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: Traversal

We traverse the array $\textit{arr}$ from the beginning. For each element $\textit{arr}[i]$, we check if $\textit{arr}[i]$ is equal to $\textit{arr}[i + \left\lfloor \frac{n}{4} \right\rfloor]$, where $n$ is the length of the array. If they are equal, then $\textit{arr}[i]$ is the element we are looking for, and we return it directly.

The time complexity is $O(n)$, where $n$ is the length of the array $\textit{arr}$. The space complexity is $O(1)$.

Python3

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

Java

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

C++

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

Go

func findSpecialInteger(arr []int) int {
	for i := 0; ; i++ {
		if arr[i] == arr[i+len(arr)/4] {
			return arr[i]
		}
	}
}

TypeScript

function findSpecialInteger(arr: number[]): number {
    const n = arr.length;
    for (let i = 0; ; ++i) {
        if (arr[i] === arr[i + (n >> 2)]) {
            return arr[i];
        }
    }
}

JavaScript

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

PHP

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