Majority Element - The Coding Shala

Home >> Interview Prep >> Majority Element

 In this post, we will learn how to find Majority Element in the array and will implement its solution in Java.

Majority Element Problem

Given the 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.

Example 1:
Input: nums = [3,2,3]
Output: 3

Example 2:
Input: nums = [2,2,1,1,1,2,2]
Output: 2

Majority Element Java Solution

Approach 1

First sort the array then find the frequency of elements, if the count is greater than n/2 return that.

Time Complexity: O(nlogn)  [sorting]

We can use HashMap to store the counts of elements that will take additional O(n) space.

Java Program: 

class Solution {
    public int majorityElement(int[] nums) {
        int check = nums.length/2;
        Arrays.sort(nums);
        int count = 1;
        for(int i=1; i<nums.length; i++) {
            if(nums[i] == nums[i-1]) count++;
            else count = 1;
            if(count > check) {
                return nums[i];
            }
        }
        return nums[0];
    }
}

Approach 2

Modification in Sorting approach.

We know that the Majority element comes more than n/2 time. After sorting the array the middle element is always the Majority element.

Java Program: 

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

Approach 3

Boyer-Moore Voting Algorithm. [Read Here]

Time Complexity: O(n)

Space Complexity: O(1)

Java Program: 

class Solution {
    public int majorityElement(int[] nums) {
        int count = 1;
        int majority = nums[0];
        for(int i=1; i<nums.length; i++) {
            if(count == 0) {
                majority = nums[i];
                count = 1;
            } else if(majority == nums[i]) count++;
            else count--;
        }
        return majority;
    }
}


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