[LeetCode 169] Majority Element

Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.

You may assume that the array is non-empty and the majority element always exist in the array.

Example 1:

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

Example 2:

Input: [2,2,1,1,1,2,2]
Output: 2

 

 

The sorting solution or O(N) space solution using hash map is trivial. There is another O(N) algorithm called Boyer-Moore Majority Vote algorithm that only uses O(1) space. We need two passes over the input array. In the 1st pass, we generate a single candidate value which is the majority value if there is a majority. The 2nd pass simply counts the frequency of that value to confirm.

In the 1st pass, we need 2 values: a candidate value, initially set to any value; a count, initially set to 0. Then for each element, do the following.

1. if current element == candidate, count++;

2. else if current element != candidate: if count is 0, set current element as the new candidate and count to 1; if count is > 0, count--;

Proof of correctness:  

1. if we first pick a majority number as candidate, then say after consuming k numbers its count is decreased to 0, this means we've consumed k / 2 majority numbers and k / 2 non-majority numbers. The remaining array has > n / 2 - k / 2 majority numbers and < n - k - n / 2 + k / 2 = n / 2 - k / 2 non-majority numbers. The majority number is still the majority number. 

2. if we first pick a non-majority number C as candidate, then say after consuming k numbers its count is decreased to 0, we've consumed at most k / 2 majority numbers to de-throne C's candidancy. This means the remaining array has > n / 2 - k / 2 majority numbers and < n - k - n / 2 + k / 2 = n / 2 - k / 2 non-majority numbers. Exactly the same with case 1.

For a more detailed explanation, refer to Majority Voting Algorithm.

class Solution {
    public int majorityElement(int[] nums) {
        int candidate = nums[0], cnt = 1;
        for(int i = 1; i < nums.length; i++) {
            if(nums[i] == candidate) {
                cnt++;
            }
            else {
                if(cnt == 0) {
                    candidate = nums[i];
                    cnt = 1;
                }
                else {
                    cnt--;
                }
            }
        }
        // cnt = 0;
        // for(int i = 0; i < nums.length; i++) {
        //     if(nums[i] == candidate) {
        //         cnt++;
        //     }
        // }
        // return cnt > nums.length / 2 ? candidate : -1;
        return candidate;
    }
}

 

 

Related Problems

Majority Element II

Majority Element III

Single Number 

Single Number II

Single Number III