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

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

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

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

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

Insight 1: The core of the problem lies in understanding the bitwise OR operation and how adding 1 affects the bits of a number. Insight 2: Analyzing the bit patterns of prime numbers and their relationship to the possible values of 'ans' is crucial for finding an efficient solution.

Which companies ask LeetCode 3315. Construct the Minimum Bitwise Array II in interviews?

This problem is frequently asked at Aon.

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Construct the Minimum Bitwise Array II - LeetCode 3315 Solution

Construct the Minimum Bitwise Array II - Complete Solution Guide

Construct the Minimum Bitwise Array II is LeetCode problem 3315, 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 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 us to construct an array `ans` of the same length as the input array `nums`. For each index `i`, the bitwise OR of `ans[i]` and `ans[i] + 1` must be equal to `nums[i]`. We need to find the *smallest* possible value for `ans[i]` that satisfies this condition. If no such value exists, we set `ans[i]` to -1. The input array `nums` contains prime numbers. The core task is, for each `num` in `nums`, find the smallest `ans` such that `ans | (ans + 1) == num`, or determine if no such `ans` exists.

Solution Approach

The solution iterates through the `nums` array and, for each number, attempts to find the smallest `ans` that satisfies the condition `ans | (ans + 1) == num`. The approach involves checking for special cases like num=2 and (num & 3) == 1, and then using bit manipulation to construct the `ans` when it does not fall into the special cases. The bit manipulation part figures out how many bits can be set when you add 1, then sets the highest bits to match num and fills out the rest with low bits, as well.

Step-by-Step Algorithm

Step 1: Initialize an empty array `ans` to store the results.
Step 2: Iterate through the `nums` array.
Step 3: For each `num` in `nums`, check if `num` is equal to 2. If it is, append -1 to `ans` and continue to the next iteration.
Step 4: If (num & 3) == 1 then simply ans.append(num - 1).
Step 5: If `num` is not 2, calculate `v = num + 1`.
Step 6: Find the position of the least significant bit of `v` using `k = (v & -v).bit_length() - 1` (or equivalent bit manipulation). This represents the power of 2 that determines the range of bits that will flip when 1 is added to `ans`.
Step 7: Create a mask `mask = (1 << k) - 1` to isolate the lower `k` bits.
Step 8: Calculate the high bits of `num` by clearing the lower `k` bits using `high_bits = num & ~mask`.
Step 9: Calculate the low bits of `ans` by setting the lower `k-1` bits to 1: `low_bits = (1 << (k - 1)) - 1`.
Step 10: Combine the high bits and low bits to get `ans[i] = high_bits | low_bits`.
Step 11: Append `ans[i]` to the `ans` array.
Step 12: After iterating through all numbers in `nums`, return the `ans` array.

Key Insights

Insight 1: The core of the problem lies in understanding the bitwise OR operation and how adding 1 affects the bits of a number.
Insight 2: Analyzing the bit patterns of prime numbers and their relationship to the possible values of 'ans' is crucial for finding an efficient solution.
Insight 3: The case where nums[i] is 2 is a special case that requires explicit handling, as no 'ans' value exists that satisfies the given condition. Similarly, (num & 3) == 1 handles the nums = 3, 5, 7 cases effectively using num - 1.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Bit Manipulation.

Companies

Asked at: Aon.

Construct the Minimum Bitwise Array II | LeetCode 3315

Medium

Array Bit Manipulation

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

Solution Code

Available in:

import java.util.List;
import java.util.ArrayList;

class Solution {
    public int[] minBitwiseArray(List<Integer> nums) {
        int[] ans = new int[nums.size()];
        for (int i = 0; i < nums.size(); i++) {
            int num = nums.get(i);
            if (num == 2) {
                ans[i] = -1;
            } else if ((num & 3) == 1) {
                ans[i] = num - 1;
            } else {
                int v = num + 1;
                int k = Integer.toBinaryString(v & -v).length() - 1;
                int mask = (1 << k) - 1;
                int high_bits = num & ~mask;
                int low_bits = (1 << (k - 1)) - 1;
                ans[i] = high_bits | low_bits;
            }
        }
        return ans;
    }
}

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Problem Info

Description

Problem 3315: Construct the Minimum Bitwise Array II - Detailed explanation and solution approaches.

