Interpolation Search

Given a sorted array of n uniformly distributed values arr[], write a function to search for a particular element x in the array.

Linear Search finds the element in O(n) time, Jump Search takes O(√ n) time and Binary Search take O(Log n) time.
The Interpolation Search is an improvement over Binary Search for instances, where the values in a sorted array are uniformly distributed. Binary Search always goes to middle element to check. On the other hand interpolation search may go to different locations according the value of key being searched. For example if the value of key is closer to the last element, interpolation search is likely to start search toward the end side.

To find the position to be searched, it uses following formula.

// The idea of formula is to return higher value of pos
// when element to be searched is closer to arr[hi]. And
// smaller value when closer to arr[lo]
pos = lo + [ (x-arr[lo])*(hi-lo) / (arr[hi]-arr[Lo]) ]

arr[] ==> Array where elements need to be searched
x     ==> Element to be searched
lo    ==> Starting index in arr[]
hi    ==> Ending index in arr[]

Algorithm
Rest of the Interpolation algorithm is same except the above partition logic.

Step1: In a loop, calculate the value of “pos” using the probe position formula.
Step2: If it is a match, return the index of the item, and exit.
Step3: If the item is less than arr[pos], calculate the probe position of the left sub-array. Otherwise calculate the same in the right sub-array.
Step4: Repeat until a match is found or the sub-array reduces to zero.

Below is C implementation of algorithm.

C

// C program to implement interpolation search
#include<stdio.h>

// If x is present in arr[0..n-1], then returns
// index of it, else returns -1.
int interpolationSearch(int arr[], int n, int x)
{
    // Find indexes of two corners
    int lo = 0, hi = (n - 1);

    // Since array is sorted, an element present
    // in array must be in range defined by corner
    while (lo <= hi && x >= arr[lo] && x <= arr[hi])
    {
        // Probing the position with keeping
        // uniform distribution in mind.
        int pos = lo + (((double)(hi-lo) /
              (arr[hi]-arr[lo]))*(x - arr[lo]));

        // Condition of target found
        if (arr[pos] == x)
            return pos;

        // If x is larger, x is in upper part
        if (arr[pos] < x)
            lo = pos + 1;

        // If x is smaller, x is in lower part
        else
            hi = pos - 1;
    }
    return -1;
}

// Driver Code
int main()
{
    // Array of items on which search will
    // be conducted.
    int arr[] =  {10, 12, 13, 16, 18, 19, 20, 21, 22, 23,
                  24, 33, 35, 42, 47};
    int n = sizeof(arr)/sizeof(arr[0]);

    int x = 18; // Element to be searched
    int index = interpolationSearch(arr, n, x);

    // If element was found
    if (index != -1)
        printf("Element found at index %d", index);
    else
        printf("Element not found.");
    return 0;
}

Java

// Java program to implement interpolation search

class Test
{
	// Array of items on which search will
    // be conducted.
	static int arr[] = new int[]{10, 12, 13, 16, 18, 19, 20, 21, 22, 23,
                                         24, 33, 35, 42, 47};
	
	// If x is present in arr[0..n-1], then returns
	// index of it, else returns -1.
	static int interpolationSearch(int x)
	{
	    // Find indexes of two corners
	    int lo = 0, hi = (arr.length - 1);
	 
	    // Since array is sorted, an element present
	    // in array must be in range defined by corner
	    while (lo <= hi && x >= arr[lo] && x <= arr[hi])
	    {
	        // Probing the position with keeping
	        // uniform distribution in mind.
	        int pos = lo + (((hi-lo) /
	              (arr[hi]-arr[lo]))*(x - arr[lo]));
	 
	        // Condition of target found
	        if (arr[pos] == x)
	            return pos;
	 
	        // If x is larger, x is in upper part
	        if (arr[pos] < x)
	            lo = pos + 1;
	 
	        // If x is smaller, x is in lower part
	        else
	            hi = pos - 1;
	    }
	    return -1;
	}
  
    // Driver method 
    public static void main(String[] args) 
    {
    	 int x = 18; // Element to be searched
    	 int index = interpolationSearch(x);
    	 
         // If element was found
         if (index != -1)
   	        System.out.println("Element found at index " + index);
   	     else
   	        System.out.println("Element not found.");
    }
}

Python

# Python program to implement interpolation search

# If x is present in arr[0..n-1], then returns
# index of it, else returns -1
def interpolationSearch(arr, n, x):
    # Find indexs of two corners
    lo = 0
    hi = (n - 1)
 
    # Since array is sorted, an element present
    # in array must be in range defined by corner
    while lo <= hi and x >= arr[lo] and x <= arr[hi]:
        # Probing the position with keeping
        # uniform distribution in mind.
        pos  = lo + int(((float(hi - lo) / 
            ( arr[hi] - arr[lo])) * ( x - arr[lo])))

        # Condition of target found
        if arr[pos] == x:
            return pos
 
        # If x is larger, x is in upper part
        if arr[pos] < x:
            lo = pos + 1;
 
        # If x is smaller, x is in lower part
        else:
            hi = pos - 1;
    
    return -1

# Driver Code
# Array of items oin which search will be conducted
arr = [10, 12, 13, 16, 18, 19, 20, 21, \
                22, 23, 24, 33, 35, 42, 47]
n = len(arr)

x = 18 # Element to be searched
index = interpolationSearch(arr, n, x)

if index != -1:
    print "Element found at index",index
else:
    print "Element not found"

# This code is contributed by Harshit Agrawal


Output :
Element found at index 4

Time Complexity : If elements are uniformly distributed, then O (log log n)). In worst case it can take upto O(n).
Auxiliary Space : O(1)

This article is contributed by Aayu sachdev. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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