Welcome to Subscribe On Youtube
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 <= 1040 <= 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; } } -
function findSpecialInteger(arr: number[]): number { const n = arr.length; for (let i = 0; ; ++i) { if (arr[i] === arr[i + (n >> 2)]) { return arr[i]; } } }