Dada una array de enteros ordenados. Necesitamos encontrar el valor más cercano al número dado. La array puede contener valores duplicados y números negativos.
Ejemplos:
Input : arr[] = {1, 2, 4, 5, 6, 6, 8, 9}
Target number = 11
Output : 9
9 is closest to 11 in given array
Input :arr[] = {2, 5, 6, 7, 8, 8, 9};
Target number = 4
Output : 5
Una solución simple es recorrer la array dada y realizar un seguimiento de la diferencia absoluta del elemento actual con cada elemento. Finalmente devuelve el elemento que tiene diferencia de absolución mínima.
Una solución eficiente es usar Binary Search .
C++
// CPP program to find element
// closest to given target.
#include <bits/stdc++.h>
using namespace std;
int getClosest(int, int, int);
// Returns element closest to target in arr[]
int findClosest(int arr[], int n, int target)
{
// Corner cases
if (target <= arr[0])
return arr[0];
if (target >= arr[n - 1])
return arr[n - 1];
// Doing binary search
int i = 0, j = n, mid = 0;
while (i < j) {
mid = (i + j) / 2;
if (arr[mid] == target)
return arr[mid];
/* If target is less than array element,
then search in left */
if (target < arr[mid]) {
// If target is greater than previous
// to mid, return closest of two
if (mid > 0 && target > arr[mid - 1])
return getClosest(arr[mid - 1],
arr[mid], target);
/* Repeat for left half */
j = mid;
}
// If target is greater than mid
else {
if (mid < n - 1 && target < arr[mid + 1])
return getClosest(arr[mid],
arr[mid + 1], target);
// update i
i = mid + 1;
}
}
// Only single element left after search
return arr[mid];
}
// Method to compare which one is the more close.
// We find the closest by taking the difference
// between the target and both values. It assumes
// that val2 is greater than val1 and target lies
// between these two.
int getClosest(int val1, int val2,
int target)
{
if (target - val1 >= val2 - target)
return val2;
else
return val1;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 4, 5, 6, 6, 8, 9 };
int n = sizeof(arr) / sizeof(arr[0]);
int target = 11;
cout << (findClosest(arr, n, target));
}
// This code is contributed bu Smitha Dinesh Semwal
Java
// Java program to find element closest to given target.
import java.util.*;
import java.lang.*;
import java.io.*;
class FindClosestNumber {
// Returns element closest to target in arr[]
public static int findClosest(int arr[], int target)
{
int n = arr.length;
// Corner cases
if (target <= arr[0])
return arr[0];
if (target >= arr[n - 1])
return arr[n - 1];
// Doing binary search
int i = 0, j = n, mid = 0;
while (i < j) {
mid = (i + j) / 2;
if (arr[mid] == target)
return arr[mid];
/* If target is less than array element,
then search in left */
if (target < arr[mid]) {
// If target is greater than previous
// to mid, return closest of two
if (mid > 0 && target > arr[mid - 1])
return getClosest(arr[mid - 1],
arr[mid], target);
/* Repeat for left half */
j = mid;
}
// If target is greater than mid
else {
if (mid < n-1 && target < arr[mid + 1])
return getClosest(arr[mid],
arr[mid + 1], target);
i = mid + 1; // update i
}
}
// Only single element left after search
return arr[mid];
}
// Method to compare which one is the more close
// We find the closest by taking the difference
// between the target and both values. It assumes
// that val2 is greater than val1 and target lies
// between these two.
public static int getClosest(int val1, int val2,
int target)
{
if (target - val1 >= val2 - target)
return val2;
else
return val1;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 2, 4, 5, 6, 6, 8, 9 };
int target = 11;
System.out.println(findClosest(arr, target));
}
}
Python3
# Python3 program to find element # closest to given target. # Returns element closest to target in arr[] def findClosest(arr, n, target): # Corner cases if (target <= arr[0]): return arr[0] if (target >= arr[n - 1]): return arr[n - 1] # Doing binary search i = 0; j = n; mid = 0 while (i < j): mid = (i + j) // 2 if (arr[mid] == target): return arr[mid] # If target is less than array # element, then search in left if (target < arr[mid]) : # If target is greater than previous # to mid, return closest of two if (mid > 0 and target > arr[mid - 1]): return getClosest(arr[mid - 1], arr[mid], target) # Repeat for left half j = mid # If target is greater than mid else : if (mid < n - 1 and target < arr[mid + 1]): return getClosest(arr[mid], arr[mid + 1], target) # update i i = mid + 1 # Only single element left after search return arr[mid] # Method to compare which one is the more close. # We find the closest by taking the difference # between the target and both values. It assumes # that val2 is greater than val1 and target lies # between these two. def getClosest(val1, val2, target): if (target - val1 >= val2 - target): return val2 else: return val1 # Driver code arr = [1, 2, 4, 5, 6, 6, 8, 9] n = len(arr) target = 11 print(findClosest(arr, n, target)) # This code is contributed by Smitha Dinesh Semwal
C#
// C# program to find element
// closest to given target.
