Dada una array arr[] de N enteros, la tarea es encontrar la diferencia máxima entre dos elementos cualquiera de la array.
Ejemplos:
Entrada: arr[] = {2, 1, 5, 3}
Salida: 4
|5 – 1| = 4
Entrada: arr[] = {-10, 4, -9, -5}
Salida: 14
Enfoque: La diferencia absoluta máxima en la array siempre será la diferencia absoluta entre el elemento mínimo y máximo de la array.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the maximum
// absolute difference between
// any two elements of the array
int maxAbsDiff(int arr[], int n)
{
// To store the minimum and the maximum
// elements from the array
int minEle = arr[0];
int maxEle = arr[0];
for (int i = 1; i < n; i++) {
minEle = min(minEle, arr[i]);
maxEle = max(maxEle, arr[i]);
}
return (maxEle - minEle);
}
// Driver code
int main()
{
int arr[] = { 2, 1, 5, 3 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << maxAbsDiff(arr, n);
return 0;
}
Java
// Java implementation of the approach
class GFG {
// Function to return the maximum
// absolute difference between
// any two elements of the array
static int maxAbsDiff(int arr[], int n)
{
// To store the minimum and the maximum
// elements from the array
int minEle = arr[0];
int maxEle = arr[0];
for (int i = 1; i < n; i++) {
minEle = Math.min(minEle, arr[i]);
maxEle = Math.max(maxEle, arr[i]);
}
return (maxEle - minEle);
}
// Driver code
public static void main(String[] args)
{
int[] arr = { 2, 1, 5, 3 };
int n = arr.length;
System.out.print(maxAbsDiff(arr, n));
}
}
Python3
# Python3 implementation of the approach # Function to return the maximum # absolute difference between # any two elements of the array def maxAbsDiff(arr, n): # To store the minimum and the maximum # elements from the array minEle = arr[0] maxEle = arr[0] for i in range(1, n): minEle = min(minEle, arr[i]) maxEle = max(maxEle, arr[i]) return (maxEle - minEle) # Driver code arr = [2, 1, 5, 3] n = len(arr) print(maxAbsDiff(arr, n)) # This code is contributed # by mohit kumar
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the maximum
// absolute difference between
// any two elements of the array
static int maxAbsDiff(int []arr, int n)
{
// To store the minimum and the maximum
// elements from the array
int minEle = arr[0];
int maxEle = arr[0];
for (int i = 1; i < n; i++)
{
minEle = Math.Min(minEle, arr[i]);
maxEle = Math.Max(maxEle, arr[i]);
}
return (maxEle - minEle);
}
// Driver code
public static void Main()
{
int[] arr = { 2, 1, 5, 3 };
int n = arr.Length;
Console.WriteLine(maxAbsDiff(arr, n));
}
}
// This code is contributed by Ryuga
PHP
<?php
// PHP implementation of the approach
// Function to return the maximum
// absolute difference between
// any two elements of the array
function maxAbsDiff($arr, $n)
{
// To store the minimum and the maximum
// elements from the array
$minEle = $arr[0];
$maxEle = $arr[0];
for ($i = 1; $i < $n; $i++)
{
$minEle = min($minEle, $arr[$i]);
$maxEle = max($maxEle, $arr[$i]);
}
return ($maxEle - $minEle);
}
// Driver code
$arr = array(2, 1, 5, 3);
$n = sizeof($arr);
echo maxAbsDiff($arr, $n);
// This code is contributed
// by Akanksha Rai
Javascript
<script>
// JavaScript implementation of the approach
// Function to return the maximum
// absolute difference between
// any two elements of the array
function maxAbsDiff(arr, n)
{
// To store the minimum and the maximum
// elements from the array
let minEle = arr[0];
let maxEle = arr[0];
for (let i = 1; i < n; i++) {
minEle = Math.min(minEle, arr[i]);
maxEle = Math.max(maxEle, arr[i]);
}
return (maxEle - minEle);
}
// Driver code
let arr = [ 2, 1, 5, 3 ];
let n = arr.length;
document.write(maxAbsDiff(arr, n));
</script>
Producción
4
Complejidad de tiempo : O(n)
Espacio Auxiliar : O(1)
Otro enfoque (usando STL) : la diferencia absoluta máxima en la array siempre será la diferencia absoluta entre el elemento mínimo y máximo de la array. A continuación se muestra la implementación del enfoque anterior:
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the maximum
// absolute difference between
// any two elements of the array
int maxAbsDiff(int arr[], int n)
{
// To find the minimum and the maximum element
// using stl
int maxele = *max_element(arr, arr + n);
int minele = *min_element(arr, arr + n);
// make variable to store answer
int ans = abs(maxele - minele);
return ans;
}
// Driver code
int main()
{
int arr[] = { -10, 4, -9, -5 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << maxAbsDiff(arr, n);
return 0;
}
Java
//Java program for the above approach
import java.util.Arrays;
class GFG {
// Function to return the maximum
// absolute difference between
// any two elements of the array
public static int maxAbsDiff(int arr[], int n)
{
// To find the minimum and the maximum element
// using stl
int maxele = Arrays.stream(arr).max().getAsInt();
int minele = Arrays.stream(arr).min().getAsInt();
// make variable to store answer
int ans = Math.abs(maxele - minele);
return ans;
}
// Driver code
public static void main (String[] args)
{
int arr[] = { -10, 4, -9, -5 };
int n = arr.length;
System.out.print(maxAbsDiff(arr, n));
}
}
// This code is contributed by Shubham Singh
Python3
# Python program for the above approach # Function to return the maximum # absolute difference between # any two elements of the array def maxAbsDiff(arr, n): # To find the minimum and the maximum element maxele = max(arr) minele = min(arr) # make variable to store answer ans = abs(maxele - minele) return ans # Driver code arr = [ -10, 4, -9, -5 ] n = len(arr) print(maxAbsDiff(arr, n)) # This code is contributed by Shubham Singh
C#
// C# program for the above approach
using System;
using System.Linq;
public class GFG{
// Function to return the maximum
// absolute difference between
// any two elements of the array
public static int maxAbsDiff(int[] arr, int n)
{
// To find the minimum and the maximum element
int maxele = arr.Max();
int minele = arr.Min();
// make variable to store answer
int ans = Math.Abs(maxele - minele);
return ans;
}
// Driver code
static public void Main ()
{
int[] arr = { -10, 4, -9, -5 };
int n = arr.Length;
Console.Write(maxAbsDiff(arr, n));
}
}
// This code is contributed by Shubham Singh
Javascript
<script>
// Javascript program for the above approach
// Function to return the maximum
// absolute difference between
// any two elements of the array
function maxAbsDiff(arr, n)
{
// To find the minimum and the maximum element
var maxele = Math.max(...arr);
var minele = Math.min(...arr);
// make variable to store answer
var ans = Math.abs(maxele - minele);
return ans;
}
// Driver code
var arr = [ -10, 4, -9, -5 ];
var n = arr.Length;
document.write(maxAbsDiff(arr, n));
// This code is contributed by Shubham Singh
</script>
Producción
14
Complejidad de tiempo : O(n)
Publicación traducida automáticamente
Artículo escrito por facebookruppal y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA