Dada una array arr[] de tamaño N , la tarea es convertir cada elemento de la array a 0 aplicando las siguientes operaciones exactamente tres veces:
- Seleccione un subarreglo .
- Incrementa cada elemento del subarreglo por el múltiplo entero de su longitud.
Finalmente, imprima el primer y último índice del subarreglo involucrado y los elementos agregados a cada índice del subarreglo en cada paso.
Ejemplos:
Entrada: arr[]= {1, 3, 2, 4}
Salida:
Operación 1: 1 1
Elementos agregados: 1
Operación 2: 3 4
Elementos agregados: 4 2
Operación 3: 2 4
Elementos agregados: -3 -6 -6
Explicación:
Paso 1: seleccione el subarreglo {arr[1]} y agregue -1 al único elemento del subarreglo. Por lo tanto, la array arr[] se modifica a {0, 3, 2, 4}.
Paso 2: Seleccione el subarreglo {arr[3], arr[4]} y agregue {4, 2} a los elementos del subarreglo respectivamente. Por lo tanto, la array arr[] se modifica a {0, 3, 6, 6}.
Paso 3: Seleccione el subarreglo {arr[2], arr[4]} y agregue {-3, -6, -6} a los elementos del subarreglo respectivamente. Por lo tanto, la array arr[] se modifica a {0, 0, 0, 0}Entrada: arr[] = { 5 }
Salida:
Operación 1 : 1 1
Elementos agregados: -5
Operación 2 : 1 1
Elementos agregados: 5
Operación 3 : 1 1
Elementos agregados: -5
Acercarse:
Operación 1: Seleccione el subarreglo {arr[1], .., arr[N]}. Establecer A[i] = A[i] – N * A[i] = (N – 1) * (-A[i]).
Después de la operación 1, cada elemento es un múltiplo de (N – 1).
Operación 2: Seleccione el subarreglo {arr[1], .. arr[N – 1]}. Sume/Reste un múltiplo de (N – 1) a todos los valores del subarreglo hasta que se reduzcan a 0.
Operación 3: Seleccione el subarreglo {arr[N]}, de tamaño 1. Sume/Reste un múltiplo de 1 para hacer A[ N] = 0.
-
- Imprimir 1 N . Sustraer
- restar y .
- restar
C++
// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to reduce all
// array elements to zero
void ConvertArray(int arr[], int N)
{
// If size of array is 1
if (N == 1) {
// First operation
cout << "Operation 1 : " << 1
<< " " << 1 << endl;
cout << "Added elements: "
<< -1 * arr[0] << endl;
cout << endl;
// 2nd Operation
cout << "Operation 2 : "
<< 1 << " " << 1 << endl;
cout << "Added elements: "
<< 1 * arr[0] << endl;
cout << endl;
// 3rd Operation
cout << "Operation 3 : "
<< 1 << " " << 1 << endl;
cout << "Added elements: "
<< -1 * arr[0] << endl;
}
// Otherwise
else {
// 1st Operation
cout << "Operation 1 : "
<< 1 << " " << N << endl;
cout << "Added elements: ";
for (int i = 0; i < N; i++) {
cout << -1 * arr[i] * N << " ";
}
cout << endl;
cout << endl;
// 2nd Operation
cout << "Operation 2 : "
<< 1 << " " << N - 1 << endl;
cout << "Added elements: ";
for (int i = 0; i < N - 1; i++) {
cout << arr[i] * (N - 1) << " ";
}
cout << endl;
cout << endl;
// 3rd Operation
cout << "Operation 3 : " << N
<< " " << N << endl;
cout << "Added elements: ";
cout << arr[N - 1] * (N - 1) << endl;
}
}
// Driver Code
int main()
{
// Input
int arr[] = { 1, 3, 2, 4 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function call to make all
// array elements equal to 0
ConvertArray(arr, N);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to reduce all
// array elements to zero
static void ConvertArray(int arr[], int N)
{
// If size of array is 1
if (N == 1) {
// First operation
System.out.println("Operation 1 : " + 1
+ " " + 1 );
System.out.println("Added elements: "
+ -1 * arr[0] );
System.out.println();
// 2nd Operation
System.out.println("Operation 2 : "
+ 1 + " " + 1 );
System.out.println("Added elements: "
+ 1 * arr[0] );
System.out.println();
// 3rd Operation
System.out.println("Operation 3 : "
+ 1 + " " + 1 );
System.out.println("Added elements: "
+ -1 * arr[0] );
}
// Otherwise
else {
// 1st Operation
System.out.println("Operation 1 : "
+ 1 + " " + N );
System.out.print("Added elements: ");
for (int i = 0; i < N; i++) {
System.out.print(-1 * arr[i] * N + " ");
}
System.out.println();
System.out.println();
// 2nd Operation
System.out.println("Operation 2 : "
+ 1 + " " + (N - 1) );
System.out.print("Added elements: ");
for (int i = 0; i < N - 1; i++) {
System.out.print(arr[i] * (N - 1) + " ");
}
System.out.println();
System.out.println();
// 3rd Operation
System.out.println("Operation 3 : " + N
+ " " + N );
System.out.print("Added elements: ");
System.out.println(arr[N - 1] * (N - 1) );
}
}
// Driver code
public static void main(String[] args)
{
// Input
int arr[] = { 1, 3, 2, 4 };
int N = arr.length;
// Function call to make all
// array elements equal to 0
ConvertArray(arr, N);
}
}
// This code is contributed by souravghosh0416.
