What is the time complexity of Construct the Minimum Bitwise Array I?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Construct the Minimum Bitwise Array I?

This problem uses Array, Bit Manipulation. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Construct the Minimum Bitwise Array I?

Insight 1: The bitwise OR operation (`|`) combines bits. If `x | (x + 1) == y`, then `y` must contain all the bits that are set in either `x` or `x + 1`. This means that `x` must be cleverly chosen to satisfy the equation. A simple example: 1|2 = 3. Insight 2: Since the numbers in `nums` are prime, it's unlikely a simple formula will directly compute `ans[i]`. A brute force search checking values from 0 up to `nums[i]-1` is a straightforward approach.

Which companies ask LeetCode 3314. Construct the Minimum Bitwise Array I in interviews?

This problem is frequently asked at Aon.

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Construct the Minimum Bitwise Array I - LeetCode 3314 Solution

Construct the Minimum Bitwise Array I - Complete Solution Guide

Construct the Minimum Bitwise Array I is LeetCode problem 3314, a Easy 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 consisting of n prime integers. You need to construct an array ans of length n , such that, for each index i , the bitwise OR of ans[i] and ans[i] + 1 is equal to nums[i] , i.e. ans[i] OR (ans[i] + 1) == nums[i] . Additionally, you must minimize each value of ans[i] in the resulting array. If it is not possible to find such a value for ans[i] that satisfies the condition , then set ans[i] = -1 . Example 1: Input: nums = [2,3,5,7] Output: [-1,1,4,3] Explanation: For i

Detailed Explanation

The problem asks you to create an array `ans` of the same length as the input array `nums`. For each element `nums[i]`, you need to find the smallest non-negative integer `ans[i]` such that the bitwise OR of `ans[i]` and `ans[i] + 1` equals `nums[i]`. If no such `ans[i]` exists, you should set `ans[i]` to -1. The input `nums` contains only prime numbers. Each `nums[i]` must be greater than 1.

Solution Approach

The solutions provided all employ a brute-force approach. They iterate through possible values of `ans[i]` from 0 up to `nums[i] - 1`, checking if the bitwise OR condition is met. If a suitable `ans[i]` is found, it's added to the `ans` array. If no such value is found, `ans[i]` is set to -1.

Step-by-Step Algorithm

Step 1: Initialize an empty array `ans` with the same length as `nums`.
Step 2: Iterate through each element `num` in the input array `nums`.
Step 3: Iterate from `i = 0` to `num - 1`. Check if `(i | (i + 1)) == num`.
Step 4: If the condition in Step 3 is true, append `i` to `ans`. Break the inner loop because we've found the minimum solution.
Step 5: If the inner loop completes without finding a solution (Step 3 is never true), append -1 to `ans`.
Step 6: Return the `ans` array.

Key Insights

Insight 1: The bitwise OR operation (`|`) combines bits. If `x | (x + 1) == y`, then `y` must contain all the bits that are set in either `x` or `x + 1`. This means that `x` must be cleverly chosen to satisfy the equation. A simple example: 1|2 = 3.
Insight 2: Since the numbers in `nums` are prime, it's unlikely a simple formula will directly compute `ans[i]`. A brute force search checking values from 0 up to `nums[i]-1` is a straightforward approach.
Insight 3: Optimizations are limited because of the requirement to find the *minimum* value of `ans[i]`. While a clever bit manipulation approach might exist for specific patterns in prime numbers, it's not generally applicable, and brute force is relatively efficient for the problem constraints (small input size).

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, Bit Manipulation.

Companies

Asked at: Aon.

Construct the Minimum Bitwise Array I | LeetCode 3314

Easy

Array Bit Manipulation

Time: O(n^2)Space: O(n)

Solution Code

Available in:

import java.util.*;
class Solution {
    public int[] minBitwiseArray(List<Integer> nums) {
        int n = nums.size();
        int[] ans = new int[n];
        for (int i = 0; i < n; i++) {
            int num = nums.get(i);
            int x = -1;
            for (int j = 0; j < num; j++) {
                if ((j | (j + 1)) == num) {
                    x = j;
                    break;
                }
            }
            ans[i] = x;
        }
        return ans;
    }
}

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Problem Info

Description

Problem 3314: Construct the Minimum Bitwise Array I - Detailed explanation and solution approaches.

Topics

Array Bit Manipulation

Companies

Aon

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3314

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

6 steps

1
Step 1: Initialize an empty array `ans` with the same length as `nums`.
2
Step 2: Iterate through each element `num` in the input array `nums`.
3
Step 3: Iterate from `i = 0` to `num - 1`. Check if `(i | (i + 1)) == num`.
4
Step 4: If the condition in Step 3 is true, append `i` to `ans`. Break the inner loop because we've found the minimum solution.
5
Step 5: If the inner loop completes without finding a solution (Step 3 is never true), append -1 to `ans`.
6
Step 6: Return the `ans` array.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The outer loop iterates `n` times. The inner loop iterates up to `nums[i]` times for each element. The space used is proportional to the size of the output array `ans`, which is `n`.

📖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

Optimized Brute Force (Potentially)

Instead of iterating up to `nums[i] -1`, we could potentially stop earlier under certain conditions. For instance, if we find a solution `k`, any larger `k'` where `k' > k` and `k' | (k' + 1) == nums[i]` could also work. But it wouldn't give us the *minimum* value. Therefore there's not significant benefit.

Time: O(n*m)Space: O(n)

Trade-offs: This potential optimization would likely not provide substantial improvement in runtime due to the nature of the problem and the constraint of finding the minimum `ans[i]`.

✨

Key Insights

1Insight 1: The bitwise OR operation (`|`) combines bits. If `x | (x + 1) == y`, then `y` must contain all the bits that are set in either `x` or `x + 1`. This means that `x` must be cleverly chosen to satisfy the equation. A simple example: 1|2 = 3.
2Insight 2: Since the numbers in `nums` are prime, it's unlikely a simple formula will directly compute `ans[i]`. A brute force search checking values from 0 up to `nums[i]-1` is a straightforward approach.
3Insight 3: Optimizations are limited because of the requirement to find the *minimum* value of `ans[i]`. While a clever bit manipulation approach might exist for specific patterns in prime numbers, it's not generally applicable, and brute force is relatively efficient for the problem constraints (small input size).

📚

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: If `nums[i]` is a number such that no value of `ans[i]` will satisfy the condition (e.g., if it's an even number greater than 2 or it violates other conditions), the code correctly sets `ans[i]` to -1.
Edge case 2: If `nums[i]` is 3, the algorithm correctly identifies `ans[i]` as 1 because 1 | (1 + 1) = 1 | 2 = 3

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly implementing the bitwise OR operation or making a logic error in the conditional statement checking the bitwise OR result against `nums[i]`.

✗Common mistake 2: Forgetting to handle the case where no solution exists for a given `nums[i]` and failing to set `ans[i]` to -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: This type of bit manipulation problem could be relevant in areas like digital circuit design, where you might need to find combinations of bit patterns that produce a desired output.

🏦Real-world scenario 2: Low-level programming tasks involving bitwise operations often require similar logical thinking and analysis to find solutions that satisfy particular bit patterns. This problem helps improve those analytical skills.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ 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 14, 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 should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Construct the Minimum Bitwise Array I"?

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

🤖 Need help with this problem?