Topics

Array Bit Manipulation

Companies

Aon

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3315

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: Initialize an empty array `ans` to store the results.
2
Step 2: Iterate through the `nums` array.
3
Step 3: For each `num` in `nums`, check if `num` is equal to 2. If it is, append -1 to `ans` and continue to the next iteration.
4
Step 4: If (num & 3) == 1 then simply ans.append(num - 1).
5
Step 5: If `num` is not 2, calculate `v = num + 1`.
6
Step 6: Find the position of the least significant bit of `v` using `k = (v & -v).bit_length() - 1` (or equivalent bit manipulation). This represents the power of 2 that determines the range of bits that will flip when 1 is added to `ans`.
7
Step 7: Create a mask `mask = (1 << k) - 1` to isolate the lower `k` bits.
8
Step 8: Calculate the high bits of `num` by clearing the lower `k` bits using `high_bits = num & ~mask`.
9
Step 9: Calculate the low bits of `ans` by setting the lower `k-1` bits to 1: `low_bits = (1 << (k - 1)) - 1`.
10
Step 10: Combine the high bits and low bits to get `ans[i] = high_bits | low_bits`.
11
Step 11: Append `ans[i]` to the `ans` array.
12
Step 12: After iterating through all numbers in `nums`, 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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is linear because we iterate through the input array `nums` once. Within each iteration, the bitwise operations and calculations take constant time, independent of the input size. The space complexity is linear because we create a new array `ans` to store the results, which has the same size as the input array `nums`.

📖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 Search

For each number `num` in `nums`, iterate through possible values of `ans` starting from 0 until a value is found that satisfies `ans | (ans + 1) == num`. This would involve a while loop inside the main for loop.

Time: O(n * m), where n is the length of nums, and m is the average number of iterations needed to find the ans for each num. In the worst case, m could be close to num itself.Space: O(n), similar to the main solution

Trade-offs: The brute force approach is much slower than the bit manipulation solution, especially for larger numbers. It's easier to understand initially but significantly less efficient. The bit manipulation approach is more complex but offers a substantial performance improvement.

✨

Key Insights

1Insight 1: The core of the problem lies in understanding the bitwise OR operation and how adding 1 affects the bits of a number.
2Insight 2: Analyzing the bit patterns of prime numbers and their relationship to the possible values of 'ans' is crucial for finding an efficient solution.
3Insight 3: The case where nums[i] is 2 is a special case that requires explicit handling, as no 'ans' value exists that satisfies the given condition. Similarly, (num & 3) == 1 handles the nums = 3, 5, 7 cases effectively using num - 1.

📚

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[i] = 2`. In this case, there is no value for `ans[i]` that satisfies the condition. The solution correctly handles this by setting `ans[i]` to -1.
Edge case 2: Prime numbers that fit (num & 3) == 1. nums = [3, 5, 7] all get solved using num - 1. Ex: 3 & 3 = 3; 5 & 3 = 1; 7 & 3 = 3.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where `nums[i] = 2`. Forgetting this leads to incorrect results.

✗Common mistake 2: Incorrectly implementing the bit manipulation logic. The position of the least significant bit and the correct high and low bits need to be calculated precisely to arrive at the minimal `ans` value. Failing to do this leads to incorrect and larger `ans`.

✅

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 memory usage where certain bit patterns must be achieved. This problem explores how specific binary patterns can be obtained by minimal adjustments to a base value, which is important in resource-constrained environments.

🏦Real-world scenario 2: Networking protocols where specific control bits need to be set or modified efficiently. Understanding bitwise operations is fundamental for manipulating network packets and control signals.

💡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.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 15, 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

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

@DevExpert92Jul 13

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

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 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.

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 "Construct the Minimum Bitwise Array II"?

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

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