Solution Approach
Solution Overview
There are 2 main approaches to solve this problem:
- Recursive Binary Search - Complex approach
- Iterative Binary Search - Complex approach
Approach 1: Recursive Binary Search
The recursive binary search approach is a classic example of the Divide & Conquer pattern. We repeatedly divide the search space in half by comparing the target value with the middle element of the current subarray.
This approach is particularly efficient for large datasets because it eliminates half of the remaining elements in each step, resulting in a logarithmic time complexity.
Algorithm:
- Define a recursive function binarySearch(bookCodes, target, left, right) that takes the array of book codes, the target code, and the left and right boundaries of the current search space.
- Base case: If left > right, the target is not in the array, so return -1.
- Calculate the middle index as mid = Math.floor((left + right) / 2).
- If bookCodes[mid] equals the target, return mid (we found the book).
- If bookCodes[mid] > target, recursively search in the left half by calling binarySearch(bookCodes, target, left, mid - 1).
- If bookCodes[mid] < target, recursively search in the right half by calling binarySearch(bookCodes, target, mid + 1, right).
- The main function should initialize left = 0 and right = bookCodes.length - 1, then call the recursive function.
Time Complexity:
In each recursive call, we eliminate half of the remaining elements, resulting in a logarithmic time complexity.
Space Complexity:
The recursion stack will have at most log n frames, corresponding to the maximum depth of the recursion.
Approach 2: Iterative Binary Search
While the problem asks for a recursive solution, it's worth mentioning the iterative approach as well. This approach uses a loop instead of recursion to implement binary search.
The iterative approach has the same time complexity as the recursive approach but uses constant space, making it more efficient in terms of space complexity.
Algorithm:
- Initialize left = 0 and right = bookCodes.length - 1.
- While left <= right:
- - Calculate the middle index as mid = Math.floor((left + right) / 2).
- - If bookCodes[mid] equals the target, return mid (we found the book).
- - If bookCodes[mid] > target, update right = mid - 1 (search in the left half).
- - If bookCodes[mid] < target, update left = mid + 1 (search in the right half).
- If the loop ends without finding the target, return -1.
Time Complexity:
Like the recursive approach, we eliminate half of the remaining elements in each iteration.
Space Complexity:
The iterative approach uses a constant amount of extra space, regardless of the input size.
Pseudocode
123456789101112131415161718192021function binarySearch(bookCodes, targetCode): return binarySearchRecursive(bookCodes, targetCode, 0, bookCodes.length - 1) function binarySearchRecursive(bookCodes, targetCode, left, right): // Base case: target not found if left > right: return -1 // Calculate middle index mid = floor((left + right) / 2) // Check if we found the target if bookCodes[mid] == targetCode: return mid // If target is smaller, search in the left half if bookCodes[mid] > targetCode: return binarySearchRecursive(bookCodes, targetCode, left, mid - 1) // If target is larger, search in the right half return binarySearchRecursive(bookCodes, targetCode, mid + 1, right)String Problems
Learning Objective
Understand different approaches to solve the library book finder problem and analyze their efficiency.
Solution Approaches
Approach 1: Recursive Binary Search
The recursive binary search approach is a classic example of the Divide & Conquer pattern. We repeatedly divide the search space in half by comparing the target value with the middle element of the current subarray.
This approach is particularly efficient for large datasets because it eliminates half of the remaining elements in each step, resulting in a logarithmic time complexity.
Algorithm:
- Define a recursive function binarySearch(bookCodes, target, left, right) that takes the array of book codes, the target code, and the left and right boundaries of the current search space.
- Base case: If left > right, the target is not in the array, so return -1.
- Calculate the middle index as mid = Math.floor((left + right) / 2).
- If bookCodes[mid] equals the target, return mid (we found the book).
- If bookCodes[mid] > target, recursively search in the left half by calling binarySearch(bookCodes, target, left, mid - 1).
- If bookCodes[mid] < target, recursively search in the right half by calling binarySearch(bookCodes, target, mid + 1, right).
- The main function should initialize left = 0 and right = bookCodes.length - 1, then call the recursive function.
Approach 2: Iterative Binary Search
While the problem asks for a recursive solution, it's worth mentioning the iterative approach as well. This approach uses a loop instead of recursion to implement binary search.
The iterative approach has the same time complexity as the recursive approach but uses constant space, making it more efficient in terms of space complexity.
Algorithm:
- Initialize left = 0 and right = bookCodes.length - 1.
- While left <= right:
- - Calculate the middle index as mid = Math.floor((left + right) / 2).
- - If bookCodes[mid] equals the target, return mid (we found the book).
- - If bookCodes[mid] > target, update right = mid - 1 (search in the left half).
- - If bookCodes[mid] < target, update left = mid + 1 (search in the right half).
- If the loop ends without finding the target, return -1.
Complexity Analysis
Recursive Binary Search Approach
Time Complexity
In each recursive call, we eliminate half of the remaining elements, resulting in a logarithmic time complexity.
Space Complexity
The recursion stack will have at most log n frames, corresponding to the maximum depth of the recursion.
Iterative Binary Search Approach
Time Complexity
Like the recursive approach, we eliminate half of the remaining elements in each iteration.
Space Complexity
The iterative approach uses a constant amount of extra space, regardless of the input size.
Comparison:
Both the recursive and iterative approaches have the same time complexity of O(log n), which is optimal for searching in a sorted array. The main difference is in space complexity: the recursive approach uses O(log n) space for the recursion stack, while the iterative approach uses O(1) space. In practice, for most inputs, this difference is negligible, and both approaches are very efficient. The recursive approach is often more intuitive and aligns well with the divide-and-conquer paradigm, making it a good choice for educational purposes. However, for very large inputs or in memory-constrained environments, the iterative approach might be preferred due to its constant space complexity.
Pseudocode
Recursive Binary Search Approach
123456789101112131415161718192021function binarySearch(bookCodes, targetCode): return binarySearchRecursive(bookCodes, targetCode, 0, bookCodes.length - 1) function binarySearchRecursive(bookCodes, targetCode, left, right): // Base case: target not found if left > right: return -1 // Calculate middle index mid = floor((left + right) / 2) // Check if we found the target if bookCodes[mid] == targetCode: return mid // If target is smaller, search in the left half if bookCodes[mid] > targetCode: return binarySearchRecursive(bookCodes, targetCode, left, mid - 1) // If target is larger, search in the right half return binarySearchRecursive(bookCodes, targetCode, mid + 1, right)Iterative Binary Search Approach
12345678910111213141516171819202122function binarySearchIterative(bookCodes, targetCode): left = 0 right = bookCodes.length - 1 while left <= right: // Calculate middle index mid = floor((left + right) / 2) // Check if we found the target if bookCodes[mid] == targetCode: return mid // If target is smaller, search in the left half if bookCodes[mid] > targetCode: right = mid - 1 // If target is larger, search in the right half else: left = mid + 1 // Target not found return -1