Given an array arr[] of size N, the task is to find the bitwise AND of all possible unordered pairs present in the given array.
Examples:
Input: arr[] = {1, 5, 3, 7}
Output: 1
Explanation:
All possible unordered pairs are (1, 5), (1, 3), (1, 7), (5, 3), (5, 7), (3, 7).
Bitwise AND of all possible pairs = ( 1 & 5 ) & ( 1 & 3 ) & ( 1 & 7 ) & ( 5 & 3 ) & ( 5 & 7 ) & ( 3 & 7 )
= 1 & 1 & 1 & 1 & 5 & 3
= 1
Therefore, the required output is 1.Input: arr[] = {4, 5, 12, 15}
Output: 4
Naive approach: The idea is to traverse the array and generate all possible pairs of the given array. Finally, print Bitwise AND of each element present in these pairs of the given array. Follow the steps below to solve the problem:
- Initialize a variable, say totalAND, to store Bitwise AND of each element from these pairs.
- Iterate over the array and generate all possible pairs (arr[i], arr[j]) from the given array.
- For each pair (arr[i], arr[j]), update the value of totalAND = (totalAND & arr[i] & arr[j]).
- Finally, print the value of totalAND.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate bitwise AND// of all pairs from the given arrayint TotalAndPair(int arr[], int N){ // Stores bitwise AND // of all possible pairs int totalAND = (1 << 30) - 1; // Generate all possible pairs for (int i = 0; i < N; i++) { for (int j = i + 1; j < N; j++) { // Calculate bitwise AND // of each pair totalAND &= arr[i] & arr[j]; } } return totalAND;}// Driver Codeint main(){ int arr[] = { 4, 5, 12, 15 }; int N = sizeof(arr) / sizeof(arr[0]); cout << TotalAndPair(arr, N);} |
Java
// Java program to implement// the above approachimport java.util.*;class GFG{// Function to calculate bitwise AND// of all pairs from the given arraystatic int TotalAndPair(int arr[], int N){ // Stores bitwise AND // of all possible pairs int totalAND = (1 << 30) - 1; // Generate all possible pairs for (int i = 0; i < N; i++) { for (int j = i + 1; j < N; j++) { // Calculate bitwise AND // of each pair totalAND &= arr[i] & arr[j]; } } return totalAND;}// Driver Codepublic static void main(String[] args){ int arr[] = { 4, 5, 12, 15 }; int N = arr.length; System.out.print(TotalAndPair(arr, N));}}// This code is contributed by shikhasingrajput |
Python3
# Python3 program to implement# the above approach# Function to calculate bitwise AND# of all pairs from the given arraydef TotalAndPair(arr, N): # Stores bitwise AND # of all possible pairs totalAND = (1 << 30) - 1 # Generate all possible pairs for i in range(N): for j in range(i + 1, N): # Calculate bitwise AND # of each pair totalAND &= (arr[i] & arr[j]) return totalAND# Driver Codeif __name__ == '__main__': arr=[4, 5, 12, 15] N = len(arr) print(TotalAndPair(arr, N)) # This code is contributed by mohit kumar 29 |
C#
// C# program to implement// the above approach using System; class GFG{ // Function to calculate bitwise AND// of all pairs from the given arraystatic int TotalAndPair(int[] arr, int N){ // Stores bitwise AND // of all possible pairs int totalAND = (1 << 30) - 1; // Generate all possible pairs for(int i = 0; i < N; i++) { for(int j = i + 1; j < N; j++) { // Calculate bitwise AND // of each pair totalAND &= arr[i] & arr[j]; } } return totalAND;} // Driver Codepublic static void Main(){ int[] arr = { 4, 5, 12, 15 }; int N = arr.Length; Console.Write(TotalAndPair(arr, N));}}// This code is contributed by sanjoy_62 |
Javascript
<script>// JavaScript program to implement// the above approach// Function to calculate bitwise AND// of all pairs from the given arrayfunction TotalAndPair(arr, N){ // Stores bitwise AND // of all possible pairs let totalAND = (1 << 30) - 1; // Generate all possible pairs for (let i = 0; i < N; i++) { for (let j = i + 1; j < N; j++) { // Calculate bitwise AND // of each pair totalAND &= arr[i] & arr[j]; } } return totalAND;}// Driver Code let arr = [ 4, 5, 12, 15 ]; let N = arr.length; document.write(TotalAndPair(arr, N));// This code is contributed by Surbhi Tyagi</script> |
Output:
4
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized based on the following observations:
- Considering an array of 4 elements, required Bitwise AND is as follows:
(arr[0] & arr[1]) & (arr[0] & arr[2]) & (arr[0] & arr[3]) & (arr[1] & arr[2]) & (arr[1] & arr[3]) & (arr[2] & arr[3])
- Since Bitwise AND follows Associative property, the above expression can be rearranged as:
(arr[0] & arr[0] & arr[0]) & (arr[1] & arr[1] & arr[1]) & (arr[2] & arr[2] & arr[2]) & (arr[3] & arr[3] & arr[3])
- It can be observed that each array element occurs exactly (N – 1) times in all possible pairs.
