What is the time complexity of Single Number II?

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

What data structures are used to solve Single Number 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 Single Number II?

Insight 1: Bit manipulation provides an efficient way to track the count of each bit without using extra space for a hash map or similar data structure. Insight 2: The key idea is to simulate a ternary (base-3) counter for each bit position. Since each number appears three times, the bits for those numbers will effectively cancel out when counted modulo 3, leaving only the bits of the single number.

Which companies ask LeetCode 137. Single Number II in interviews?

This problem is frequently asked at Accenture, Yahoo.

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Single Number II - LeetCode 137 Solution

Q: How to solve Single Number II?

The solution uses bit manipulation to keep track of the number of times each bit has appeared. The `ones` and `twos` variables are used to simulate a base-3 counter. For each number in the array, we update `ones` and `twos` based on the current number and their previous values. The final value of `ones` will hold the single number, because it represents the bits that have appeared only once (mod 3).

Single Number II - Complete Solution Guide

Single Number II is LeetCode problem 137, 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

Given an integer array nums where every element appears three times except for one, which appears exactly once . Find the single element and return it . You must implement a solution with a linear runtime complexity and use only constant extra space. Example 1: Input: nums = [2,2,3,2] Output: 3 Example 2: Input: nums = [0,1,0,1,0,1,99] Output: 99 Constraints: 1 <= nums.length <= 3 * 10 4 -2 31 <= nums[i] <= 2 31 - 1 Each element in nums appears exactly three times except for one element which ap

Detailed Explanation

The problem states that we are given an array of integers, `nums`, where every number appears exactly three times, except for one number which appears only once. The goal is to find and return that single number that occurs only once. The challenge is to achieve this with a linear time complexity (O(n)) and constant extra space (O(1)).

Solution Approach

The solution uses bit manipulation to keep track of the number of times each bit has appeared. The `ones` and `twos` variables are used to simulate a base-3 counter. For each number in the array, we update `ones` and `twos` based on the current number and their previous values. The final value of `ones` will hold the single number, because it represents the bits that have appeared only once (mod 3).

Step-by-Step Algorithm

Step 1: Initialize two integer variables, `ones` and `twos`, to 0. These will store the bits that have appeared once and twice, respectively.
Step 2: Iterate through each number `num` in the input array `nums`.
Step 3: Update `ones` as follows: `ones = (ones ^ num) & ~twos`. This XORs `ones` with the current number `num`. If a bit appears for the first time, it sets the corresponding bit in `ones`. The `& ~twos` part ensures that if a bit already exists in `twos` (meaning it has appeared twice), it's removed from `ones` to avoid counting it as a single occurrence.
Step 4: Update `twos` as follows: `twos = (twos ^ num) & ~ones`. This XORs `twos` with the current number `num`. If a bit appears for the second time, it sets the corresponding bit in `twos`. The `& ~ones` part ensures that if a bit already exists in `ones` (meaning it has appeared once), it's removed from `twos` to avoid double-counting.
Step 5: After iterating through all numbers, the value of `ones` will contain the single number that appears only once. Return `ones`.

Key Insights

Insight 1: Bit manipulation provides an efficient way to track the count of each bit without using extra space for a hash map or similar data structure.
Insight 2: The key idea is to simulate a ternary (base-3) counter for each bit position. Since each number appears three times, the bits for those numbers will effectively cancel out when counted modulo 3, leaving only the bits of the single number.
Insight 3: Using two variables, `ones` and `twos`, we can keep track of the count of each bit modulo 3. `ones` represents the bits that have appeared once, and `twos` represents the bits that have appeared twice.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Bit Manipulation.

Companies

Asked at: Accenture, Yahoo.

Single Number II | LeetCode 137

Medium

Array Bit Manipulation

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int singleNumber(int[] nums) {
        int ones = 0;
        int twos = 0;
        for (int num : nums) {
            ones = (ones ^ num) & ~twos;
            twos = (twos ^ num) & ~ones;
        }
        return ones;
    }
}

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Problem Info

Description

Problem 137: Single Number II - Detailed explanation and solution approaches.

