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

Solution 1

Python Code
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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

Java Code
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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;
    }
}

C++ Code
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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;
    }
};

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

JavaScript Code
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/**
 * @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;
};

PHP Code
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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;
    }
}