Dados dos enteros N y K , la tarea es encontrar una array de tamaño K que contenga solo elementos pares o impares donde la suma de todos los elementos de la array sea N. Si no existe tal array, escriba «No».
Ejemplos:
Entrada: N = 18, K = 3
Salida: 6 6 6
Entrada: N = 19, K = 5
Salida: 3 3 3 3 7
Enfoque: La idea es elegir el número par o impar más pequeño K-1 veces y finalmente, calcular el último número con la ayuda de la suma total. Si el último número también es par para el número par e impar para el número impar más pequeño. Entonces es posible elegir tal array. De lo contrario, no existe tal array posible.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation to find an
// array of size K with all the
// even or odd elements in the array
#include <bits/stdc++.h>
using namespace std;
// Function to find the array with
// all the even / odd elements
void getArrayOfSizeK(int n, int k)
{
vector<int> ans;
// If array could be constructed
// by adding odd elements
// we need to find the kth
// element is odd or even
// if array of odd elements
// would be the answer then
// k-1 elements would be 1
// First let's check
// kth is odd or even
int odd = n - ((k - 1) * 1);
// if last element is also
// an odd number then
// we can choose odd
// elements for our answer
if (odd > 0 && odd % 2 != 0) {
// Add 1 in the array (k-1) times
for (int i = 0; i < k - 1; i++) {
ans.push_back(1);
}
// Add last odd element
ans.push_back(odd);
}
// If array of even elements
// would be the answer then
// k-1 elements would be 2
int even = n - ((k - 1) * 2);
// if last element is also
// an even number then
// we can choose even
// elements for our answer
if (even > 0 && even % 2 == 0 && ans.size() == 0) {
// Add 2 in the array (k-1) times
for (int i = 0; i < k - 1; i++) {
ans.push_back(2);
}
// Add last even element
ans.push_back(even);
}
// Printing the array
if (ans.size() > 0) {
for (int i = 0; i < k; i++) {
cout << ans[i] << " ";
}
}
else {
cout << "NO" << endl;
}
}
// Driver Code
int main()
{
int n = 10, k = 3;
getArrayOfSizeK(n, k);
return 0;
}
Java
// Java implementation to find an
// array of size K with all the
// even or odd elements in the array
import java.util.*;
class GFG {
// Function to find the array with
// all the even / odd elements
static void getArrayOfSizeK(int n, int k)
{
Vector<Integer> ans = new Vector<Integer>();
// If array could be constructed
// by adding odd elements
// we need to find the kth
// element is odd or even
// If array of odd elements
// would be the answer then
// k-1 elements would be 1
// First let's check
// kth is odd or even
int odd = n - ((k - 1) * 1);
// If last element is also
// an odd number then
// we can choose odd
// elements for our answer
if (odd > 0 && odd % 2 != 0) {
// Add 1 in the array (k-1) times
for (int i = 0; i < k - 1; i++) {
ans.add(1);
}
// Add last odd element
ans.add(odd);
}
// If array of even elements
// would be the answer then
// k-1 elements would be 2
int even = n - ((k - 1) * 2);
// If last element is also
// an even number then
// we can choose even
// elements for our answer
if (even > 0 && even % 2 == 0 && ans.size() == 0) {
// Add 2 in the array (k-1) times
for (int i = 0; i < k - 1; i++) {
ans.add(2);
}
// Add last even element
ans.add(even);
}
// Printing the array
if (ans.size() > 0) {
for (int i = 0; i < k; i++) {
System.out.print(ans.get(i) + " ");
}
}
else {
System.out.println("NO");
}
}
// Driver code
public static void main(String args[])
{
int n = 10, k = 3;
getArrayOfSizeK(n, k);
}
}
// This code is contributed by Surendra_Gangwar
Python3
# Python 3 implementation to find an
# array of size K with all the
# even or odd elements in the array
# Function to find the array with
# all the even / odd elements
def getArrayOfSizeK(n, k):
ans = []
# If array could be constructed
# by adding odd elements
# we need to find the kth
# element is odd or even
# if array of odd elements
# would be the answer then
# k-1 elements would be 1
# First let's check
# kth is odd or even
odd = n - ((k - 1) * 1)
# if last element is also
# an odd number then
# we can choose odd
# elements for our answer
if (odd > 0
and odd % 2 != 0):
# Add 1 in the array (k-1) times
for i in range(k - 1):
ans.append(1)
# Add last odd element
ans.append(odd)
# If array of even elements
# would be the answer then
# k-1 elements would be 2
even = n - ((k - 1) * 2)
# if last element is also
# an even number then
# we can choose even
# elements for our answer
if (even > 0
and even % 2 == 0
and len(ans) == 0):
# Add 2 in the array (k-1) times
for i in range(k - 1):
ans.append(2)
# Add last even element
ans.append(even)