Topics

Array Bit Manipulation

Companies

AccentureYahoo

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:137

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

5 steps

1
Step 1: Initialize two integer variables, `ones` and `twos`, to 0. These will store the bits that have appeared once and twice, respectively.
2
Step 2: Iterate through each number `num` in the input array `nums`.
3
Step 3: Update `ones` as follows: `ones = (ones ^ num) & ~twos`. This XORs `ones` with the current number `num`. If a bit appears for the first time, it sets the corresponding bit in `ones`. The `& ~twos` part ensures that if a bit already exists in `twos` (meaning it has appeared twice), it's removed from `ones` to avoid counting it as a single occurrence.
4
Step 4: Update `twos` as follows: `twos = (twos ^ num) & ~ones`. This XORs `twos` with the current number `num`. If a bit appears for the second time, it sets the corresponding bit in `twos`. The `& ~ones` part ensures that if a bit already exists in `ones` (meaning it has appeared once), it's removed from `twos` to avoid double-counting.
5
Step 5: After iterating through all numbers, the value of `ones` will contain the single number that appears only once. Return `ones`.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The solution iterates through the input array `nums` once. Therefore, the time complexity is linear, O(n), where n is the number of elements in `nums`. The solution uses only two integer variables, `ones` and `twos`, which occupy constant space. Thus, the space complexity is O(1).

📖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

Hash Map

Use a hash map to store the count of each number in the array. Iterate through the array and update the count for each number. Finally, iterate through the hash map and return the number with a count of 1.

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

Trade-offs: While it's a simpler approach to understand, it requires O(n) extra space, violating the constraint of constant extra space. Use this only if space complexity is not a concern.

Sorting

Sort the array. Then, iterate through the sorted array. If nums[i] != nums[i+1] and nums[i] != nums[i-1], nums[i] is the number we are looking for. If nums[i] == nums[i+1] == nums[i+2], we increment i by 3.

Time: O(n log n)Space: O(1) or O(n) depending on sorting algorithm

Trade-offs: Sorting has a higher time complexity than the bit manipulation solution, and also might need additional space depending on the sorting method used. This is not optimal.

✨

Key Insights

1Insight 1: Bit manipulation provides an efficient way to track the count of each bit without using extra space for a hash map or similar data structure.
2Insight 2: The key idea is to simulate a ternary (base-3) counter for each bit position. Since each number appears three times, the bits for those numbers will effectively cancel out when counted modulo 3, leaving only the bits of the single number.
3Insight 3: Using two variables, `ones` and `twos`, we can keep track of the count of each bit modulo 3. `ones` represents the bits that have appeared once, and `twos` represents the bits that have appeared twice.

📚

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: Input array is empty. While the constraints specify at least one element, it's good to consider. The code will return 0, which may or may not be correct based on how the problem is framed.
Edge case 2: The single number is negative. The bit manipulation operations work correctly with negative numbers due to how integers are represented in two's complement.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly implementing the bit manipulation logic, leading to incorrect updates of `ones` and `twos`. The order of operations and the `& ~` part are crucial.

✗Common mistake 2: Failing to handle the edge cases where the input array is empty or the single number is at the beginning or end of the array (if using sorting).

✗Common mistake 3: Using extra space by creating a hash map or similar data structure when the problem requires constant space.

✅

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: Error detection and correction codes. This bit manipulation technique is analogous to detecting single-bit errors in data transmission or storage. By representing data as bits and applying similar logic, errors can be identified and corrected.

🏦Real-world scenario 2: Data analysis and filtering. If you have a large dataset where most values appear a specific number of times (e.g., log entries), you can use this type of bit manipulation to identify unique or infrequent occurrences without requiring excessive memory.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: October 16, 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

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

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

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

@TechLeadJun 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 "Single Number 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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