Single Number II - Practice Coding Problems

Single Number II - Problem

Imagine you're analyzing data from a survey where almost every response appears exactly three times in your dataset, but there's one special response that appears only once. Your task is to efficiently identify this unique response.

Given an integer array nums where every element appears three times except for one element which appears exactly once, find and return the single element.

Challenge: Your solution must run in O(n) time and use only O(1) extra space - no sorting allowed!

Example: In array [2,2,3,2], the number 3 appears once while 2 appears three times, so return 3.

Input & Output

example_1.py — Basic Case

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

› Output: 3

💡 Note: Element 2 appears 3 times, element 3 appears 1 time. The single number is 3.

example_2.py — Negative Numbers

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

› Output: 99

💡 Note: Elements 0 and 1 each appear 3 times, element 99 appears 1 time. The single number is 99.

example_3.py — Single Element Array

$ Input: [1]

› Output: 1

💡 Note: Only one element in the array, so it's the single number by definition.

Visualization

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Understanding the Visualization

Initialize Counters

Set up two binary counters: 'ones' for first occurrence, 'twos' for second occurrence

Process Each Number

For each number, update both counters using bit manipulation rules

Auto-Reset on Triplets

When a bit appears 3 times, both counters automatically reset that bit to 0

Extract Result

The 'ones' counter contains only bits from the single number

Key Takeaway

🎯 Key Insight: Use mathematical properties of XOR and bit tracking to eliminate elements that appear in multiples of 3, leaving only the single element

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass to build hash map + single pass to find answer

✓ Linear Growth

Space Complexity

O(n)

Hash map stores at most n/3 unique elements in worst case

⚡ Linearithmic Space

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

Asked in

G Google 45 ⊞ Microsoft 38 a Amazon 32 Apple 28

The optimal solution uses bit manipulation with two variables (ones and twos) to track bit occurrences. When any bit appears 3 times, it gets eliminated from both trackers, leaving only the single number's bits in O(n) time and O(1) space.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Nested Loops)	O(n²)	O(1)	For each element, count its occurrences in the entire array
Hash Table Counting	O(n)	O(n)	Count frequencies using hash map, then find element with count 1
Bit Manipulation (Optimal)	O(n)	O(1)	Use bit manipulation to track occurrences and cancel out triplets

Brute Force (Nested Loops) — Algorithm Steps

Iterate through each element in the array
For current element, scan entire array to count occurrences
If count equals 1, return this element
Continue until single element is found

Visualization

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Step-by-Step Walkthrough

Pick First Element

Select nums[0] and prepare to count

Count Occurrences

Scan entire array counting how many times nums[0] appears

Check Count

If count = 1, found answer. Otherwise, move to next element

Repeat Process

Continue with next elements until single number found

Code -

solution.c — C

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

int singleNumber(int* nums, int numsSize) {
    for (int i = 0; i < numsSize; i++) {
        int count = 0;
        for (int j = 0; j < numsSize; j++) {
            if (nums[j] == nums[i]) {
                count++;
            }
        }
        if (count == 1) {
            return nums[i];
        }
    }
    return -1;
}

int main() {
    char input[1000];
    fgets(input, sizeof(input), stdin);
    int nums[100];
    int count = 0;
    char* token = strtok(input, "[,]");
    while (token != NULL) {
        nums[count++] = atoi(token);
        token = strtok(NULL, "[,]");
    }
    printf("%d\n", singleNumber(nums, count));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n elements, we scan through n elements to count occurrences

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables for counting, no extra data structures

✓ Linear Space

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

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Nested Loops) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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