Interpolation Search

For the binary search technique, the lists are divided into equal parts. For the interpolation searching technique, the procedure will try to locate the exact position using interpolation formula. After finding the estimated location, it can separate the list using that location. As it tries to find exact location every time, so the searching time reduces. This technique can find items easily if the items are uniformly distributed.

The complexity of Interpolation Search Technique

• Time Complexity: O(log2(log2 n)) for the average case, and O(n) for the worst case (when items are distributed exponentially)
• Space Complexity: O(1)

Input and Output

Input:
A sorted list of data:
10 13 15 26 28 50 56 88 94 127 159 356 480 567 689 699 780 850 956 995
The search key 780
Output:
Item found at location: 16

Algorithm

interpolationSearch(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
   while start <= end AND key >= array[start] AND key <= array[end] do
      dist := key – array[start]
      valRange := array[end] – array[start]
      fraction := dist / valRange
      indexRange := end – start
      estimate := start + (fraction * indexRange)
      if array[estimate] = key then
         return estimate position
      if array[estimate] < key then
         start := estimate + 1
      else
         end = estimate -1
   done
   return invalid position
End

Example

#include<iostream>
using namespace std;
int interpolationSearch(int array[], int start, int end, int key) {
   int dist, valRange, indexRange, estimate;
   float fraction;

   while(start <= end && key >= array[start] && key <= array[end]) {
      dist = key - array[start];
      valRange = array[end] - array[start]; //range of value
      fraction = dist / valRange;
      indexRange = end - start;
      estimate = start + (fraction * indexRange); //estimated position of the key

      if(array[estimate] == key)
         return estimate;
      if(array[estimate] < key)
         start = estimate +1;
      else
         end = estimate - 1;
   }
   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 = interpolationSearch(arr, 0, n-1, searchKey)) >= 0)
      cout << "Item found at location: " << loc << endl;
   else
      cout << "Item is not found in the list." << endl;
}

Output

Enter number of items: 20
Enter items:
10 13 15 26 28 50 56 88 94 127 159 356 480 567 689 699 780 850 956 995
Enter search key to search in the list: 780
Item found at location: 16