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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
Updated on: 15-Jun-2020
2K+ Views
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