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Interpolation Search

  • Difficulty Level : Easy
  • Last Updated : 28 Jun, 2021

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 the middle element to check. On the other hand, interpolation search may go to different locations according to the value of the key being searched. For example, if the value of the 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[]

The formula for pos can be derived as follows.

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Let's assume that the elements of the array are linearly distributed. 

General equation of line : y = m*x + c.
y is the value in the array and x is its index.

Now putting value of lo,hi and x in the equation
arr[hi] = m*hi+c ----(1)
arr[lo] = m*lo+c ----(2)
x = m*pos + c     ----(3)

m = (arr[hi] - arr[lo] )/ (hi - lo)

subtracting eqxn (2) from (3)
x - arr[lo] = m * (pos - lo)
lo + (x - arr[lo])/m = pos
pos = lo + (x - arr[lo]) *(hi - lo)/(arr[hi] - arr[lo])

Algorithm 
Rest of the Interpolation algorithm is the 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 the implementation of algorithm. 

C++




// C++ program to implement interpolation search
#include<bits/stdc++.h>
using namespace std;
 
// 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])
    {
        if (lo == hi)
        {
            if (arr[lo] == x) return lo;
            return -1;
        }
        // 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 the 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)
        cout << "Element found at index " << index;
    else
        cout << "Element not found.";
    return 0;
}
 
// This code is contributed by Mukul Singh.

C++




// C++ program to implement interpolation
// search with recursion
#include <bits/stdc++.h>
using namespace std;
 
// If x is present in arr[0..n-1], then returns
// index of it, else returns -1.
int interpolationSearch(int arr[], int lo, int hi, int x)
{
    int pos;
 
    // Since array is sorted, an element present
    // in array must be in range defined by corner
    if (lo <= hi && x >= arr[lo] && x <= arr[hi]) {
 
        // Probing the position with keeping
        // uniform distribution in mind.
        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 right sub array
        if (arr[pos] < x)
            return interpolationSearch(arr, pos + 1, hi, x);
 
        // If x is smaller, x is in left sub array
        if (arr[pos] > x)
            return interpolationSearch(arr, lo, pos - 1, x);
    }
    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]);
 
    // Element to be searched
    int x = 18;
    int index = interpolationSearch(arr, 0, n - 1, x);
 
    // If element was found
    if (index != -1)
        cout << "Element found at index " << index;
    else
        cout << "Element not found.";
 
    return 0;
}
 
// This code is contributed by equbalzeeshan

C




// C program to implement interpolation search
// with recursion
#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 lo, int hi, int x)
{
    int pos;
    // Since array is sorted, an element present
    // in array must be in range defined by corner
    if (lo <= hi && x >= arr[lo] && x <= arr[hi]) {
        // Probing the position with keeping
        // uniform distribution in mind.
        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 right sub array
        if (arr[pos] < x)
            return interpolationSearch(arr, pos + 1, hi, x);
 
        // If x is smaller, x is in left sub array
        if (arr[pos] > x)
            return interpolationSearch(arr, lo, pos - 1, x);
    }
    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, 0, n - 1, 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 with recursion
import java.util.*;
 
class GFG {
 
    // If x is present in arr[0..n-1], then returns
    // index of it, else returns -1.
    public static int interpolationSearch(int arr[], int lo,
                                          int hi, int x)
    {
        int pos;
 
        // Since array is sorted, an element
        // present in array must be in range
        // defined by corner
        if (lo <= hi && x >= arr[lo] && x <= arr[hi]) {
 
            // Probing the position with keeping
            // uniform distribution in mind.
            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 right sub array
            if (arr[pos] < x)
                return interpolationSearch(arr, pos + 1, hi,
                                           x);
 
            // If x is smaller, x is in left sub array
            if (arr[pos] > x)
                return interpolationSearch(arr, lo, pos - 1,
                                           x);
        }
        return -1;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        // 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 = arr.length;
 
        // Element to be searched
        int x = 18;
        int index = interpolationSearch(arr, 0, n - 1, x);
 
