LeetCode 1287: Element Appearing More Than 25% In Sorted Array Solution

Master LeetCode problem 1287 (Element Appearing More Than 25% In Sorted Array), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1287. Element Appearing More Than 25% In Sorted Array

Easy

Array

Problem Explanation

Explanation

To solve this problem, we can iterate through the array and keep track of the count of each element. If the count of any element becomes greater than 25% of the array size, we return that element as it is the desired integer that occurs more than 25% of the time.

Algorithm:

Initialize a variable threshold as arr.length / 4.
Iterate through the array and count the occurrences of each element.
If the count of any element exceeds the threshold, return that element.

Time Complexity:

The time complexity of this algorithm is O(n), where n is the size of the input array.

Space Complexity:

The space complexity of this algorithm is O(n) to store the counts of each element.

Solution Code

class Solution {
    public int findSpecialInteger(int[] arr) {
        int threshold = arr.length / 4;
        Map<Integer, Integer> countMap = new HashMap<>();
        
        for (int num : arr) {
            countMap.put(num, countMap.getOrDefault(num, 0) + 1);
            if (countMap.get(num) > threshold) {
                return num;
            }
        }
        
        return -1; // No element found
    }
}

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What is the time complexity of LeetCode 1287 (Element Appearing More Than 25% In Sorted Array)?

The time complexity for LeetCode 1287 (Element Appearing More Than 25% In Sorted Array) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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