package com.thealgorithms.searches; import com.thealgorithms.devutils.searches.SearchAlgorithm; /** * Binary Search Algorithm Implementation * *

Binary search is one of the most efficient searching algorithms for finding a target element * in a SORTED array. It works by repeatedly dividing the search space in half, eliminating half of * the remaining elements in each step. * *

IMPORTANT: This algorithm ONLY works correctly if the input array is sorted in ascending * order. * *

Algorithm Overview: 1. Start with the entire array (left = 0, right = array.length - 1) 2. * Calculate the middle index 3. Compare the middle element with the target: - If middle element * equals target: Found! Return the index - If middle element is less than target: Search the right * half - If middle element is greater than target: Search the left half 4. Repeat until element is * found or search space is exhausted * *

Performance Analysis: - Best-case time complexity: O(1) - Element found at middle on first * try - Average-case time complexity: O(log n) - Most common scenario - Worst-case time * complexity: O(log n) - Element not found or at extreme end - Space complexity: O(1) - Only uses * a constant amount of extra space * *

Example Walkthrough: Array: [1, 3, 5, 7, 9, 11, 13, 15, 17, 19] Target: 7 * *

Step 1: left=0, right=9, mid=4, array[4]=9 (9 > 7, search left half) Step 2: left=0, * right=3, mid=1, array[1]=3 (3 < 7, search right half) Step 3: left=2, right=3, mid=2, * array[2]=5 (5 < 7, search right half) Step 4: left=3, right=3, mid=3, array[3]=7 (Found! * Return index 3) * * @author Varun Upadhyay (https://github.com/varunu28) * @author Podshivalov Nikita (https://github.com/nikitap492) * @see SearchAlgorithm * @see IterativeBinarySearch */ class BinarySearch implements SearchAlgorithm { /** * Generic method to perform binary search on any comparable type. This is the main entry point * for binary search operations. * *

Example Usage: *

     * Integer[] numbers = {1, 3, 5, 7, 9, 11};
     * int result = new BinarySearch().find(numbers, 7);
     * // result will be 3 (index of element 7)
     *
     * int notFound = new BinarySearch().find(numbers, 4);
     * // notFound will be -1 (element 4 does not exist)
     * 
* * @param The type of elements in the array (must be Comparable) * @param array The sorted array to search in (MUST be sorted in ascending order) * @param key The element to search for * @return The index of the key if found, -1 if not found or if array is null/empty */ @Override public > int find(T[] array, T key) { // Handle edge case: empty array if (array == null || array.length == 0) { return -1; } // Delegate to the core search implementation return search(array, key, 0, array.length - 1); } /** * Core recursive implementation of binary search algorithm. This method divides the problem * into smaller subproblems recursively. * *

How it works: *

    *
  1. Calculate the middle index to avoid integer overflow
  2. *
  3. Check if middle element matches the target
  4. *
  5. If not, recursively search either left or right half
  6. *
  7. Base case: left > right means element not found
  8. *
* *

Time Complexity: O(log n) because we halve the search space each time. * Space Complexity: O(log n) due to recursive call stack. * * @param The type of elements (must be Comparable) * @param array The sorted array to search in * @param key The element we're looking for * @param left The leftmost index of current search range (inclusive) * @param right The rightmost index of current search range (inclusive) * @return The index where key is located, or -1 if not found */ private > int search(T[] array, T key, int left, int right) { // Base case: Search space is exhausted // This happens when left pointer crosses right pointer if (right < left) { return -1; // Key not found in the array } // Calculate middle index // Using (left + right) / 2 could cause integer overflow for large arrays // So we use: left + (right - left) / 2 which is mathematically equivalent // but prevents overflow int median = (left + right) >>> 1; // Unsigned right shift is faster division by 2 // Get the value at middle position for comparison int comp = key.compareTo(array[median]); // Case 1: Found the target element at middle position if (comp == 0) { return median; // Return the index where element was found } // Case 2: Target is smaller than middle element // This means if target exists, it must be in the LEFT half else if (comp < 0) { // Recursively search the left half // New search range: [left, median - 1] return search(array, key, left, median - 1); } // Case 3: Target is greater than middle element // This means if target exists, it must be in the RIGHT half else { // Recursively search the right half // New search range: [median + 1, right] return search(array, key, median + 1, right); } } }