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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 \(O(1)\).

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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

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

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

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

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

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/**
 * @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];
        }
    }
};

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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];
            }
        }
    }
}

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