non-Zero Product of the Array Elements?

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

Minimum Non-Zero Product of the Array Elements - Complete Solution Guide

Minimum Non-Zero Product of the Array Elements is LeetCode problem 1969, 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 positive integer p . Consider an array nums ( 1-indexed ) that consists of the integers in the inclusive range [1, 2 p - 1] in their binary representations. You are allowed to do the following operation any number of times: Choose two elements x and y from nums . Choose a bit in x and swap it with its corresponding bit in y . Corresponding bit refers to the bit that is in the same position in the other integer. For example, if x = 11 0 1 and y = 00 1 1 , after swapping the 2 nd b

Detailed Explanation

The problem asks us to find the minimum non-zero product of an array `nums` consisting of integers in the range `[1, 2^p - 1]`. We are allowed to swap bits between any two elements of the array any number of times. The goal is to minimize the product of all elements in `nums` after these swaps, and return this minimum product modulo `10^9 + 7`. The crucial point is that we can redistribute the bits to any element, effectively allowing us to rearrange the numbers.

Solution Approach

The solution exploits the bit swapping property to transform the initial array. The approach is to realize that the minimum product is achieved by creating as many 1s as possible. Since the range is `[1, 2^p - 1]`, there are `2^p - 1` numbers in the array. We can make `(2^p - 1) - 1` values equal to 1 and one value equal to `2^p - 1`, and all other values equal to `2^p - 2`. We calculate the product of these values modulo `10^9 + 7`.

Step-by-Step Algorithm

Step 1: Calculate `val1 = 2^p - 1`. This represents the largest number in the original array.
Step 2: Calculate `val2 = 2^p - 2`. This is the value that will be repeated most often in the optimized product.
Step 3: Calculate `exp = 2^(p-1) - 1`. This represents the number of times `val2` appears in the product.
Step 4: Calculate `term1 = val1 % MOD`. This calculates the modulo of the maximum value.
Step 5: Calculate `term2 = pow(val2, exp, MOD)`. This uses modular exponentiation to efficiently compute `val2^exp % MOD`.
Step 6: Calculate `result = (term1 * term2) % MOD`. This combines all modulo calculations to get our final answer.
Step 7: Return the `result`.

Key Insights

Insight 1: The key is realizing that you can redistribute the bits of the numbers in the array `nums`. This means you can effectively decide which numbers have which bits set.
Insight 2: To minimize the product, you want as many numbers as possible to be as small as possible. Since the product must be non-zero, we aim to make as many numbers as 1 as possible.
Insight 3: The optimal arrangement is to have as many 1s as possible, and the rest of the numbers as close to the upper bound as possible, i.e., equal to `2^p - 2`. There will always be a single `2^p - 1` as well.

Complexity Analysis

Time Complexity: O(log(p))

Space Complexity: O(1)

Topics

This problem involves: Math, Greedy, Recursion.

Companies

Asked at: PayPal.

Minimum Non-Zero Product of the Array Elements | LeetCode 1969

Medium

Math Greedy Recursion

Time: O(log(p))Space: O(1)

Solution Code

Available in:

class Solution {
    public int minNonZeroProduct(int p) {
        final int MOD = (int)1e9 + 7;
        
        // The numbers are in the range [1, 2^p - 1].
        // The key insight is that we can redistribute bits.
        // To minimize the product, we want to make as many numbers as possible equal to 1.
        // The rest of the numbers will become large.
        // This leads to a configuration with:
        // - (2^(p-1) - 1) numbers equal to 1.
        // - (2^(p-1) - 1) numbers equal to (2^p - 2).
        // - 1 number equal to (2^p - 1).
        //
        // The product is (2^p - 1) * (2^p - 2)^(2^(p-1) - 1).
        // We compute this product modulo 10^9 + 7.
        //
        // For p=1, the formula gives 1 * 0^0. The product of zero terms (empty product)
        // is 1. Python's pow(0,0) is 1, so the formula holds for p=1 as well.
        
        // val1 = 2^p - 1
        long val1 = (1L << p) - 1;
        
        // val2 = 2^p - 2
        long val2 = (1L << p) - 2;
        
        // exp = 2^(p-1) - 1
        long exp = (1L << (p - 1)) - 1;
        
        // Compute (val1 % MOD * pow(val2, exp, MOD)) % MOD
        
        long term1 = val1 % MOD;
        
        // Calculate (val2^exp) % MOD using modular exponentiation
        long term2 = power(val2, exp, MOD);
        
        long result = (term1 * term2) % MOD;
        
        return (int) result;
    }

    private long power(long base, long exp, int mod) {
        long res = 1;
        base %= mod;
        while (exp > 0) {
            if (exp % 2 == 1) res = (res * base) % mod;
            base = (base * base) % mod;
            exp >>= 1;
        }
        return res;
    }
}

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Problem Info

Description

Problem 1969: Minimum Non-Zero Product of the Array Elements - Detailed explanation and solution approaches.

