Dada una array, la tarea es encontrar los elementos mínimo y máximo en la array que son divisibles por un número dado k.
Ejemplos:
Input: arr[] = {12, 1235, 45, 67, 1}, k=5
Output: Minimum = 45, Maximum = 1235
Input: arr[] = {10, 1230, 45, 67, 1}, k=10
Output: Minimum = 10, Maximum = 1230
Acercarse:
- Tome una variable mínima que almacene el elemento mínimo e inicialícela con INT_MAX y compárela con cada elemento de la array y actualice el siguiente elemento mínimo que es divisible por k.
- Tome una variable máxima que almacene el elemento máximo e inicialícela con INT_MIN y compárela con cada elemento de la array y actualice el siguiente elemento máximo que es divisible por k.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the minimum element
int getMin(int arr[], int n, int k)
{
int res = INT_MAX;
for (int i = 0; i < n; i++) {
if (arr[i] % k == 0)
res = min(res, arr[i]);
}
return res;
}
// Function to find the maximum element
int getMax(int arr[], int n, int k)
{
int res = INT_MIN;
for (int i = 1; i < n; i++) {
if (arr[i] % k == 0)
res = max(res, arr[i]);
}
return res;
}
// Driver code
int main()
{
int arr[] = { 10, 1230, 45, 67, 1 };
int k = 10;
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Minimum element of array which is divisible by k: "
<< getMin(arr, n, k) << "\n";
cout << "Maximum element of array which is divisible by k: "
<< getMax(arr, n, k);
return 0;
}
C
// C implementation of the above approach
#include <stdio.h>
#include <limits.h>
int max(int a, int b)
{
int max = a;
if(max < b)
max = b;
return max;
}
int min(int a,int b)
{
int min = a;
if(min > b)
min = b;
return min;
}
// Function to find the minimum element
int getMin(int arr[], int n, int k)
{
int res = INT_MAX;
for (int i = 0; i < n; i++) {
if (arr[i] % k == 0)
res = min(res, arr[i]);
}
return res;
}
// Function to find the maximum element
int getMax(int arr[], int n, int k)
{
int res = INT_MIN;
for (int i = 1; i < n; i++) {
if (arr[i] % k == 0)
res = max(res, arr[i]);
}
return res;
}
// Driver code
int main()
{
int arr[] = { 10, 1230, 45, 67, 1 };
int k = 10;
int n = sizeof(arr) / sizeof(arr[0]);
printf("Minimum element of array which is divisible by k: %d\n",getMin(arr, n, k));
printf("Maximum element of array which is divisible by k: %d\n",getMax(arr, n, k));
return 0;
}
// This code is contributed by kothavvsaakash.
Java
//Java implementation of the above approach
class GFG {
// Function to find the minimum element
static int getMin(int arr[], int n, int k) {
int res = Integer.MAX_VALUE;
for (int i = 0; i < n; i++) {
if (arr[i] % k == 0) {
res = Math.min(res, arr[i]);
}
}
return res;
}
// Function to find the maximum element
static int getMax(int arr[], int n, int k) {
int res = Integer.MIN_VALUE;
for (int i = 1; i < n; i++) {
if (arr[i] % k == 0) {
res = Math.max(res, arr[i]);
}
}
return res;
}
// Driver code
public static void main(String[] args) {
int arr[] = {10, 1230, 45, 67, 1};
int k = 10;
int n = arr.length;
System.out.println("Minimum element of array which is divisible by k: "
+ getMin(arr, n, k));
System.out.println("Maximum element of array which is divisible by k: "
+ getMax(arr, n, k));
}
}
//This code contribute by Shikha Singh
Python 3
# Python 3 implementation of the
# above approach
import sys
# Function to find the minimum element
def getMin(arr, n, k):
res = sys.maxsize
for i in range(n):
if (arr[i] % k == 0):
res = min(res, arr[i])
return res
# Function to find the maximum element
def getMax(arr, n, k):
res = 0
for i in range(1, n):
if (arr[i] % k == 0):
res = max(res, arr[i])
return res
# Driver code
if __name__ == "__main__":
arr = [ 10, 1230, 45, 67, 1 ]
k = 10
n = len(arr)
print("Minimum element of array which",
"is divisible by k: ", getMin(arr, n, k))
print( "Maximum element of array which",
"is divisible by k: ", getMax(arr, n, k))
# This code is contributed
# by ChitraNayal
C#
// C# implementation of the above approach
using System;
class GFG
{
// Function to find the minimum element
static int getMin(int []arr, int n, int k)
{
int res = int.MaxValue;
for (int i = 0; i < n; i++)
{
if (arr[i] % k == 0)
{
res = Math.Min(res, arr[i]);
}
}
return res;
}
// Function to find the maximum element
static int getMax(int []arr, int n, int k)
{
int res = int.MinValue;
for (int i = 1; i < n; i++)
{
if (arr[i] % k == 0)
{
res = Math.Max(res, arr[i]);
}
}
return res;
}
// Driver code
static public void Main ()
{
int []arr = {10, 1230, 45, 67, 1};
int k = 10;
int n = arr.Length;
Console.WriteLine("Minimum element of array " +
"which is divisible by k: " +
getMin(arr, n, k));
Console.WriteLine("Maximum element of array " +
"which is divisible by k: " +
getMax(arr, n, k));
}
}
// This code is contributes by ajit
PHP
<?php
// PHP implementation of the above approach
// Function to find the minimum element
function getMin($arr, $n, $k)
{
$res = PHP_INT_MAX;
for ($i = 0; $i < $n; $i++)
{
if ($arr[$i] % $k == 0)
$res = min($res, $arr[$i]);
}
return $res;
}
// Function to find the maximum element
function getMax($arr, $n, $k)
{
$res = PHP_INT_MIN;
for ($i = 1; $i < $n; $i++)
{
if ($arr[$i] % $k == 0)
$res = max($res, $arr[$i]);
}
return $res;
}
// Driver code
$arr = array( 10, 1230, 45, 67, 1 );
$k = 10;
$n = sizeof($arr);
echo "Minimum element of array which is " .
"divisible by k: ", getMin($arr, $n, $k) , "\n";
echo "Maximum element of array which is " .
"divisible by k: ", getMax($arr, $n, $k);
// This code is contributed by akt_mit
?>
Javascript
<script>
// Javascript implementation of the above approach
// Function to find the minimum element
function getMin(arr, n, k)
{
let res = Number.MAX_VALUE;
for (let i = 0; i < n; i++) {
if (arr[i] % k == 0)
res = Math.min(res, arr[i]);
}
return res;
}
// Function to find the maximum element
function getMax(arr, n, k)
{
let res = Number.MIN_VALUE;
for (let i = 1; i < n; i++) {
if (arr[i] % k == 0)
res = Math.max(res, arr[i]);
}
return res;
}
// Driver code
let arr = [ 10, 1230, 45, 67, 1 ];
let k = 10;
let n = arr.length;
document.write("Minimum element of array which is divisible by k: "
+ getMin(arr, n, k) + "<br>");
document.write("Maximum element of array which is divisible by k: "
+ getMax(arr, n, k));
</script>
Producción:
Minimum element of array which is divisible by k: 10 Maximum element of array which is divisible by k: 1230
Publicación traducida automáticamente
Artículo escrito por VishalBachchas y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA