169. Majority Element -- Easy

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

 

M1: sort, time: O(nlogn), space: O(logn)

class Solution {
    public int majorityElement(int[] nums) {
        Arrays.sort(nums);
        return nums[nums.length / 2];
    }
}

M2: bit manipulation, time: O(n), space: O(1)

class Solution {
    public int majorityElement(int[] nums) {
        int res = 0;
        for(int i = 0; i < 32; i++) {
            int ones = 0, zeros = 0;
            for(int n : nums) {
                if(ones > nums.length / 2 || zeros > nums.length / 2) {
                    break;
                }
                if((n & (1 << i)) != 0) {
                    ones++;
                } else {
                    zeros++;
                }
            }
            if(ones > zeros) {
                res |= (1 << i);
            }
        }
        return res;
    }
}

M3: Boyer-Moore Voting Algorithm, time: O(n), space: O(1)

class Solution {
    public int majorityElement(int[] nums) {
        Integer candidate = null;
        int cnt = 0;
        for(int n : nums) {
            if(cnt == 0) {
                candidate = n;
            }
            cnt += (candidate == n) ? 1 : -1;
            
        }
        return candidate;
    }
}

M4: hash table, time: O(n), space: O(n)

class Solution {
    public int majorityElement(int[] nums) {
        Map<Integer, Integer> map = new HashMap<>();
        for(int n : nums) {
            map.put(n, map.getOrDefault(n, 0) + 1);
        }
        
        int res = 0;
        for(Map.Entry<Integer, Integer> entry : map.entrySet()) {
            if(entry.getValue() > nums.length / 2) {
                res = entry.getKey();
                break;
            }
        }
        return res;
    }
}