1150. Check If a Number Is Majority Element in a Sorted Array π
Description
Given an integer array nums
sorted in non-decreasing order and an integer target
, return true
if target
is a majority element, or false
otherwise.
A majority element in an array nums
is an element that appears more than nums.length / 2
times in the array.
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], target <= 109
nums
is sorted in non-decreasing order.
Solutions
Solution 1: Binary Search
We notice that the elements in the array $nums$ are non-decreasing, that is, the elements in the array $nums$ are monotonically increasing. Therefore, we can use the method of binary search to find the index $left$ of the first element in the array $nums$ that is greater than or equal to $target$, and the index $right$ of the first element in the array $nums$ that is greater than $target$. If $right - left > \frac{n}{2}$, it means that the number of occurrences of the element $target$ in the array $nums$ exceeds half of the length of the array, so return $true$, otherwise return $false$.
The time complexity is $O(\log n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the array $nums$.
1 2 3 4 5 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 |
|
1 2 3 4 5 6 7 8 |
|
1 2 3 4 5 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 |
|
Solution 2: Binary Search (Optimized)
In Solution 1, we used binary search twice to find the index $left$ of the first element in the array $nums$ that is greater than or equal to $target$, and the index $right$ of the first element in the array $nums$ that is greater than $target$. However, we can use binary search once to find the index $left$ of the first element in the array $nums$ that is greater than or equal to $target$, and then judge whether $nums[left + \frac{n}{2}]$ is equal to $target$. If they are equal, it means that the number of occurrences of the element $target$ in the array $nums$ exceeds half of the length of the array, so return $true$, otherwise return $false$.
The time complexity is $O(\log n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the array $nums$.