Leetcode 229

This solution is pretty simple:

We loop through nums
We add 1 to a map of integers called m, so we add 1 to m[num] where num is the current number in nums.
Then we check whether m[num] == len(nums) / 3 + 1 this is checking whether the element is a majority element.

Note: We can’t do m[num] >= len(nums) / 3 + 1. This can be shown with an example:

If we have the example nums = [1, 2, 1, 1, 1], and we have the expected output of res = [1], it won’t work because we can:

First add one to m[1] so m = [1 : 1]
Then add one to m[2] so m = [1 : 1, 2 : 1]
Then add one to m[1] so m = [1 : 2, 2 : 1], since 5 / 3 == 1, m[1] = 2, and 2 > 1. We can append 1 to res, so res = [1].
Then add one to m[1] so m = [1 : 3, 2 : 1], since 5 / 3 == 1, m[1] = 3, and 3 > 1. We can append 1 to res, so res = [1, 1].
Then add one to m[1] so m = [1 : 4, 2 : 1], since 5 / 3 == 1, m[1] = 4, and 4 > 1. We can append 1 to res, so res = [1, 1, 1].

Since res = [1, 1, 1] we can see how it is not the expected output of res = [1].

The Code:

func majorityElement(nums []int) []int {
    m := make(map[int] int)
    res := []int{}

    for _, num := range nums {
        m[num]++
        if m[num] == len(nums)/3 + 1 {
            res = append(res, num)
        }
    }
    return res
}