Dada una array arr[] , la tarea es incrementar todos los elementos impares en 1 y disminuir todos los pares en 1 .
Ejemplos:
Entrada: arr[] = {3, 6, 8}
Salida: 4 5 9Entrada: arr[] = {9, 7, 3}
Salida: 10 6 4
Enfoque: recorra la array elemento por elemento y, si la posición del elemento actual es impar, increméntela en 1; de lo contrario, disminuya en 1. Imprima el contenido de la array actualizada al final.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Utility function to print // the contents of an array void printArr(int arr[], int n) { for (int i = 0; i < n; i++) cout << arr[i] << " "; } // Function to increment all the odd // positioned elements by 1 and decrement // all the even positioned elements by 1 void updateArr(int arr[], int n) { for (int i = 0; i < n; i++) // If current element is odd positioned if ((i + 1) % 2 == 1) arr[i]++; // If even positioned else arr[i]--; // Print the updated array printArr(arr, n); } // Driver code int main() { int arr[] = { 3, 6, 8 }; int n = sizeof(arr) / sizeof(arr[0]); updateArr(arr, n); return 0; }
Java
// Java implementation of the approach class GfG { // Utility function to print // the contents of an array static void printArr(int arr[], int n) { for (int i = 0; i < n; i++) System.out.print(arr[i] + " "); } // Function to increment all the odd // positioned elements by 1 and decrement // all the even positioned elements by 1 static void updateArr(int arr[], int n) { for (int i = 0; i < n; i++) // If current element is odd positioned if ((i + 1) % 2 == 1) arr[i]++; // If even positioned else arr[i]--; // Print the updated array printArr(arr, n); } // Driver code public static void main(String[] args) { int arr[] = { 3, 6, 8 }; int n = arr.length; updateArr(arr, n); } } // This code is contributed by Prerna Saini
Python3
# Python3 implementation of the approach # Utility function to print # the contents of an array def printArr(arr, n): for i in range(0, n): print(arr[i], end = " "); # Function to increment all the odd # positioned elements by 1 and decrement # all the even positioned elements by 1 def updateArr(arr, n): for i in range(0, n): # If current element is odd positioned if ((i + 1) % 2 == 1): arr[i] += 1; # If even positioned else: arr[i] -= 1; # Print the updated array printArr(arr, n); # Driver code if __name__ == '__main__': arr = [3, 6, 8]; n = len(arr); updateArr(arr, n); # This code contributed by PrinciRaj1992
C#
// C# implementation of the approach class GfG { // Utility function to print // the contents of an array static void printArr(int []arr, int n) { for (int i = 0; i < n; i++) System.Console.Write(arr[i] + " "); } // Function to increment all the odd // positioned elements by 1 and decrement // all the even positioned elements by 1 static void updateArr(int []arr, int n) { for (int i = 0; i < n; i++) // If current element is odd positioned if ((i + 1) % 2 == 1) arr[i]++; // If even positioned else arr[i]--; // Print the updated array printArr(arr, n); } // Driver code static void Main() { int []arr = { 3, 6, 8 }; int n = arr.Length; updateArr(arr, n); } } // This code is contributed by mits
PHP
<?php // PHP implementation of the approach // Utility function to print // the contents of an array function printArr($arr, $n) { for ($i = 0; $i < $n; $i++) echo $arr[$i] . " "; } // Function to increment all the odd // positioned elements by 1 and decrement // all the even positioned elements by 1 function updateArr($arr, $n) { for ($i = 0; $i < $n; $i++) // If current element is odd positioned if (($i + 1) % 2 == 1) $arr[$i]++; // If even positioned else $arr[$i]--; // Print the updated array printArr($arr, $n); } // Driver code $arr = array( 3, 6, 8 ); $n = count($arr); updateArr($arr, $n); // This code is contributed by mits ?>
Javascript
<script> // javascript implementation of the approach // Utility function to print // the contents of an array function printArr(arr, n) { var i; for (i = 0; i < n; i++) document.write(arr[i] + " "); } // Function to increment all the odd // positioned elements by 1 and decrement // all the even positioned elements by 1 function updateArr(arr, n) { var i; for (i = 0; i < n; i++) // If current element is odd positioned if ((i + 1) % 2 == 1) arr[i]++; // If even positioned else arr[i]--; // Print the updated array printArr(arr, n); } // Driver code var arr = [3, 6, 8]; var n = arr.length; updateArr(arr, n); </script>
Producción:
4 5 9
Complejidad de tiempo: O(n)
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por ShobhangiSarkar y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA