A simple approach is to do linear search.The time complexity of above algorithm is O(n). Another approach to perform the same task is using Binary Search.
Binary Search: Search a sorted array by repeatedly dividing the search interval in half. Begin with an interval covering the whole array. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. Otherwise narrow it to the upper half. Repeatedly check until the value is found or the interval is empty.
Example:
Image Source : http://www.nyckidd.com/bob/Linear%20Search%20and%20Binary%20Search_WorkingCopy.pdf
The idea of binary search is to use the information that the array is sorted and reduce the time complexity to O(Logn).
We basically ignore half of the elements just after one comparison.
- Compare x with the middle element.
- If x matches with middle element, we return the mid index.
- Else If x is greater than the mid element, then x can only lie in right half subarray after the mid element. So we recur for right half.
- Else (x is smaller) recur for the left half.
Recursive implementation of Binary Search
// Java implementation of recursive Binary Searchclass BinarySearch{ // Returns index of x if it is present in arr[l..r], else // return -1 int binarySearch(int arr[], int l, int r, int x) { if (r>=l) { int mid = l + (r - l)/2; // If the element is present at the middle itself if (arr[mid] == x) return mid; // If element is smaller than mid, then it can only // be present in left subarray if (arr[mid] > x) return binarySearch(arr, l, mid-1, x); // Else the element can only be present in right // subarray return binarySearch(arr, mid+1, r, x); } // We reach here when element is not present in array return -1; } // Driver method to test above public static void main(String args[]) { BinarySearch ob = new BinarySearch(); int arr[] = {2,3,4,10,40}; int n = arr.length; int x = 10; int result = ob.binarySearch(arr,0,n-1,x); if (result == -1) System.out.println("Element not present"); else System.out.println("Element found at index "+result); }}/* This code is contributed by Rajat Mishra */ |
Output:
Element is present at index 3
Iterative implementation of Binary Search
// Java implementation of iterative Binary Searchclass BinarySearch{ // Returns index of x if it is present in arr[], else // return -1 int binarySearch(int arr[], int x) { int l = 0, r = arr.length - 1; while (l <= r) { int m = l + (r-l)/2; // Check if x is present at mid if (arr[m] == x) return m; // If x greater, ignore left half if (arr[m] < x) l = m + 1; // If x is smaller, ignore right half else r = m - 1; } // if we reach here, then element was not present return -1; } // Driver method to test above public static void main(String args[]) { BinarySearch ob = new BinarySearch(); int arr[] = {2, 3, 4, 10, 40}; int n = arr.length; int x = 10; int result = ob.binarySearch(arr, x); if (result == -1) System.out.println("Element not present"); else System.out.println("Element found at index "+result); }} |
Output:
Element is present at index 3Time Complexity:
The time complexity of Binary Search can be written as
T(n) = T(n/2) + cThe above recurrence can be solved either using Recurrence T ree method or Master method. It falls in case II of Master Method and solution of the recurrence is
.Auxiliary Space: O(1) in case of iterative implementation. In case of recursive implementation, O(Logn) recursion call stack space.
Algorithmic Paradigm: Divide and Conquer
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