Topics

Math Greedy Recursion

Companies

PayPal

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1969

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

7 steps

1
Step 1: Calculate `val1 = 2^p - 1`. This represents the largest number in the original array.
2
Step 2: Calculate `val2 = 2^p - 2`. This is the value that will be repeated most often in the optimized product.
3
Step 3: Calculate `exp = 2^(p-1) - 1`. This represents the number of times `val2` appears in the product.
4
Step 4: Calculate `term1 = val1 % MOD`. This calculates the modulo of the maximum value.
5
Step 5: Calculate `term2 = pow(val2, exp, MOD)`. This uses modular exponentiation to efficiently compute `val2^exp % MOD`.
6
Step 6: Calculate `result = (term1 * term2) % MOD`. This combines all modulo calculations to get our final answer.
7
Step 7: Return the `result`.

💡

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(log(p))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant operation in the provided solutions is the modular exponentiation, specifically the `pow(val2, exp, MOD)` operation. Modular exponentiation generally takes O(log(exp)) time, where exp is the exponent. In this case, `exp = 2^(p-1) - 1`, so the time complexity is effectively O(log(2^(p-1) - 1)), which simplifies to O(p). However, since `p` is the input, we consider p as the size of the input in binary, so log(2^p) ~ p which turns into O(log N) where N= 2^p, meaning the complexity is O(p), can also be written O(log(N)). But since the number `p` (the input to the function), the number of bits, determines the time the solution takes, then the time complexity can also be considered to be O(log(p)). All other operations are constant time. The space complexity is O(1) because we only use a constant number of variables regardless of the input size.

📖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

Iterative Modular Exponentiation

Instead of using the built-in `pow` function (in Python) or relying on libraries, implement modular exponentiation iteratively. This is done in the Java and C implementations.

Time: O(log(p))Space: O(1)

Trade-offs: This can be more efficient in certain environments where the built-in power function is not optimized for modular arithmetic. It can also offer better control over the calculation process, and may be necessary in environments where a built-in power function is not available.

✨

Key Insights

1Insight 1: The key is realizing that you can redistribute the bits of the numbers in the array `nums`. This means you can effectively decide which numbers have which bits set.
2Insight 2: To minimize the product, you want as many numbers as possible to be as small as possible. Since the product must be non-zero, we aim to make as many numbers as 1 as possible.
3Insight 3: The optimal arrangement is to have as many 1s as possible, and the rest of the numbers as close to the upper bound as possible, i.e., equal to `2^p - 2`. There will always be a single `2^p - 1` as well.

📚

Key Terms Explained

For Students

Recursion

A function that calls itself to solve smaller versions of the same problem.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: p = 1. The range is [1]. The result should be 1. The provided code handles this correctly because the formula reduces to (1) * (0)^0 % MOD which is 1. pow(0, 0) in python and the C++ implementation via a loop yields 1.
Edge case 2: Large values of p. Because 2^p can be very large, we need to use long long (or long in Java) to prevent integer overflow during calculations. The code uses `long long` appropriately in C++ and C and `long` in Java.
Edge case 3: The modulo operation is crucial to prevent overflow, especially when p is large. The solution consistently applies the modulo operation to intermediate results.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow. Failing to use `long long` or `long` to store intermediate values will cause incorrect results for larger values of `p`.

✗Common mistake 2: Not applying the modulo operator frequently enough. The modulo operator must be applied after each multiplication to prevent overflow.

✗Common mistake 3: Forgetting to calculate the exponent correctly. The exponent should be `2^(p-1) - 1`, representing the number of times `2^p - 2` appears in the minimized product.

✅

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: Cryptography. Modular arithmetic and exponentiation are fundamental to many cryptographic algorithms, such as RSA. Understanding these concepts is essential for working with secure communication and data encryption.

🏦Real-world scenario 2: Hashing algorithms. Modular arithmetic is often used in hash functions to ensure that the hash values fall within a specific range, which is useful in data storage and retrieval.

💡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

Always test your solution with edge cases: empty input, single element, and boundary values.

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: November 17, 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 14

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

@DevExpert92Jul 7

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

@AlgoMasterJun 30

❓

Frequently Asked Questions

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

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 "Minimum Non-Zero Product of the Array Elements"?

For this medium-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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