# Printing the array
if (len(ans) > 0):
for i in range(k):
print(ans[i], end=" ")
else:
print("NO")
# Driver Code
if __name__ == "__main__":
n, k = 10, 3
getArrayOfSizeK(n, k)
# This code is contributed by chitranayal
C#
// C# implementation to find an
// array of size K with all the
// even or odd elements in the array
using System;
using System.Collections.Generic;
class GFG {
// Function to find the array with
// all the even / odd elements
static void getArrayOfSizeK(int n, int k)
{
List<int> ans = new List<int>();
// If array could be constructed
// by adding odd elements
// we need to find the kth
// element is odd or even
// If array of odd elements
// would be the answer then
// k-1 elements would be 1
// First let's check
// kth is odd or even
int odd = n - ((k - 1) * 1);
// If last element is also
// an odd number then
// we can choose odd
// elements for our answer
if (odd > 0 && odd % 2 != 0) {
// Add 1 in the array (k-1) times
for (int i = 0; i < k - 1; i++) {
ans.Add(1);
}
// Add last odd element
ans.Add(odd);
}
// If array of even elements
// would be the answer then
// k-1 elements would be 2
int even = n - ((k - 1) * 2);
// If last element is also
// an even number then
// we can choose even
// elements for our answer
if (even > 0 && even % 2 == 0 && ans.Count == 0) {
// Add 2 in the array (k-1) times
for (int i = 0; i < k - 1; i++) {
ans.Add(2);
}
// Add last even element
ans.Add(even);
}
// Printing the array
if (ans.Count > 0) {
for (int i = 0; i < k; i++) {
Console.Write(ans[i] + " ");
}
}
else {
Console.WriteLine("NO");
}
}
// Driver code
public static void Main(String[] args)
{
int n = 10, k = 3;
getArrayOfSizeK(n, k);
}
}
// This code is contributed by amal kumar choubey
Javascript
<script>
// Javascript implementation to find an
// array of size K with all the
// even or odd elements in the array
// Function to find the array with
// all the even / odd elements
function getArrayOfSizeK(n, k)
{
let ans = [];
// If array could be constructed
// by adding odd elements
// we need to find the kth
// element is odd or even
// If array of odd elements
// would be the answer then
// k-1 elements would be 1
// First let's check
// kth is odd or even
let odd = n - ((k - 1) * 1);
// If last element is also
// an odd number then
// we can choose odd
// elements for our answer
if (odd > 0 && odd % 2 != 0)
{
// Add 1 in the array (k-1) times
for(let i = 0; i < k - 1; i++)
{
ans.push(1);
}
// Add last odd element
ans.push(odd);
}
// If array of even elements
// would be the answer then
// k-1 elements would be 2
let even = n - ((k - 1) * 2);
// If last element is also
// an even number then
// we can choose even
// elements for our answer
if (even > 0 && even % 2 == 0 &&
ans.length == 0)
{
// Add 2 in the array (k-1) times
for(let i = 0; i < k - 1; i++)
{
ans.push(2);
}
// Add last even element
ans.push(even);
}
// Printing the array
if (ans.length > 0)
{
for(let i = 0; i < k; i++)
{
document.write(ans[i] + " ");
}
}
else
{
document.write("NO");
}
}
// Driver Code
let n = 10, k = 3;
getArrayOfSizeK(n, k);
// This code is contributed by target_2.
</script>
Go
// Go implementation to find an
// array of size K with all the
// even or odd elements in the array
package main
import "fmt"
// Function to find the array with
// all the even / odd elements
func getArrayOfSizeK(n, k int) {
ans := make([]int, 0)
// If array could be constructed
// by adding odd elements
// we need to find the kth
// element is odd or even
// If array of odd elements
// would be the answer then
// k-1 elements would be 1
// First let's check
// kth is odd or even
odd := n - ((k - 1) * 1)
if odd > 0 && odd % 2 != 0 {
// Add 1 in the array (k-1) times
for i := 0; i < k - 1; i++ {
ans = append(ans, 1)
}
// Add last odd element
ans = append(ans, odd)
}
// If array of even elements
// would be the answer then
// k-1 elements would be 2
even := n - ((k - 1) * 2)
// if last element is also
// an even number then
// we can choose even
// elements for our answer
if even > 0 && even % 2 == 0 && len(ans) == 0 {
// Add 2 in the array (k-1) times
for i := 0; i < k - 1; i++ {
ans = append(ans, 2)
}
// Add last even element
ans = append(ans, even)
}
// Printing the array
if len(ans) > 0 {
for i := 0; i < k; i++ {
fmt.Print(ans[i], " ")
}
} else {
fmt.Println("NO")
}
}
// Driver Code
func main() {
n, k := 10, 3;
getArrayOfSizeK(n, k);
}
Producción:
2 2 6
Complejidad de tiempo: O(k), donde k representa el entero dado.
Espacio Auxiliar: O(k), donde k representa el entero dado.
Publicación traducida automáticamente
Artículo escrito por raghav9352 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA