README_EN.md

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

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Description

Given an array nums sorted in non-decreasing order, and a number target, return True if and only if target is a majority element.

A majority element is an element that appears more than N/2 times in an array of length N.

 

Example 1:

Input: nums = [2,4,5,5,5,5,5,6,6], target = 5
Output: true
Explanation: 
The value 5 appears 5 times and the length of the array is 9.
Thus, 5 is a majority element because 5 > 9/2 is true.

Example 2:

Input: nums = [10,100,101,101], target = 101
Output: false
Explanation: 
The value 101 appears 2 times and the length of the array is 4.
Thus, 101 is not a majority element because 2 > 4/2 is false.

 

Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 10^9
• 1 <= target <= 10^9

Solutions

Python3

class Solution:
    def isMajorityElement(self, nums: List[int], target: int) -> bool:
        def bsearch_left(nums, target, left, right):
            while left < right:
                mid = (left + right) >> 1
                if nums[mid] >= target:
                    right = mid
                else:
                    left = mid + 1
            return left if nums[left] == target else -1

        def bsearch_right(nums, target, left, right):
            while left < right:
                mid = (left + right + 1) >> 1
                if nums[mid] <= target:
                    left = mid
                else:
                    right = mid - 1
            return left if nums[left] == target else -1

        n = len(nums)
        left = bsearch_left(nums, target, 0, n - 1)
        if left == -1:
            return False
        right = bsearch_right(nums, target, left, n - 1)
        if right == -1:
            return False
        return right - left + 1 > n >> 1

class Solution:
    def isMajorityElement(self, nums: List[int], target: int) -> bool:
        left, right = bisect.bisect_left(nums, target), bisect.bisect_right(nums, target)
        return right - left > (len(nums) >> 1)

Java

class Solution {
    public boolean isMajorityElement(int[] nums, int target) {
        int n = nums.length;
        int left = bsearchLeft(nums, target, 0, n - 1);
        if (left == -1) {
            return false;
        }
        int right = bsearchRight(nums, target, left, n - 1);
        if (right == -1) {
            return false;
        }
        return right - left + 1 > (n >> 1);
    }

    private int bsearchLeft(int[] nums, int target, int left, int right) {
        while (left < right) {
            int mid = (left + right) >> 1;
            if (nums[mid] >= target) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return nums[left] == target ? left : -1;
    }

    private int bsearchRight(int[] nums, int target, int left, int right) {
        while (left < right) {
            int mid = (left + right + 1) >> 1;
            if (nums[mid] <= target) {
                left = mid;
            } else {
                right = mid - 1;
            }
        }
        return nums[left] == target ? left : -1;
    }
}

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