g0101_0200.s0169_majority_element.Solution Maven / Gradle / Ivy

Go to download

Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java21 Show documentation

Java-based LeetCode algorithm problem solutions, regularly updated

There is a newer version: 1.38

package g0101_0200.s0169_majority_element;

// #Easy #Top_100_Liked_Questions #Top_Interview_Questions #Array #Hash_Table #Sorting #Counting
// #Divide_and_Conquer #Data_Structure_II_Day_1_Array #Udemy_Famous_Algorithm
// #Big_O_Time_O(n)_Space_O(1) #2022_06_25_Time_1_ms_(100.00%)_Space_45.5_MB_(97.51%)

/**
 * 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?
**/
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;
        }
    }
}