- Based on the X & X = X property of Bitwise AND operators, the above expression can be rearranged to arr[0] & arr[1] & arr[2] & arr[3], which is equal to the Bitwise AND of all elements of original array.
Follow the steps below to solve the problem:
- Initialize a variable totalAND to store the result.
- Traverse the given array.
- Calculate Bitwise AND of all unordered pairs by updating totalAND = totalAND & arr[i].
Below is the implementation of the above approach
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate bitwise AND// of all pairs from the given arrayint TotalAndPair(int arr[], int N){ // Stores bitwise AND // of all possible pairs int totalAND = (1 << 30) - 1; // Iterate over the array arr[] for (int i = 0; i < N; i++) { // Calculate bitwise AND // of each array element totalAND &= arr[i]; } return totalAND;}// Driver Codeint main(){ int arr[] = { 4, 5, 12, 15 }; int N = sizeof(arr) / sizeof(arr[0]); cout << TotalAndPair(arr, N);} |
Java
// Java program to implement// the above approachimport java.util.*;class GFG{// Function to calculate bitwise AND// of all pairs from the given arraystatic int TotalAndPair(int arr[], int N){ // Stores bitwise AND // of all possible pairs int totalAND = (1 << 30) - 1; // Iterate over the array arr[] for (int i = 0; i < N; i++) { // Calculate bitwise AND // of each array element totalAND &= arr[i]; } return totalAND;}// Driver Codepublic static void main(String[] args){ int arr[] = { 4, 5, 12, 15 }; int N = arr.length; System.out.print(TotalAndPair(arr, N));}}// This code is contributed by 29AjayKumar |
Python3
# Python program to implement# the above approach# Function to calculate bitwise AND# of all pairs from the given arraydef TotalAndPair(arr, N): # Stores bitwise AND # of all possible pairs totalAND = (1 << 30) - 1; # Iterate over the array arr for i in range(N): # Calculate bitwise AND # of each array element totalAND &= arr[i]; return totalAND;# Driver Codeif __name__ == '__main__': arr = [4, 5, 12, 15]; N = len(arr); print(TotalAndPair(arr, N)); # This code is contributed by 29AjayKumar |
C#
// C# program to implement// the above approachusing System;class GFG{// Function to calculate bitwise AND// of all pairs from the given arraystatic int TotalAndPair(int []arr, int N){ // Stores bitwise AND // of all possible pairs int totalAND = (1 << 30) - 1; // Iterate over the array arr[] for(int i = 0; i < N; i++) { // Calculate bitwise AND // of each array element totalAND &= arr[i]; } return totalAND;}// Driver Codepublic static void Main(String[] args){ int []arr = { 4, 5, 12, 15 }; int N = arr.Length; Console.Write(TotalAndPair(arr, N));}}// This code is contributed by AnkThon |
Javascript
<script>// JavaScript program to implement// the above approach// Function to calculate bitwise AND// of all pairs from the given arrayfunction TotalAndPair(arr, N){ // Stores bitwise AND // of all possible pairs let totalAND = (1 << 30) - 1; // Iterate over the array arr[] for (let i = 0; i < N; i++) { // Calculate bitwise AND // of each array element totalAND &= arr[i]; } return totalAND;}// Driver Code let arr = [ 4, 5, 12, 15 ]; let N = arr.length; document.write(TotalAndPair(arr, N));// This code is contributed by Surbhi Tyagi.</script> |
Output:
4
Time Complexity: O(N)
Auxiliary Space: O(1)
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