using System;
class GFG
{
// Returns element closest
// to target in arr[]
public static int findClosest(int []arr,
int target)
{
int n = arr.Length;
// Corner cases
if (target <= arr[0])
return arr[0];
if (target >= arr[n - 1])
return arr[n - 1];
// Doing binary search
int i = 0, j = n, mid = 0;
while (i < j)
{
mid = (i + j) / 2;
if (arr[mid] == target)
return arr[mid];
/* If target is less
than array element,
then search in left */
if (target < arr[mid])
{
// If target is greater
// than previous to mid,
// return closest of two
if (mid > 0 && target > arr[mid - 1])
return getClosest(arr[mid - 1],
arr[mid], target);
/* Repeat for left half */
j = mid;
}
// If target is
// greater than mid
else
{
if (mid < n-1 && target < arr[mid + 1])
return getClosest(arr[mid],
arr[mid + 1], target);
i = mid + 1; // update i
}
}
// Only single element
// left after search
return arr[mid];
}
// Method to compare which one
// is the more close We find the
// closest by taking the difference
// between the target and both
// values. It assumes that val2 is
// greater than val1 and target
// lies between these two.
public static int getClosest(int val1, int val2,
int target)
{
if (target - val1 >= val2 - target)
return val2;
else
return val1;
}
// Driver code
public static void Main()
{
int []arr = {1, 2, 4, 5,
6, 6, 8, 9};
int target = 11;
Console.WriteLine(findClosest(arr, target));
}
}
// This code is contributed by anuj_67.
PHP
<?php
// PHP program to find element closest
// to given target.
// Returns element closest to target in arr[]
function findClosest($arr, $n, $target)
{
// Corner cases
if ($target <= $arr[0])
return $arr[0];
if ($target >= $arr[$n - 1])
return $arr[$n - 1];
// Doing binary search
$i = 0;
$j = $n;
$mid = 0;
while ($i < $j)
{
$mid = ($i + $j) / 2;
if ($arr[$mid] == $target)
return $arr[$mid];
/* If target is less than array element,
then search in left */
if ($target < $arr[$mid])
{
// If target is greater than previous
// to mid, return closest of two
if ($mid > 0 && $target > $arr[$mid - 1])
return getClosest($arr[$mid - 1],
$arr[$mid], $target);
/* Repeat for left half */
$j = $mid;
}
// If target is greater than mid
else
{
if ($mid < $n - 1 &&
$target < $arr[$mid + 1])
return getClosest($arr[$mid],
$arr[$mid + 1], $target);
// update i
$i = $mid + 1;
}
}
// Only single element left after search
return $arr[$mid];
}
// Method to compare which one is the more close.
// We find the closest by taking the difference
// between the target and both values. It assumes
// that val2 is greater than val1 and target lies
// between these two.
function getClosest($val1, $val2, $target)
{
if ($target - $val1 >= $val2 - $target)
return $val2;
else
return $val1;
}
// Driver code
$arr = array( 1, 2, 4, 5, 6, 6, 8, 9 );
$n = sizeof($arr);
$target = 11;
echo (findClosest($arr, $n, $target));
// This code is contributed bu Sachin.
?>
Javascript
<script>
// JavaScript program to find element
// closest to given target.
// Returns element closest to target in arr[]
function findClosest(arr, target)
{
let n = arr.length;
// Corner cases
if (target <= arr[0])
return arr[0];
if (target >= arr[n - 1])
return arr[n - 1];
// Doing binary search
let i = 0, j = n, mid = 0;
while (i < j)
{
mid = (i + j) / 2;
if (arr[mid] == target)
return arr[mid];
// If target is less than array
// element,then search in left
if (target < arr[mid])
{
// If target is greater than previous
// to mid, return closest of two
if (mid > 0 && target > arr[mid - 1])
return getClosest(arr[mid - 1],
arr[mid], target);
// Repeat for left half
j = mid;
}
// If target is greater than mid
else
{
if (mid < n - 1 && target < arr[mid + 1])
return getClosest(arr[mid],
arr[mid + 1],
target);
i = mid + 1; // update i
}
}
// Only single element left after search
return arr[mid];
}
// Method to compare which one is the more close
// We find the closest by taking the difference
// between the target and both values. It assumes
// that val2 is greater than val1 and target lies
// between these two.
function getClosest(val1, val2, target)
{
if (target - val1 >= val2 - target)
return val2;
else
return val1;
}
// Driver Code
let arr = [ 1, 2, 4, 5, 6, 6, 8, 9 ];
let target = 11;
document.write(findClosest(arr, target));
// This code is contributed by code_hunt
</script>
Producción:
9
Complejidad de tiempo: O(log n) (debido a la búsqueda binaria)
Espacio auxiliar: O(log n) (se crea una pila implícita debido a la recursividad)
Publicación traducida automáticamente
Artículo escrito por smarakchopdar y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA