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 Kvoid 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 codeint 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 approachimport 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 Kdef 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 codearr = [3, 4]K = 2multiplyAllByK(arr, K)# This code is contributed by Samim Hossain Mondal. |
C#
// C# code to implement above approachusing 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 Kvoid 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 codeint 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 Kvoid 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 codeint 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 approachimport 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 Kdef multiplyAllByK(arr, K): lambda_func = lambda n: n * K for i in range(len(arr)): print(lambda_func(arr[i]), end = ' ')# Driver codearr = [3, 4]K = 2multiplyAllByK(arr, K)# This code is contributed by Samim Hossain Mondal. |
C#
// C# code to implement above approachusing 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 Kfunction multiplyAllByK(arr, K) { arr = arr.map(k => { return k * K }); for (x of arr) document.write(x + " ");}// Driver codelet 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)
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