# 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 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 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);       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 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);       // 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 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

## 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 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

 

## 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, $x);   // If element was found if ($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 #include using 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 function int 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);         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 approach def 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 function if __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 corners let low = 0; let 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.     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 function let 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 searched let index = interpolationSearch(arr, n, x);   // If element was found if (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 approach using 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 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;           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)

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