# 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 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[]
```

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 ` `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); ` ` `  `    ``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 ` `#include ` ` `  `// 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); ` ` `  `    ``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]) ` `        ``{         ` ` `  `            ``if` `(lo == hi) ` `            ``{ ` `                ``if` `(arr[lo] == x) ``return` `lo; ` `                ``return` `-``1``; ` `            ``} ` `        `  `            ``// 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 the 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]: ` `        ``if` `lo ``=``=` `hi: ` `            ``if` `arr[lo] ``=``=` `x:  ` `                ``return` `lo; ` `            ``return` `-``1``; ` `         `  `        ``# 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 `

## C#

 `// C# program to implement  ` `// interpolation search ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``// 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]) ` `        ``{ ` `            ``if` `(lo == hi) ` `            ``{ ` `                ``if` `(arr[lo] == x) ``return` `lo; ` `                ``return` `-1; ` `            ``} ` ` `  `            ``// 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 the lower part ` `            ``else` `                ``hi = pos - 1; ` `        ``} ` `        ``return` `-1; ` `    ``} ` ` `  `    ``// Driver Code  ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int` `x = 18; ``// Element to be searched ` `        ``int` `index = interpolationSearch(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 anuj_67. `

## PHP

 `= ``\$arr``[``\$l``] ``and` `                        ``\$x` `<= ``\$arr``[``\$h``]) ` `    ``{ ` `        ``if` `(``\$l` `== ``\$h``) ` `        ``{ ` `              ``if` `(``\$arr``[``\$l``] == ``\$x``) ``return` `\$l``; ` `              ``return` `-1; ` `        ``} ` ` `  `        ``// Probing the position with keeping ` `        ``// uniform distribution in mind. ` `        ``\$m` `= ``intval``(``\$l` `+ ((``\$x` `- ``\$arr``[``\$l``]) * (``\$h` `- ``\$l``) /  ` `                               ``(``\$arr``[``\$h``] - ``\$arr``[``\$l``]))); ` `         `  `        ``// Condition of target found ` `        ``if` `(``\$arr``[``\$m``] == ``\$x``)  ` `        ``{ ` `            ``return` `\$m``; ` `        ``} ` `         `  `        ``// If x is larger, x is in upper part ` `        ``elseif` `(``\$arr``[``\$m``] < ``\$x``) ` `        ``{ ` `            ``\$l` `= ``\$m` `+ 1; ` `        ``}  ` `         `  `        ``// If x is smaller, x is in the lower part ` `        ``else` `        ``{ ` `            ``\$h` `= ``\$m` `- 1; ` `        ``} ` `    ``} ` `     `  `    ``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` `= ``count``(``\$arr``);  ` `\$x` `= 18; ``// Element to be searched  ` `\$index` `= interpolationSearch(``\$arr``, ``\$x``, ``\$n``);  ` ` `  `// If element was found  ` `if` `(``\$index` `!= -1)  ` `    ``echo` `"Element found at index "` `. ``\$index``;  ` `else` `    ``echo` `"Element not found."``;  ` ` `  `// This code is contributed by Deepika Pathak ` `?> `

Output:

```Element found at index 4
```

Recursive Solution :

## 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 `

## 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;  ` `}  `

## 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 `

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)

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