LeetCode-in-All

169. Majority Element

Easy

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.

Example 1:

Input: nums = [3,2,3]

Output: 3

Example 2:

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

Output: 2

Constraints:

• n == nums.length
• 1 <= n <= 5 * 104
• -231 <= nums[i] <= 231 - 1

Follow-up: Could you solve the problem in linear time and in O(1) space?

Solution

public class Solution {
    public int majorityElement(int[] arr) {
        int count = 1;
        int majority = arr[0];
        // For Potential Majority Element
        for (int i = 1; i < arr.length; i++) {
            if (arr[i] == majority) {
                count++;
            } else {
                if (count > 1) {
                    count--;
                } else {
                    majority = arr[i];
                }
            }
        }
        // For Confirmation
        count = 0;
        for (int j : arr) {
            if (j == majority) {
                count++;
            }
        }
        if (count >= (arr.length / 2) + 1) {
            return majority;
        } else {
            return -1;
        }
    }
}

Time Complexity (Big O Time):

The time complexity of this program is O(n), where n is the length of the input array arr. Here’s why:

1. The first loop iterates through the entire array once (for (int i = 1; i < arr.length; i++)).
   • Inside this loop, there are constant-time operations (assignments and comparisons) for each element of the array.
   • Therefore, this part has a time complexity of O(n).
2. The second loop also iterates through the entire array once (for (int j : arr)).
   • Inside this loop, there are constant-time operations (comparisons) for each element of the array.
   • Therefore, this part also has a time complexity of O(n).

In summary, both loops together contribute to the linear time complexity O(n).

Space Complexity (Big O Space):

The space complexity of this program is O(1), which means it uses a constant amount of extra space. Regardless of the size of the input array, the program only uses a fixed number of variables (count and majority) to keep track of the majority element.

In summary, the time complexity is O(n), where n is the length of the input array, and the space complexity is O(1), indicating a constant amount of additional memory usage.

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