# Single Number II

## June 22, 2020

## Introduction

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

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Example 1:

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

Example 2:

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

## Solution

Very simple solution would be to cache all counters.

```
func singleNumber(nums []int) int {
cache := make(map[int]int)
for _, num := range nums {
cache[num]++
}
for k, v := range cache {
if v == 1 {
return k
}
}
return -1
}
```

Performance of this solution is:

```
Runtime: 4 ms
Memory Usage: 4.4 MB
```

But let’s solve it differently, we already know that all numbers have 3 occurrences, therefore if we summarize distinct numbers and multiply them by 3, we can have ideal sum, and we also can have real sum.

```
func singleNumber(nums []int) int {
visited := make(map[int]bool)
realSum := 0
for _, num := range nums {
realSum += num
visited[num] = true
}
idealSum := 0
for num := range visited {
idealSum += 3 * num
}
return (idealSum - realSum) / 2
}
```

Performance of this solution is:

```
Runtime: 4 ms
Memory Usage: 4.1 MB
```