Majority Element

Given an array nums of size n, return the majority element.

The majority element is the element that appears more than ⌊n / 2⌋ times. You may assume that the majority element always exists in the array.

O(N^2) Brute force

• For every number in array we count its occurrences.
• If count is greater than n/2, then we return the number.

O(N)Time with extra space

• we Can take a vector or unordered_map to store the frequency of each element.
• We traverse the hash map/ vector to find the n/2 frequency.
• if we found we return the element.
• SC: O(N)
• TC:O(N)/O(NlogN) - Yes if we use unordered_map it’s worst case time complexity is O(NlogN), which occurs when all elements are divisible by prime number and result in collision. But if we use frequency vector it’s worst case time complexity is O(N).

Code

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        unordered_map<int,int> mp;
        for(int x: nums) mp[x]++;
        int n = nums.size()/2;
        int ans = 0;
        for(auto x:mp){
            if(x.second>n){
                ans = x.first;
                return ans;
            }
        }
        return ans;
    }
};

Moore’s Voting Algorithm

• TC:O(N)
• SC:O(1)
• We maintain 2 variables count and candidate.
• We traverse the array and for every element we do following:
  • If count is 0, then we set candidate as current element.
  • If current element is same as candidate, then we increment count by 1.
  • else we decrement count by 1.
• run loop to check the frequency of candidate i strictly greater than N/2 or not.
• if no return -1, else return candidate.

class Solution {
public:
    // moore's voting algorithm
    int majorityElement(vector<int>& nums)
    {
        int count = 0;
        int candidate = 0;
        for (int x : nums) {
            if (count == 0) {
                candidate = x;
            }
            // count += (x == candidate) ? 1 : -1;
            if (x == candidate) count++;
            else count--;
        }
        return candidate;
    }
};