Dada una array, invierta cada sub-array formada por k elementos consecutivos.
Ejemplos:
Entrada: arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3.
Salida: [3, 2, 1, 6, 5, 4, 9, 8, 7 , 10]
Entrada: arr = [1, 2, 3, 4, 5, 6, 7], k = 5.
Salida: [5, 4, 3, 2, 1, 7, 6]
Acercarse:
- Usaremos la técnica de dos punteros para resolver este problema.
- Primero, inicializaremos nuestro primer puntero d con valor k-1 ( d = k-1 ) y una variable m con valor 2 ( m = 2 ).
- Ahora, iteraremos la array con nuestro segundo puntero i y verificaremos
- Si i < d , Intercambiar (arr[i], arr[d]) y disminuir d en 1. De lo contrario,
- Haz d = k * m – 1 , i = k * (m – 1) – 1 y m = m + 1.
A continuación se muestra la implementación del enfoque anterior.
C++
// C++ program to reverse every sub-array
// formed by consecutive k elements
#include<bits/stdc++.h>
using namespace std;
// Function to reverse every sub-array
// formed by consecutive k elements
void ReverseInGroup(int arr[], int n, int k)
{
if(n < k)
{
k = n;
}
// Initialize variables
int d = k - 1, m = 2;
int i = 0;
for(i = 0; i < n; i++)
{
if (i >= d)
{
// Update the variables
d = k * (m);
if(d >= n)
{
d = n;
}
i = k * (m - 1) - 1;
m++;
}
else
{
int t = arr[i];
arr[i] = arr[d];
arr[d] = t;
}
d = d - 1;
}
return;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int k = 3;
int n = sizeof(arr) / sizeof(arr[0]);
ReverseInGroup(arr, n, k);
for(int i = 0; i < n; i++)
cout << arr[i] << " ";
return 0;
}
// This code is contributed by Code_Mech
C
// C program to reverse every sub-array
// formed by consecutive k elements
#include<stdio.h>
// Function to reverse every sub-array
// formed by consecutive k elements
void ReverseInGroup(int arr[], int n, int k)
{
if(n<k)
{
k=n;
}
// Initialize variables
int d = k-1, m=2;
int i = 0;
for (i = 0; i < n; i++)
{
if (i >= d)
{
// Update the variables
d = k * (m);
if(d>=n)
{
d = n;
}
i = k * (m - 1)-1;
m++;
}
else
{
int t = arr[i];
arr[i] = arr[d];
arr[d] = t;
}
d = d - 1;
}
return;
}
// Driver code
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 7};
int k = 3;
int n = sizeof(arr) / sizeof(arr[0]);
ReverseInGroup(arr, n, k);
for (int i = 0; i < n; i++)
printf("%d ", arr[i]);
return 0;
}
Java
// Java program to reverse every sub-array
// formed by consecutive k elements
class GFG{
// Function to reverse every sub-array
// formed by consecutive k elements
static void ReverseInGroup(int arr[],
int n, int k)
{
if(n < k)
{
k = n;
}
// Initialize variables
int d = k - 1, m = 2;
int i = 0;
for(i = 0; i < n; i++)
{
if (i >= d)
{
// Update the variables
d = k * (m);
if(d >= n)
{
d = n;
}
i = k * (m - 1) - 1;
m++;
}
else
{
int t = arr[i];
arr[i] = arr[d];
arr[d] = t;
}
d = d - 1;
}
return;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int k = 3;
int n = arr.length;
ReverseInGroup(arr, n, k);
for(int i = 0; i < n; i++)
System.out.printf("%d ", arr[i]);
}
}
// This code is contributed by sapnasingh4991
Python3
# Python3 program to reverse # every sub-array formed by # consecutive k elements # Function to reverse every # sub-array formed by consecutive # k elements def ReverseInGroup(arr, n, k): if(n < k): k = n # Initialize variables d = k - 1 m = 2 i = 0 while i < n: if (i >= d): # Update the # variables d = k * (m) if(d >= n): d = n i = k * (m - 1) - 1 m += 1 else: arr[i], arr[d] = (arr[d], arr[i]) d = d - 1 i += 1 return # Driver code if __name__ == "__main__": arr = [1, 2, 3, 4, 5, 6, 7] k = 3 n = len(arr) ReverseInGroup(arr, n, k) for i in range(n): print(arr[i], end = " ") # This code is contributed by Chitranayal
C#
// C# program to reverse every sub-array
// formed by consecutive k elements
using System;
class GFG{
// Function to reverse every sub-array
// formed by consecutive k elements
static void ReverseInGroup(int []arr,
int n, int k)
{
if(n < k)
{
k = n;
}
// Initialize variables
int d = k - 1, m = 2;
int i = 0;
for(i = 0; i < n; i++)
{
if (i >= d)
{
// Update the variables
d = k * (m);
if(d >= n)
{
d = n;
}
i = k * (m - 1) - 1;
m++;
}
else
{
int t = arr[i];
arr[i] = arr[d];
arr[d] = t;
}
d = d - 1;
}
return;
}
// Driver code
public static void Main()
{
int []arr = { 1, 2, 3, 4, 5, 6, 7 };
int k = 3;
int n = arr.Length;
ReverseInGroup(arr, n, k);
for(int i = 0; i < n; i++)
Console.Write(arr[i] + " ");
}
}
// This code is contributed by Code_Mech
Javascript
<script>
// Javascript program to reverse every sub-array
// formed by consecutive k elements
// Function to reverse every sub-array
// formed by consecutive k elements
function ReverseInGroup(arr, n, k)
{
if(n < k)
{
k = n;
}
// Initialize variables
let d = k - 1, m = 2;
let i = 0;
for(i = 0; i < n; i++)
{
if (i >= d)
{
// Update the variables
d = k * (m);
if(d >= n)
{
d = n;
}
i = k * (m - 1) - 1;
m++;
}
else
{
let t = arr[i];
arr[i] = arr[d];
arr[d] = t;
}
d = d - 1;
}
return;
}
// Driver code
let arr = [ 1, 2, 3, 4, 5, 6, 7 ];
let k = 3;
let n = arr.length;
ReverseInGroup(arr, n, k);
for(let i = 0; i < n; i++)
document.write(arr[i] + " ");
// This code is contributed by _saurabh_jaiswal
</script>
Producción:
3 2 1 6 5 4 7
Complejidad temporal: O(N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por vaibhavgopanpalli18 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA