# Easy: Single Number

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Section titled Problem%20Statement

## Problem Statement

In a non-empty array of integers, every number appears twice except for one, find that single number.

Example 1:

Input: 1, 4, 6, 1, 3, 6, 3
Output: 4


Example 2:

Input: 1, 9, 1
Output: 9

Section titled Solution

## Solution

The simple solution is using a hashmap and iterate through the input:

• if number already in the hashmap, remove it
• if number is not present in hashmap, add it
• in the end, only left 1 data in the hashmap

Time and space Complexity: Time Complexity of the above solution will be $O(n)$ and space complexity will also be $O(n)$.

Section titled Can%20we%20do%20better%20than%20this%20using%20the%20XOR%20Pattern%3F

### Can we do better than this using the XOR Pattern?

Recall the following two properties of XOR:

• It returns zero if we take XOR of two same numbers.
• It returns the same number if we XOR with zero. So we can XOR all the numbers in the input; duplicate numbers will zero out each other and we will be left with the single number.

Try it yourself https://go.dev/play/p/XcOwP-nIMDO

Section titled Code

## Code

package main

import (
"testing"
)

func FindSingleNumber(list []int) int {
num := 0
for i := range list {
num ^= list[i]
}

return num
}

func TestLastIndex(t *testing.T) {
tests := []struct {
list []int
want int
}{
{list: []int{1, 4, 2, 1, 3, 2, 3}, want: 4},
{list: []int{1, 9, 1}, want: 9},
}
for _, tt := range tests {
if got := FindSingleNumber(tt.list); got != tt.want {
t.Errorf("LastIndex(%v) = %v, want %v", tt.list, got, tt.want)
}
}
}


Time Complexity: Time complexity of this solution is $O(n)$ as we iterate through all numbers of the input once.

Space Complexity: The algorithm runs in constant space $O(1)$.

grokking algorithm go pattern xor