Búsqueda de interpolación

Dada una array ordenada de n valores uniformemente distribuidos arr[], escriba una función para buscar un elemento x en particular en la array. 
La búsqueda lineal encuentra el elemento en el tiempo O(n), la búsqueda por salto toma el tiempo O(√ n) y la búsqueda binaria toma el tiempo O(log n). 
La búsqueda de interpolación es una mejora con respecto a la búsqueda binariapor ejemplo, donde los valores en una array ordenada se distribuyen uniformemente. La interpolación construye nuevos puntos de datos dentro del rango de un conjunto discreto de puntos de datos conocidos. La búsqueda binaria siempre va al elemento central para verificar. Por otro lado, la búsqueda por interpolación puede ir a diferentes ubicaciones según el valor de la clave que se busca. Por ejemplo, si el valor de la clave está más cerca del último elemento, es probable que la búsqueda por interpolación comience la búsqueda hacia el lado final.
Para encontrar la posición a buscar, utiliza la siguiente fórmula. 

// 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[]

Hay muchos métodos de interpolación diferentes y uno de ellos se conoce como interpolación lineal. La interpolación lineal toma dos puntos de datos que asumimos como (x1,y1) y (x2,y2) y la fórmula es: en el punto (x,y).

Este algoritmo funciona de la misma manera que buscamos una palabra en un diccionario. El algoritmo de búsqueda por interpolación mejora el algoritmo de búsqueda binaria. La fórmula para encontrar un valor es: K = datos-bajo/alto-bajo.

K es una constante que se utiliza para reducir el espacio de búsqueda. En el caso de búsqueda binaria, el valor de esta constante es: K=(bajo+alto)/2.

La fórmula para pos se puede derivar de la siguiente manera.

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])

Algoritmo 
El resto del algoritmo de interpolación es el mismo excepto por la lógica de partición anterior. 
Paso 1: en un ciclo, calcule el valor de «pos» utilizando la fórmula de posición de la sonda. 
Paso 2: si es una coincidencia, devuelva el índice del elemento y salga. 
Paso 3: si el elemento es menor que arr[pos], calcule la posición de la sonda del subarreglo izquierdo. De lo contrario, calcule lo mismo en el subarreglo derecho. 
Paso 4: repita hasta que se encuentre una coincidencia o el subarreglo se reduzca a cero.
A continuación se muestra la implementación del algoritmo. 

// 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++ 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 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 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 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# 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

<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>

Element found at index 4

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