How do you solve Element Appearing More Than 25% In Sorted Array on LeetCode?

Element Appearing More Than 25% In Sorted Array (LeetCode #1287) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The array is sorted, allowing for efficient checks at specific intervals.

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

The optimal solution for Element Appearing More Than 25% In Sorted Array runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only traverse the array a constant number of times based on the threshold, making it efficient.

What is the brute force solution for Element Appearing More Than 25% In Sorted Array?

The brute force approach for Element Appearing More Than 25% In Sorted Array is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will check each element in the array and count how many times it appears. If it appears more than 25% of the total elements, we return that element. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Element Appearing More Than 25% In Sorted Array?

Element Appearing More Than 25% In Sorted Array uses the following concepts: Array. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Element Appearing More Than 25% In Sorted Array?

Always consider the properties of the input data (sorted, unique, etc.). Think about how you can reduce the number of checks needed. Practice identifying patterns in problems to apply efficient algorithms.

What are common mistakes when solving Element Appearing More Than 25% In Sorted Array?

Not leveraging the sorted property of the array. Counting occurrences in a nested loop instead of checking specific intervals.

#1287

Element Appearing More Than 25% In Sorted Array

Easy

ArrayHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

Since the array is sorted, we can leverage this property to find the element that appears more than 25% of the time without counting every occurrence. We can check specific intervals based on the sorted nature of the array.

⚙️

Algorithm

4 steps

1Step 1: Calculate the length of the array.
2Step 2: Determine the threshold for 25% of the array length.
3Step 3: Check the elements at the start, 25%, 50%, and 75% positions of the array.
4Step 4: For each of these positions, verify if the element appears more than the threshold in the surrounding elements.

solution.py6 lines

1def findSpecialInteger(arr):
2    n = len(arr)
3    threshold = n // 4
4    for i in range(0, n, threshold):
5        if i + threshold < n and arr[i] == arr[i + threshold]:
6            return arr[i]

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array a constant number of times based on the threshold, making it efficient.

1The array is sorted, allowing for efficient checks at specific intervals.
2The problem guarantees one element appears more than 25%, simplifying our search.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.