Given an array of integers arr[] of N elements, the task is to generate another array having (Bitwise) AND of previous and next elements with the following exceptions.
- The first element is the bitwise AND of the first and the second element.
- The last element is the bitwise AND of the last and the second last element.
Examples:
Input: arr[] = {1, 2, 3, 4, 5, 6}
Output: 0 1 0 1 4 4
The new array will be {1 & 2, 1 & 3, 2 & 4, 3 & 5, 4 & 6, 5 & 6}Input: arr[] = {9, 8, 7}
Output: 8 1 0
Approach: The first and the second element of the new array can be calculated as arr[0] & arr[1] and arr[N – 1] & arr[N – 2] respectively. The rest of the elements can be calculated as arr[i – 1] & arr[i + 1].
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <iostream>using namespace std;// Function to generate the array that// satisfies the given conditionvoid generateArr(int arr[], int n){ // If there is only a single element // in the array if (n == 1) { cout << arr[0]; return; } // To store the generated array int barr[n]; // First element barr[0] = arr[0] & arr[1]; // Last element barr[n - 1] = arr[n - 1] & arr[n - 2]; // Rest of the elements for (int i = 1; i < n - 1; i++) barr[i] = arr[i - 1] & arr[i + 1]; // Print the generated array for (int i = 0; i < n; i++) cout << barr[i] << " ";}// Driver codeint main(){ int arr[] = { 1, 2, 3, 4, 5, 6 }; int n = sizeof(arr) / sizeof(arr[0]); generateArr(arr, n); return 0;} |
Java
// Java implementation of the approachimport java .io.*; class GFG { static void generateArr(int[] arr, int n) { // Nothing to do when array size is 1 if (n <= 1) return; // store current value of arr[0] // and update it int prev = arr[0]; arr[0] = arr[0] & arr[1]; // Update rest of the array elements for (int i = 1; i < n - 1; i++) { // Store current value of // next interaction int curr = arr[i]; // Update current value using // previous value arr[i] = prev & arr[i + 1]; // Update previous value prev = curr; } // Update last array element separately arr[n - 1] = prev & arr[n - 1]; } // Driver Code public static void main(String[] args) { int[] arr = { 1, 2, 3, 4, 5, 6 }; int n = arr.length; generateArr(arr, n); // Print the modified array for (int i = 0; i < n; i++) System.out.print(arr[i] + " "); } } // This code is contributed by Nikhil |
Python3
# Python3 implementation of the approach# Function to generate the array that# satisfies the given conditiondef generateArr(arr, n): # If there is only a single element # in the array if (n == 1): print(arr[0]); return; # To store the generated array barr = [0] * n; # First element barr[0] = arr[0] & arr[1]; # Last element barr[n - 1] = arr[n - 1] & arr[n - 2]; # Rest of the elements for i in range(1, n - 1): barr[i] = arr[i - 1] & arr[i + 1]; # Print the generated array for i in range(n): print(barr[i], end = " ");# Driver Codeif __name__ == '__main__': arr = [1, 2, 3, 4, 5, 6]; n = len(arr); generateArr(arr, n);# This code is contributed by 29AjayKumar |
C#
// C# implementation of the approachusing System;class GFG{static void generateArr(int[] arr, int n) { // Nothing to do when array size is 1 if (n <= 1) return; // store current value of arr[0] // and update it int prev = arr[0]; arr[0] = arr[0] & arr[1]; // Update rest of the array elements for (int i = 1; i < n - 1; i++) { // Store current value of // next interaction int curr = arr[i]; // Update current value using // previous value arr[i] = prev & arr[i + 1]; // Update previous value prev = curr; } // Update last array element separately arr[n - 1] = prev & arr[n - 1]; } // Driver Codestatic public void Main (){ int[] arr = { 1, 2, 3, 4, 5, 6 }; int n = arr.Length; generateArr(arr, n); // Print the modified array for (int i = 0; i < n; i++) Console.Write(arr[i] + " "); } } // This code is contributed by ajit. |
Javascript
<script> // Javascript implementation of the approach // Function to generate the array that // satisfies the given condition function generateArr(arr, n) { // If there is only a single element // in the array if (n == 1) { document.write(arr[0]); return; } // To store the generated array let barr = new Array(n); // First element barr[0] = arr[0] & arr[1]; // Last element barr[n - 1] = arr[n - 1] & arr[n - 2]; // Rest of the elements for (let i = 1; i < n - 1; i++) barr[i] = arr[i - 1] & arr[i + 1]; // Print the generated array for (let i = 0; i < n; i++) document.write(barr[i] + " "); } let arr = [ 1, 2, 3, 4, 5, 6 ]; let n = arr.length; generateArr(arr, n); </script> |
Output:
0 1 0 1 4 4
Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(N), where N is the size of the given array.
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