        // If element was found
        if (index != -1)
            System.out.println("Element found at index "
                               + index);
        else
            System.out.println("Element not found.");
    }
}
 
// This code is contributed by equbalzeeshan

Python




# Python3 program to implement
# interpolation search
# with recursion
 
# If x is present in arr[0..n-1], then
# returns index of it, else returns -1.
 
 
def interpolationSearch(arr, lo, hi, x):
 
    # Since array is sorted, an element present
    # in array must be in range defined by corner
    if (lo <= hi and x >= arr[lo] and x <= arr[hi]):
 
        # Probing the position with keeping
        # uniform distribution in mind.
        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 right subarray
        if arr[pos] < x:
            return interpolationSearch(arr, pos + 1,
                                       hi, x)
 
        # If x is smaller, x is in left subarray
        if arr[pos] > x:
            return interpolationSearch(arr, lo,
                                       pos - 1, x)
    return -1
 
# Driver code
 
 
# Array of items in which
# search will be conducted
arr = [10, 12, 13, 16, 18, 19, 20,
       21, 22, 23, 24, 33, 35, 42, 47]
n = len(arr)
 
# Element to be searched
x = 18
index = interpolationSearch(arr, 0, n - 1, x)
 
if index != -1:
    print("Element found at index", index)
else:
    print("Element not found")
 
# This code is contributed by Hardik Jain

C#




// C# program to implement
// interpolation search
using System;
 
class GFG{
 
// If x is present in
// arr[0..n-1], then
// returns index of it,
// else returns -1.
static int interpolationSearch(int []arr, int lo,
                               int hi, int x)
{
    int pos;
     
    // Since array is sorted, an element
    // present in array must be in range
    // defined by corner
    if (lo <= hi && x >= arr[lo] &&
                    x <= arr[hi])
    {
         
        // Probing the position
        // with keeping uniform
        // distribution in mind.
        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 right sub array
        if(arr[pos] < x)
            return interpolationSearch(arr, pos + 1,
                                       hi, x);
         
        // If x is smaller, x is in left sub array
        if(arr[pos] > x)
            return interpolationSearch(arr, lo,
                                       pos - 1, x);
    }
    return -1;
}
 
// Driver Code
public static void Main()
{
     
    // Array of items on which search will
    // be conducted.
    int []arr = new int[]{ 10, 12, 13, 16, 18,
                           19, 20, 21, 22, 23,
                           24, 33, 35, 42, 47 };
                            
    // Element to be searched                      
    int x = 18;
    int n = arr.Length;
    int index = interpolationSearch(arr, 0, n - 1, x);
     
    // If element was found
    if (index != -1)
        Console.WriteLine("Element found at index " +
                           index);
    else
        Console.WriteLine("Element not found.");
}
}
 
// This code is contributed by equbalzeeshan

Javascript




<script>
// Javascript program to implement Interpolation Search
 
// If x is present in arr[0..n-1], then returns
// index of it, else returns -1.
 
function interpolationSearch(arr, lo, hi, x){
  let pos;
   
  // Since array is sorted, an element present
  // in array must be in range defined by corner
   
  if (lo <= hi && x >= arr[lo] && x <= arr[hi]) {
     
    // Probing the position with keeping
    // uniform distribution in mind.
    pos = lo + Math.floor(((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 right sub array
        if (arr[pos] < x){
          return interpolationSearch(arr, pos + 1, hi, x);
        }
  
        // If x is smaller, x is in left sub array
        if (arr[pos] > x){
          return interpolationSearch(arr, lo, pos - 1, x);
        }
    }
    return -1;
}
 
// Driver Code
let arr = [10, 12, 13, 16, 18, 19, 20, 21,
           22, 23, 24, 33, 35, 42, 47];
 
let n = arr.length;
 
// Element to be searched
let x = 18
let index = interpolationSearch(arr, 0, n - 1, x);
 
// If element was found
if (index != -1){
   document.write(`Element found at index ${index}`)
}else{
   document.write("Element not found");
}
 
// This code is contributed by _saurabh_jaiswal
</script>
Output
Element found at index 4

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