Binary search works by repeatedly dividing the search interval in half. This requires knowing whether the target value is in the lower or upper half, which is only possible if the list is sorted.

Divide the list into smaller sublists, recursively sort the sublists, and then merge them back together in sorted order.

O(n) in the worst case, where n is the number of elements in the list.

O(log n), where n is the number of elements in the list.

O(n^2) in the worst and average cases, where n is the number of elements in the list.

O(n^2) in all cases, where n is the number of elements in the list.

O(n log n) in all cases, where n is the number of elements in the list.

Merge Sort requires more memory due to the creation of temporary arrays during the merge process.

Recursion can make code more concise and easier to read for certain problems.

Recursion can lead to stack overflow errors if the base case is not reached or the recursion depth is too large.

A search algorithm that finds the position of a target value within a sorted array by repeatedly dividing the search interval in half.

A divide-and-conquer sorting algorithm that divides the input array into smaller sub-arrays, recursively sorts them, and then merges the sorted sub-arrays.

A programming technique where a function calls itself in order to solve a problem.

The condition that stops a recursive function from calling itself infinitely.

A searching algorithm that sequentially checks each element of a list until a match is found or the entire list has been searched.

A sorting algorithm that builds the final sorted array one item at a time by repeatedly inserting the next element into its correct position within the already sorted portion of the array.

A sorting algorithm that repeatedly finds the minimum element from the unsorted portion of the array and places it at the beginning.

A resizable array implementation of the List interface in Java.

The position of an element within an array or ArrayList, starting from 0.

A contiguous portion of a list or array.

1. Find the middle element. 2. If the target equals the middle, return the index. 3. If the target is less than the middle, search the left half. 4. If the target is greater than the middle, search the right half. 5. Repeat until found or the search space is empty.

1. Divide the list into sublists. 2. Recursively sort each sublist. 3. Merge the sorted sublists.

1. Base case: If the index reaches the end of the list, the list is sorted. 2. Take the element at the current index. 3. Shift the element to the left until it's in the correct sorted position. 4. Recursively call the function for the next index.

1. Base case: If the index reaches the end of the list, the list is sorted. 2. Find the smallest element in the unsorted portion of the list. 3. Swap the smallest element with the element at the current index. 4. Recursively call the function for the next index.

1. Base case 1: If the element at the current index matches the target, return the index. 2. Base case 2: If the current index is the last element and no match is found, return -1. 3. Recursive step: Call the function again with the next index.

1. Assume the first element is sorted. 2. Select the next element. 3. Compare to the sorted elements and shift until correct position is found. 4. Insert the element.

1. Find the minimum element in the unsorted portion. 2. Swap it with the first unsorted element. 3. Repeat for the remaining unsorted portion.

1. Iterate through each element in the array. 2. Compare the current element with the target. 3. If a match is found, return the index. 4. If no match is found after checking all elements, return -1.

1. Initialize left and right pointers. 2. While left <= right, calculate the middle index. 3. If the middle element equals the target, return the index. 4. If the target is less than the middle element, move the right pointer. 5. If the target is greater than the middle element, move the left pointer. 6. If the target is not found, return -1.

Call recursiveBinarySearch(array, n, 0, array.size() - 1); where array is the sorted ArrayList, n is the value to search, 0 is the starting index, and array.size() - 1 is the ending index.