# 1287. Element Appearing More Than 25% In Sorted Array

## Description

Given an integer array sorted in non-decreasing order, there is exactly one integer in the array that occurs more than 25% of the time, return that integer.

Example 1:

Input: arr = [1,2,2,6,6,6,6,7,10]
Output: 6


Example 2:

Input: arr = [1,1]
Output: 1


Constraints:

• 1 <= arr.length <= 104
• 0 <= arr[i] <= 105

## Solutions

• class Solution {
public int findSpecialInteger(int[] arr) {
int n = arr.length;
for (int i = 0; i < n; ++i) {
if (arr[i] == arr[i + (n >> 2)]) {
return arr[i];
}
}
return 0;
}
}

• class Solution {
public:
int findSpecialInteger(vector<int>& arr) {
int n = arr.size();
for (int i = 0; i < n; ++i)
if (arr[i] == arr[i + (n >> 2)]) return arr[i];
return 0;
}
};

• class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
n = len(arr)
for i, val in enumerate(arr):
if val == arr[i + (n >> 2)]:
return val
return 0


• func findSpecialInteger(arr []int) int {
n := len(arr)
for i, val := range arr {
if val == arr[i+(n>>2)] {
return val
}
}
return 0
}

• /**
* @param {number[]} arr
* @return {number}
*/
var findSpecialInteger = function (arr) {
const n = arr.length;
for (let i = 0; i < n; ++i) {
if (arr[i] == arr[i + (n >> 2)]) {
return arr[i];
}
}
return 0;
};


• class Solution {
/**
* @param Integer[] $arr * @return Integer */ function findSpecialInteger($arr) {
$len = count($arr);
for ($i = 0;$i < $len;$i++) {
if ($arr[$i] == $arr[$i + ($len >> 2)]) { return$arr[\$i];
}
}
return -1;
}
}