Dada una array arr[] de tamaño N y un entero K . La tarea es multiplicar cada elemento de la array por K .
Ejemplos:
Entrada: arr[] = { 3, 4 }, K = 2
Salida: 6 8
Explicación: Los elementos se convierten en 3*2 = 6 y 4*2 = 8.Entrada: arr[] = { 0, 1, 2 }, K = 7
Salida: { 0, 7, 14 }
Enfoque: El problema dado se puede resolver siguiendo los siguientes pasos:
- Iterar a través de todos los elementos de la lista
- Multiplica cada elemento por K
- Devolvió la lista modificada
A continuación se muestra la implementación del enfoque anterior.
C++
// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to multiply all
// the elements of array by K
void multiplyAllByK(vector<int> arr, int K)
{
int N = arr.size();
// Loop to multiply all
// the array elements
for (int i = 0; i < N; i++) {
int x = arr[i];
arr[i] = K * x;
}
// Print the modified array
for (int i = 0; i < N; i++)
cout << (arr[i]) << " ";
}
// Driver code
int main()
{
vector<int> arr;
arr.push_back(3);
arr.push_back(4);
int K = 2;
multiplyAllByK(arr, K);
return 0;
}
// This code is contributed by lokeshpotta20.
Java
// Java code to implement above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to multiply all
// the elements of array by K
public static void multiplyAllByK(
ArrayList<Integer> arr, int K)
{
int N = arr.size();
// Loop to multiply all
// the array elements
for (int i = 0; i < N; i++) {
int x = arr.get(i);
arr.set(i, K * x);
}
// Print the modified array
for (int i = 0; i < N; i++)
System.out.print(arr.get(i) + " ");
}
// Driver code
public static void main(String[] args)
{
ArrayList<Integer> arr = new ArrayList<Integer>();
arr.add(3);
arr.add(4);
int K = 2;
multiplyAllByK(arr, K);
}
}
Python
# Python code to implement above approach # Function to multiply all # the elements of array by K def multiplyAllByK(arr, K): n = len(arr) for i in range(n): x = arr[i] arr[i] = K * x for i in range(n): print(arr[i], end = ' ') # Driver code arr = [3, 4] K = 2 multiplyAllByK(arr, K) # This code is contributed by Samim Hossain Mondal.
C#
// C# code to implement above approach
using System;
using System.Collections.Generic;
public class GFG {
// Function to multiply all
// the elements of array by K
public static void multiplyAllByK(
List<int> arr, int K)
{
int N = arr.Count;
// Loop to multiply all
// the array elements
for (int i = 0; i < N; i++) {
int x = arr[i];
arr[i] =( K * x);
}
// Print the modified array
for (int i = 0; i < N; i++)
Console.Write(arr[i] + " ");
}
// Driver code
public static void Main(String[] args)
{
List<int> arr = new List<int>();
arr.Add(3);
arr.Add(4);
int K = 2;
multiplyAllByK(arr, K);
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// JavaScript code for the above approach
// Function to multiply all
// the elements of array by K
const multiplyAllByK = (arr, K) => {
let N = arr.length;
// Loop to multiply all
// the array elements
for (let i = 0; i < N; i++) {
let x = arr[i];
arr[i] = K * x;
}
// Print the modified array
for (let i = 0; i < N; i++)
document.write(`${arr[i]} `);
}
// Driver code
let arr = [];
arr.push(3);
arr.push(4);
let K = 2;
multiplyAllByK(arr, K);
// This code is contributed by rakeshsahni
</script>
Producción
6 8
Complejidad temporal: O(N)
Espacio auxiliar: O(1)
Enfoque mediante la expresión lambda: esto también se puede implementar mediante la expresión lambda .
n -> n * K
donde n puede ser un elemento particular o una array completa.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ code to implement above approach
#include <iostream>
using namespace std;
// Function to multiply all
// the elements of array by K
void multiplyAllByK(int arr[], int K)
{
for(int i = 0; i < 2; i++)
arr[i] *= K;
for (int i = 0; i < 2; i++)
cout << arr[i] << " ";
}
// Driver code
int main()
{
int arr[2];
arr[0] = 3;
arr[1] = 4;
int K = 2;
multiplyAllByK(arr, K);
return 0;
}
// This code is contributed by Shubham Singh
C
// C code to implement above approach
#include <stdio.h>
// Function to multiply all
// the elements of array by K
void multiplyAllByK(int arr[], int K)
{
for(int i = 0; i < 2; i++) arr[i] *= K;
for (int i = 0; i<2; i++)
printf("%d ",arr[i]);
}
// Driver code
int main()
{
int arr[2];
arr[0] = 3;
arr[1] = 4;
int K = 2;
multiplyAllByK(arr, K);
return 0;
}
//This code is contributed by Shubham Singh
Java
// Java code to implement above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to multiply all
// the elements of array by K
public static void multiplyAllByK(
ArrayList<Integer> arr, int K)
{
arr.replaceAll(n -> n * K);
for (Integer x : arr)
System.out.print(x + " ");
}
// Driver code
public static void main(String[] args)
{
ArrayList<Integer> arr
= new ArrayList<Integer>();
arr.add(3);
arr.add(4);
int K = 2;
multiplyAllByK(arr, K);
}
}
Python3
# Python3 code to implement above approach # Function to multiply all # the elements of array by K def multiplyAllByK(arr, K): lambda_func = lambda n: n * K for i in range(len(arr)): print(lambda_func(arr[i]), end = ' ') # Driver code arr = [3, 4] K = 2 multiplyAllByK(arr, K) # This code is contributed by Samim Hossain Mondal.
C#
// C# code to implement above approach
using System;
using System.Collections.Generic;
using System.Linq;
public class GFG {
// Function to multiply all
// the elements of array by K
public static void multiplyAllByK(
List<int> arr, int K)
{
var temp = arr.Select(n => n * K);
foreach (int x in temp)
Console.Write(x + " ");
}
// Driver code
public static void Main(String[] args)
{
List<int> arr
= new List<int>();
arr.Add(3);
arr.Add(4);
int K = 2;
multiplyAllByK(arr, K);
}
}
// This code is contributed by shikhasingrajput
Javascript
<script>
// Javascript code to implement above approach
// Function to multiply all
// the elements of array by K
function multiplyAllByK(arr, K) {
arr = arr.map(k => {
return k * K
});
for (x of arr)
document.write(x + " ");
}
// Driver code
let arr = [];
arr.push(3);
arr.push(4);
let K = 2;
multiplyAllByK(arr, K);
// This code is contributed by saurabh_jaiswal.
</script>
Producción
6 8
Complejidad temporal: O(N)
Espacio auxiliar: O(1)