Reference: LeetCode

Difficulty: Medium

## Problem

Given an integer array of size

`n`

, find all elements that appear more than`⌊n/3⌋`

times.

**Note:** The algorithm should run in linear time and in $O(1)$ space.

**Example:**

1 | Input: [3,2,3] |

1 | Input: [1,1,1,3,3,2,2,2] |

## Analysis

**Methods:**

HashMap

**Time:**$O(N)$

**Space:**$O(N)$Boyer-Moore Majority Vote Algorithm

Reference: My Understanding of Boyer-Moore Majority Vote

Let’s first go back to `169. Majority Element`

and try to think about the idea of `pairing`

, which happens when `count -= 1`

.

Suppose we have `9`

items in an array:

No matter in what order we select elements from the array, we can only get two results: `Fully Paired`

and `Partially Paired`

.

1 | public int majorityElement(int[] nums) { |

So here comes the `n/3`

problem, we would only think about the fully pairing situation. If the over one third majority exists, it should be left after pairing.

And why would we use three elements as a pair? Because it makes sure that in fully pairing the count of the majority element equals `n/3`

.

**More Examples:**

1 | // [ 1 1 1 3 3 2 2 2 ] |

**Note:**

- The ordering of
`if`

is very important. Though there are many branches, but each time only one could be executed. - It is okay to initialize candidates with arbitrary numbers.

1 | public List<Integer> majorityElement(int[] nums) { |

**Time:** $O(N)$**Space:** $O(1)$