# Single Number II

## Problem Statement

Given a **non-empty** array of integers, every element appears *three* times except for one, which appears exactly once. Find that single one.

{% hint style="info" %}
Note:

Your algorithm should have a linear runtime complexity.
{% endhint %}

## Testcases

```
Example 1:

Input: [2,2,3,2]
Output: 3

Example 2:

Input: [0,1,0,1,0,1,99]
Output: 99
```

## Algorithm

### Iterative Mapping Approach

1. Initialize a map to store the value of the element and the number of occurrences of the element in the array as a key-value pair. 
2. Iterate through the array of elements and store the count of each element in the array. 
3. Find the element in the array which has `count = 1` and return the value of that element. 

{% hint style="info" %}
Time Complexity: O(n)
{% endhint %}

## Code

```cpp
int singleNumber(vector<int>& nums) {
    unordered_map<int, int> m;
    // Iterate through the vector to store the count of each element in a map
    for(int i = 0; i < nums.size(); i++) {
        if(m.find(nums[i]) != m.end()) {
            m[nums[i]]++;
        }
        else {
            m[nums[i]] = 1;
        }
    }
    // Find the element which occurs only once
    for(int i = 0; i < nums.size(); i++) {
        if(m[nums[i]] == 1) {
            // Return the value of the element
            return nums[i];
        }
    }
    return -1;
}
```


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