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

- Problem Description
- Naive Approach: Brute-force Algorithm
- Optimized Approach 1: Counting Sequences
- Testing and Validation
- Time and Space Complexity Analysis
- 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:

- Initialize a counter variable to keep track of occurrences.
- Iterate through the sorted array and count consecutive similar numbers.
- 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:

- Calculate the threshold index
`threshold_idx = len(arr) // 4`

. - 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:

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

## 5. Time and Space Complexity Analysis

**Naive Approach**:- Time Complexity: O(n)
- Space Complexity: O(1)

**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.