Single Number II

Single Number II - Problem

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 linear runtime complexity and use only constant extra space.

Input & Output

Example 1 — Basic Case

$ Input: nums = [2,2,3,2]

› Output: 3

💡 Note: Element 2 appears 3 times, element 3 appears 1 time. Return 3.

Example 2 — Larger Array

$ Input: nums = [0,1,0,1,0,1,99]

› Output: 99

💡 Note: Elements 0 and 1 each appear 3 times, element 99 appears 1 time. Return 99.

Example 3 — Single Element

$ Input: nums = [5]

› Output: 5

💡 Note: Only one element in array, it appears once. Return 5.

Constraints

1 ≤ nums.length ≤ 3 × 10⁴
-2³¹ ≤ nums[i] ≤ 2³¹ - 1
Each element in the array appears exactly three times except for one element which appears exactly once.

Visualization

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Asked in

G Google 25 M Microsoft 20 a Amazon 18 Apple 15

The key insight is to use bit manipulation to count occurrences modulo 3. The optimal digital counter approach uses two variables to track bits that appear once or twice, automatically resetting when they appear three times. Time: O(n), Space: O(1).

Common Approaches

✓ Bit Manipulation - Digital Circuit

⏱️ Time: O(n) Space: O(1)

Track each bit position across all numbers. Use two variables 'ones' and 'twos' to count bit occurrences. When a bit appears 3 times, reset counters.

Brute Force - Nested Loop

⏱️ Time: O(n²) Space: O(1)

Check every element against all other elements to count how many times it appears. The one that appears exactly once is our answer.

Hash Map - Frequency Count

⏱️ Time: O(n) Space: O(n)

Make two passes: first to build frequency map of all elements, second to find the element that appears exactly once.

Bit Manipulation - Digital Circuit — Algorithm Steps

Step 1: Initialize ones=0, twos=0
Step 2: For each number, update ones and twos
Step 3: Reset when bit count reaches 3
Step 4: Return ones (contains single number)

Visualization

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Step-by-Step Walkthrough

Initialize

ones=0, twos=0

Process Each Number

Update bit counters

Reset on 3

Clear bits that appear 3 times

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int numsSize) {
    int ones = 0;
    int twos = 0;
    
    for (int i = 0; i < numsSize; i++) {
        int num = nums[i];
        // Update twos: bits that appear twice
        twos |= ones & num;
        // Update ones: bits that appear once
        ones ^= num;
        // Clear bits that appear three times
        int common = ones & twos;
        ones &= ~common;
        twos &= ~common;
    }
    
    return ones;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[10000];
    int size = 0;
    
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != '\n') {
            nums[size++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    int result = solution(nums, size);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through array, constant work per element

✓ Linear Growth

Space Complexity

O(1)

Only using two integer variables regardless of input size

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Bit Manipulation - Digital Circuit — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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