LeetCode 1150: Check If a Number Is Majority Element in a Sorted Array Solution

Master LeetCode problem 1150 (Check If a Number Is Majority Element in a Sorted Array), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1150. Check If a Number Is Majority Element in a Sorted Array

Easy

Array Binary Search

Problem Explanation

Explanation

Given a sorted array of integers, we need to check if a specific number appears more than n/2 times, where n is the length of the array. We can use binary search to find the first and last occurrences of the number in the array. If the count of occurrences is greater than n/2, then the number is a majority element.

Algorithm:

Find the first occurrence of the number using binary search.
Find the last occurrence of the number using binary search.
Calculate the count of occurrences.
Check if the count is greater than n/2.

Time Complexity: O(log n) - Binary search Space Complexity: O(1)

Solution Code

class Solution {
    public boolean isMajorityElement(int[] nums, int target) {
        int first = findFirstOccurrence(nums, target);
        int last = findLastOccurrence(nums, target);
        
        int count = last - first + 1;
        int n = nums.length;
        
        return count > n / 2;
    }
    
    private int findFirstOccurrence(int[] nums, int target) {
        int low = 0, high = nums.length - 1;
        
        while (low < high) {
            int mid = low + (high - low) / 2;
            if (nums[mid] < target) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        
        return low;
    }
    
    private int findLastOccurrence(int[] nums, int target) {
        int low = 0, high = nums.length - 1;
        
        while (low < high) {
            int mid = low + (high - low + 1) / 2;
            if (nums[mid] > target) {
                high = mid - 1;
            } else {
                low = mid;
            }
        }
        
        return low;
    }
}

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