# Leetcode: Check If a Number Is Majority Element in a Sorted Array

Check If a Number Is Majority Element in a Sorted Array

Similar Problems:

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.
```

Note:

1. 1 <= nums.length <= 1000
2. 1 <= nums[i] <= 10^9
3. 1 <= target <= 10^9

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

• Solution:
```// Blog link: https://code.dennyzhang.com/is-a-a-majority-element
// Basic Ideas: count
// Complexity: Time O(n), Space O(1)
func isMajorityElement(nums []int, target int) bool {
cnt := len(nums)/2+1
for _, num := range nums {
if num == target { cnt-- }
if cnt == 0 { return true }
}
return false
}
```

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