Generate an Array by multiplying each element of given Array by K

Last Updated : 19 Jan, 2022

Given an array arr[] of size N and an integer K. The task is to multiply each element of the array by K.

Examples :

Input: arr[] = { 3, 4 }, K = 2
Output: 6 8
Explanation: The elements become 3*2 = 6 and 4*2 = 8.

Input: arr[] = { 0, 1, 2 }, K = 7
Output: { 0, 7, 14 }

Approach: The given problem can be solved using the following steps :

Iterate through all the elements in the list
Multiply each element by K
Returned the modified list

Below is the implementation of the above approach.

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# 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>

Output

6 8

Time Complexity: O(N)
Auxiliary Space: O(1)

Approach Using Lambda Expression: This can also be implemented using lambda expression.

n -> n * K
where n can be a particular element, or complete array.

Below is the implementation of the above approach:

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 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# 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>

Output

6 8

Time Complexity: O(N)
Auxiliary Space: O(1)

Maximize count of unique array elements by incrementing array elements by K

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Generate an Array by multiplying each element of given Array by K

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