# Interpolation Search

Last Updated : 15 May, 2023

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 takes 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. Interpolation constructs new data points within the range of a discrete set of known data points. 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 the 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]

arr[] ==> Array where elements need to be searched

x     ==> Element to be searched

lo    ==> Starting index in arr[]

hi    ==> Ending index in arr[]

There are many different interpolation methods and one such is known as linear interpolation. Linear interpolation takes two data points which we assume as (x1,y1) and (x2,y2) and the formula is :  at point(x,y).

This algorithm works in a way we search for a word in a dictionary. The interpolation search algorithm improves the binary search algorithm.  The formula for finding a value is: K = data-low/high-low.

K is a constant which is used to narrow the search space. In the case of binary search, the value for this constant is: K=(low+high)/2.

The formula for pos can be derived as follows.

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
The rest of the Interpolation algorithm is the same except for 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 the algorithm.

## C

 // C program to implement interpolation search// with recursion#include  // 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 Codeint 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;}

## C++

 // C++ program to implement interpolation// search with recursion#include 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 Codeint 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

## Java

 // Java program to implement interpolation// search with recursionimport 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

## Python3

 # 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 conductedarr = [10, 12, 13, 16, 18, 19, 20,       21, 22, 23, 24, 33, 35, 42, 47]n = len(arr) # Element to be searchedx = 18index = 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 searchusing 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

 

## PHP

 = $arr[$lo] && $x <= $arr[$hi]) { // Probing the position with keeping // uniform distribution in mind. $pos = (int)($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, $x);    }    return -1;} // Driver Code// Array of items on which search will// be conducted.$arr = array(10, 12, 13, 16, 18, 19, 20, 21, 22, 23, 24, 33, 35, 42, 47);$n = sizeof($arr); $x = 47; // Element to be searched$index = interpolationSearch($arr, 0, $n - 1, $x); // If element was foundif ($index != -1) echo "Element found at index ".$index;else    echo "Element not found.";return 0;#This code is contributed by Susobhan Akhuli?>

Output
Element found at index 4

Time Complexity: O(log2(log2 n)) for the average case, and O(n) for the worst case
Auxiliary Space Complexity: O(1)

#### Another approach:-

This is the iteration approach for the interpolation search.

• 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 the algorithm.

## C++

 // C++ program to implement interpolation search by using iteration approach#includeusing namespace std;   int interpolationSearch(int arr[], int n, int x){    // Find indexes of two corners    int low = 0, high = (n - 1);    // Since array is sorted, an element present    // in array must be in range defined by corner    while (low <= high && x >= arr[low] && x <= arr[high])    {        if (low == high)        {if (arr[low] == x) return low;        return -1;        }        // Probing the position with keeping        // uniform distribution in mind.        int pos = low + (((double)(high - low) /            (arr[high] - arr[low])) * (x - arr[low]));           // Condition of target found        if (arr[pos] == x)            return pos;        // If x is larger, x is in upper part        if (arr[pos] < x)            low = pos + 1;        // If x is smaller, x is in the lower part        else            high = pos - 1;    }    return -1;}   // Main functionint main(){    // Array of items on whighch 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 contributed by  Ajay Singh

## Python3

 # Python equivalent of above C++ code # Python program to implement interpolation search by using iteration approachdef interpolationSearch(arr, n, x):        # Find indexes of two corners     low = 0    high = (n - 1)        # Since array is sorted, an element present     # in array must be in range defined by corner     while low <= high and x >= arr[low] and x <= arr[high]:         if low == high:             if arr[low] == x:                 return low;             return -1;            # Probing the position with keeping         # uniform distribution in mind.         pos = int(low + (((float(high - low)/( arr[high] - arr[low])) * (x - arr[low]))))            # Condition of target found         if arr[pos] == x:             return pos            # If x is larger, x is in upper part         if arr[pos] < x:             low = pos + 1;            # If x is smaller, x is in lower part         else:             high = pos - 1;            return -1   # Main functionif __name__ == "__main__":    # Array of items on whighch 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 element was found    if index != -1:         print ("Element found at index",index)    else:         print ("Element not found")

## Javascript

 // JavaScript program to implement interpolation search by using iteration approach function interpolationSearch(arr, n, x) {// Find indexes of two cornerslet low = 0;let high = n - 1; // Since array is sorted, an element present// in array must be in range defined by cornerwhile (low <= high && x >= arr[low] && x <= arr[high]) {    if (low == high) {        if (arr[low] == x) {            return low;        }        return -1;    }     // Probing the position with keeping    // uniform distribution in mind.    let pos = Math.floor(low + (((high - low) / (arr[high] - arr[low])) * (x - arr[low])));     // Condition of target found    if (arr[pos] == x) {        return pos;    }     // If x is larger, x is in upper part    if (arr[pos] < x) {        low = pos + 1;    }     // If x is smaller, x is in lower part    else {        high = pos - 1;    }} return -1;}// Main functionlet arr = [10, 12, 13, 16, 18, 19, 20, 21, 22, 23, 24, 33, 35, 42, 47];let n = arr.length; let x = 18; // Element to be searchedlet index = interpolationSearch(arr, n, x); // If element was foundif (index != -1) {console.log("Element found at index", index);} else {console.log("Element not found");}

## C#

 // C# program to implement interpolation search by using// iteration approachusing System; class Program{    // Interpolation Search function    static int InterpolationSearch(int[] arr, int n, int x)    {        int low = 0;        int high = n - 1;           while (low <= high && x >= arr[low] && x <= arr[high])         {            if (low == high)             {                if (arr[low] == x)                     return low;                 return -1;             }               int pos = low + (int)(((float)(high - low) / (arr[high] - arr[low])) * (x - arr[low]));               if (arr[pos] == x)                 return pos;                if (arr[pos] < x)                 low = pos + 1;                else                high = pos - 1;         }           return -1;    }       // Main function    static void Main(string[] args)    {        int[] arr = {10, 12, 13, 16, 18, 19, 20, 21, 22, 23, 24, 33, 35, 42, 47};        int n = arr.Length;           int x = 18;        int index = InterpolationSearch(arr, n, x);           if (index != -1)             Console.WriteLine("Element found at index " + index);        else            Console.WriteLine("Element not found");    }} // This code is contributed by Susobhan Akhuli

## Java

 // Java program to implement interpolation// search with recursionimport 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;         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.");    }}

Output
Element found at index 4

Time Complexity: O(log2(log2 n)) for the average case, and O(n) for the worst case
Auxiliary Space Complexity: O(1)

Previous
Next