Leetcode – Element Appearing More Than 25% In Sorted Array Solution

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The problem titled “Element Appearing More Than 25% In Sorted Array” on Leetcode poses an interesting challenge. It may initially seem simple due to the sorted nature of the array, but the constraints related to time complexity make it a noteworthy problem to explore in detail. In this guide, we will discuss various ways to solve this problem using Python.

Table of Contents

  1. Problem Description
  2. Naive Approach: Brute-force Algorithm
  3. Optimized Approach 1: Counting Sequences
  4. Testing and Validation
  5. Time and Space Complexity Analysis
  6. Conclusion

1. Problem Description

You’re given a sorted integer array, arr, where the condition is that an integer x will appear more than 25% of the time in the array. Your task is to find and return that integer x.

Constraints:

  • The array is non-empty and has a length in the range [1,10^4].
  • The array contains integers in the range [−10^5,10^5].

Example:

Input: arr = [1, 2, 2, 6, 6, 6, 6, 7, 10]

Output: 6

2. Naive Approach: Brute-force Algorithm

Algorithm:

  1. Initialize a counter variable to keep track of occurrences.
  2. Iterate through the sorted array and count consecutive similar numbers.
  3. If a number occurs more than 25% of the time, return that number.

Python Code:

def findSpecialInteger(arr):
    threshold = len(arr) // 4
    count = 1

    for i in range(1, len(arr)):
        if arr[i] == arr[i - 1]:
            count += 1
            if count > threshold:
                return arr[i]
        else:
            count = 1

    return arr[0]

print(findSpecialInteger([1, 2, 2, 6, 6, 6, 6, 7, 10]))  # Output should be 6

3. Optimized Approach 1: Counting Sequences

Since the array is sorted, if a number occurs more than 25% of the time, then one of its instances will definitely appear at an index that is a multiple of the array’s length divided by 4.

Algorithm:

  1. Calculate the threshold index threshold_idx = len(arr) // 4.
  2. Check the numbers at indices threshold_idx, 2 * threshold_idx, 3 * threshold_idx and so on.

Python Code:

def findSpecialInteger(arr):
    threshold_idx = len(arr) // 4

    for i in range(threshold_idx, len(arr)):
        if arr[i] == arr[i - threshold_idx]:
            return arr[i]

    return -1  # Should never reach this point based on problem constraints

print(findSpecialInteger([1, 2, 2, 6, 6, 6, 6, 7, 10]))  # Output should be 6

4. Testing and Validation

While testing, ensure that:

  1. The array is sorted.
  2. Test with edge cases like minimum and maximum array lengths.
  3. Include negative numbers in your test cases.

5. Time and Space Complexity Analysis

  1. Naive Approach:
    • Time Complexity: O(n)
    • Space Complexity: O(1)
  2. Optimized Approach 1:
    • Time Complexity: O(n)
    • Space Complexity: O(1)

6. Conclusion

In this guide, we have discussed multiple approaches to solve the “Element Appearing More Than 25% In Sorted Array” problem on Leetcode. We started with a naive approach and then discussed an optimized method. It’s always crucial to consider time and space complexities to come up with the most efficient solution.

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