Count the instances of 'X' in a sorted array, limited to specific index ranges
In the realm of algorithmic problem-solving, a common task is to find the number of occurrences of a specific value x in a subarray of a given sorted array. This article outlines a method that uses binary search to achieve this efficiently, offering a significant improvement over a naive linear scan approach.
The problem at hand involves a sorted array and a list of queries, where each query is a triplet [l, r, x]. The subarray of interest is defined by the indices [l, r].
To tackle this problem, follow these steps:
- Define the subarray bounds: Determine the subarray you want to search, defined by indices .
- Find the first occurrence of x in the subarray [L, R]:
- Use a modified binary search on the subarray from to .
- Search for the first index where and no smaller index in holds .
- If not found, return .
- Find the last occurrence of x in the subarray [L, R]:
- Similarly, use a binary search within for the last index where .
- Calculate the number of occurrences:
- If either first or last occurrence is , then count is zero.
- Otherwise, count = .
Binary search is an optimal choice for this problem due to the sorted nature of the array and subarray. Binary search halves the search space on each iteration, leading to an overall time complexity of O(log n) instead of O(n).
Here's a Python code template for the function :
```python def first_occurrence(arr, x, left, right): result = -1 while left <= right: mid = left + (right - left) // 2 if arr[mid] == x: result = mid right = mid - 1 # search to the left elif arr[mid] < x: left = mid + 1 else: right = mid - 1 return result
def last_occurrence(arr, x, left, right): result = -1 while left <= right: mid = left + (right - left) // 2 if arr[mid] == x: result = mid left = mid + 1 # search to the right elif arr[mid] < x: left = mid + 1 else: right = mid - 1 return result
def count_occurrences(arr, x, L, R): first = first_occurrence(arr, x, L, R) if first == -1: return 0 last = last_occurrence(arr, x, L, R) return last - first + 1
```
This method offers a better approach than linear scanning since it works in O(log n) time, minimizing time complexity especially for large arrays.
For each query, the first position where x appears is found using . The binary search approach uses and functions.
If x does not exist in the array, 0 is stored as the answer for that query. The time complexity of the binary search approach is O(q × log n).
References: - Finding first and last occurrences using binary search: [1][2][4] - Range-based count by combining first and last occurrence indices: [3]
[1] https://www.geeksforgeeks.org/find-first-and-last-position-of-element-in-a-sorted-array/ [2] https://www.geeksforgeeks.org/find-first-and-last-occurrence-of-an-element-in-a-sorted-array/ [3] https://www.geeksforgeeks.org/count-number-occurrences-element-given-range-sorted-array/ [4] https://www.programiz.com/dsa/lower_bound
In the context of data-and-cloud-computing, technology, and algorithms, this problem involves the use of binary search to efficiently find the number of occurrences of a specific value (x) in a subarray of a given sorted array, which is a common task in math problem-solving. The triplet [l, r, x] defines the subarray of interest, and a Python code template is provided to solve this problem. This approach demonstrates an improvement over a naive linear scan method, as it has a time complexity of O(log n) instead of O(n).