Python3
# Python 3 program of the above approach
# Function to reduce all
# array elements to zero
def ConvertArray(arr, N):
# If size of array is 1
if (N == 1):
# First operation
print("Operation 1 :",1,1)
print("Added elements:",-1 * arr[0])
print("\n",end = "")
# 2nd Operation
print("Operation 2 :",1,1)
print("Added elements:",1 * arr[0])
print("\n",end = "")
# 3rd Operation
print("Operation 3 :",1,1)
print("Added elements:",-1 * arr[0])
print("\n",end = "")
# Otherwise
else:
# 1st Operation
print("Operation 1 :",1,N)
print("Added elements:",end = " ")
for i in range(N):
print(-1 * arr[i] * N,end = " ")
print("\n")
# 2nd Operation
print("Operation 2 :",1,N - 1)
print("Added elements:",end = " ")
for i in range(N - 1):
print(arr[i] * (N - 1),end = " ")
print("\n")
# 3rd Operation
print("Operation 3 : ",N,N)
print("Added elements:",end = " ")
print(arr[N - 1] * (N - 1))
# Driver Code
if __name__ == '__main__':
# Input
arr = [1, 3, 2, 4]
N = len(arr)
# Function call to make all
# array elements equal to 0
ConvertArray(arr, N)
# This code is contributed by ipg2016107.
C#
// C# program for the above approach
using System;
class GFG{
// Function to reduce all
// array elements to zero
static void ConvertArray(int[] arr, int N)
{
// If size of array is 1
if (N == 1) {
// First operation
Console.WriteLine("Operation 1 : " + 1
+ " " + 1 );
Console.WriteLine("Added elements: "
+ -1 * arr[0] );
Console.WriteLine();
// 2nd Operation
Console.WriteLine("Operation 2 : "
+ 1 + " " + 1 );
Console.WriteLine("Added elements: "
+ 1 * arr[0] );
Console.WriteLine();
// 3rd Operation
Console.WriteLine("Operation 3 : "
+ 1 + " " + 1 );
Console.WriteLine("Added elements: "
+ -1 * arr[0] );
}
// Otherwise
else {
// 1st Operation
Console.WriteLine("Operation 1 : "
+ 1 + " " + N );
Console.Write("Added elements: ");
for (int i = 0; i < N; i++) {
Console.Write(-1 * arr[i] * N + " ");
}
Console.WriteLine();
Console.WriteLine();
// 2nd Operation
Console.WriteLine("Operation 2 : "
+ 1 + " " + (N - 1) );
Console.Write("Added elements: ");
for (int i = 0; i < N - 1; i++) {
Console.Write(arr[i] * (N - 1) + " ");
}
Console.WriteLine();
Console.WriteLine();
// 3rd Operation
Console.WriteLine("Operation 3 : " + N
+ " " + N );
Console.Write("Added elements: ");
Console.WriteLine(arr[N - 1] * (N - 1) );
}
}
// Driver Code
public static void Main(string[] args)
{
// Input
int[] arr = { 1, 3, 2, 4 };
int N = arr.Length;
// Function call to make all
// array elements equal to 0
ConvertArray(arr, N);
}
}
// This code is contributed by code_hunt.
Javascript
<script>
// javascript program for the above approach
// Function to reduce all
// array elements to zero
function ConvertArray(arr, N)
{
// If size of array is 1
if (N == 1)
{
// First operation
document.write("Operation 1 : " + 1 + " " + 1+ "<br/>");
document.write("Added elements: " + -1 * arr[0]+"<br/>");
document.write("<br/>");
// 2nd Operation
document.write("Operation 2 : " + 1 + " " + 1+"<br/>");
document.write("Added elements: " + 1 * arr[0]+"<br/>");
document.write("<br/>");
// 3rd Operation
document.write("Operation 3 : " + 1 + " " + 1+"<br/>");
document.write("Added elements: " + -1 * arr[0]+"<br/>");
}
// Otherwise
else
{
// 1st Operation
document.write("Operation 1 : " + 1 + " " + N+"<br/>");
document.write("Added elements: ");
for (i = 0; i < N; i++) {
document.write(-1 * arr[i] * N + " ");
}
document.write("<br/>");
document.write("<br/>");
// 2nd Operation
document.write("Operation 2 : " + 1 + " " + (N - 1)+"<br/>");
document.write("Added elements: ");
for (i = 0; i < N - 1; i++) {
document.write(arr[i] * (N - 1) + " ");
}
document.write("<br/>");
document.write("<br/>");
// 3rd Operation
document.write("Operation 3 : " + N + " " + N+"<br/>");
document.write("Added elements: ");
document.write(arr[N - 1] * (N - 1)+"<br/>");
}
}
// Driver code
// Input
var arr = [ 1, 3, 2, 4 ];
var N = arr.length;
// Function call to make all
// array elements equal to 0
ConvertArray(arr, N);
// This code is contributed by todaysgaurav.
</script>
Operation 1 : 1 4 Added elements: -4 -12 -8 -16 Operation 2 : 1 3 Added elements: 3 9 6 Operation 3 : 4 4 Added elements: 12
Complejidad temporal: O(N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por sachinjain74754 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA