When the list is sorted we can use the binary search technique to find items on the list. In this procedure, the entire list is divided into two sub-lists. If the item is found in the middle position, it returns the location, otherwise jumps to either left or right sub-list and do the same process again until finding the item or exceed the range.
The complexity of Binary Search Technique
- Time Complexity: O(1) for the best case. O(log2 n) for average or worst case.
- Space Complexity: O(1)
Input and Output
Input: A sorted list of data: 12 25 48 52 67 79 88 93 The search key 79 Output: Item found at location: 5
Algorithm
binarySearch(array, start, end, key)
Input − An sorted array, start and end location, and the search key
Output − location of the key (if found), otherwise wrong location.
Begin if start <= end then mid := start + (end - start) /2 if array[mid] = key then return mid location if array[mid] > key then call binarySearch(array, mid+1, end, key) else when array[mid] < key then call binarySearch(array, start, mid-1, key) else return invalid location End
Example
#include<iostream>
using namespace std;
int binarySearch(int array[], int start, int end, int key) {
if(start <= end) {
int mid = (start + (end - start) /2); //mid location of the list
if(array[mid] == key)
return mid;
if(array[mid] > key)
return binarySearch(array, start, mid-1, key);
return binarySearch(array, mid+1, end, key);
}
return -1;
}
int main() {
int n, searchKey, loc;
cout << "Enter number of items: ";
cin >> n;
int arr[n]; //create an array of size n
cout << "Enter items: " << endl;
for(int i = 0; i< n; i++) {
cin >> arr[i];
}
cout << "Enter search key to search in the list: ";
cin >> searchKey;
if((loc = binarySearch(arr, 0, n, searchKey)) >= 0)
cout << "Item found at location: " << loc << endl;
else
cout << "Item is not found in the list." << endl;
}Output
Enter number of items: 8 Enter items: 12 25 48 52 67 79 88 93 Enter search key to search in the list: 79 Item found at location: 5