Construct the Array using given bitwise AND, OR and XOR Last Updated : 17 Apr, 2023 Comments Improve Suggest changes Like Article Like Report Given Bitwise AND, OR, and XOR of N elements of an array denoted by a, b, c. The task is to find the elements of the array. If there exists no such array then print "-1".Examples: Input: N = 3, a = 4, b = 6, c = 6. Output: {4, 4, 6} Explanation: Bitwise AND of array = 4 & 4 & 6 = 4 Bitwise OR of array = 4 | 4 | 6 = 6 Bitwise XOR of array = 4 ^ 4 ^ 6 = 6Input: N = 2, a = 4, b = 6, c = 6. Output: -1 Approach: For Bitwise AND, if ith bit is set in a, then in the array every element must have ith bit set because if even one element's ith bit is 0 then bitwise AND of the array will result in ith bit to be 0.Secondly, if ith bit is not set in a, then OR and XOR values need to be handled simultaneously. If ith bit is set in b, then at least one element must have ith bit set. So, set ith bit in the only first element of the array.Now, if ith bit was set in b then ith bit must be checked in c. If that bit is set in c then there's no problem as the first element's ith bit is already set so 1 ^ 0 = 1. If that bit is not set in c then set the ith bit of the second element. Now, there will not be any effect in b and for c, 1 ^ 1 will be 0.Then, just calculate Bitwise AND, OR, and XOR of the array to check if it's equal or not. If results are not equal then the array is not possible else given array is the answer. Below is the implementation of the approach: C++ // C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find the array void findArray(int n, int a, int b, int c) { int arr[n + 1] = {}; // Loop through all bits in number for (int bit = 30; bit >= 0; bit--) { // If bit is set in AND // then set it in every element // of the array int set = a & (1 << bit); if (set) { for (int i = 0; i < n; i++) arr[i] |= set; } // If bit is not set in AND else { // But set in b(OR) if (b & (1 << bit)) { // Set bit position // in first element arr[0] |= (1 << bit); // If bit is not set in c // then set it in second // element to keep xor as // zero for bit position if (!(c & (1 << bit))) { arr[1] |= (1 << bit); } } } } int aa = INT_MAX, bb = 0, cc = 0; // Calculate AND, OR // and XOR of array for (int i = 0; i < n; i++) { aa &= arr[i]; bb |= arr[i]; cc ^= arr[i]; } // Check if values are equal or not if (a == aa && b == bb && c == cc) { for (int i = 0; i < n; i++) cout << arr[i] << " "; } // If not, then array // is not possible else cout << "-1"; } // Driver Code int main() { // Given Bitwise AND, OR, and XOR int n = 3, a = 4, b = 6, c = 6; // Function Call findArray(n, a, b, c); return 0; } Java // Java program for the above approach import java.io.*; class GFG{ // Function to find the array static void findArray(int n, int a, int b, int c) { int arr[] = new int[n + 1]; // Loop through all bits in number for(int bit = 30; bit >= 0; bit--) { // If bit is set in AND // then set it in every element // of the array int set = a & (1 << bit); if (set != 0) { for(int i = 0; i < n; i++) arr[i] |= set; } // If bit is not set in AND else { // But set in b(OR) if ((b & (1 << bit)) != 0) { // Set bit position // in first element arr[0] |= (1 << bit); // If bit is not set in c // then set it in second // element to keep xor as // zero for bit position if ((c & (1 << bit)) == 0) { arr[1] |= (1 << bit); } } } } int aa = Integer.MAX_VALUE, bb = 0, cc = 0; // Calculate AND, OR // and XOR of array for(int i = 0; i < n; i++) { aa &= arr[i]; bb |= arr[i]; cc ^= arr[i]; } // Check if values are equal or not if (a == aa && b == bb && c == cc) { for(int i = 0; i < n; i++) System.out.print(arr[i] + " "); } // If not, then array // is not possible else System.out.println("-1"); } // Driver code public static void main(String[] args) { // Given Bitwise AND, OR, and XOR int n = 3, a = 4, b = 6, c = 6; // Function Call findArray(n, a, b, c); } } // This code is contributed by Pratima Pandey Python3 # Python3 program for # the above approach import sys # Function to find the array def findArray(n, a, b, c): arr = [0] * (n + 1) # Loop through all bits in number for bit in range (30, -1, -1): # If bit is set in AND # then set it in every element # of the array set = a & (1 << bit) if (set): for i in range (n): arr[i] |= set # If bit is not set in AND else : # But set in b(OR) if (b & (1 << bit)): # Set bit position # in first element arr[0] |= (1 << bit) # If bit is not set in c # then set it in second # element to keep xor as # zero for bit position if (not (c & (1 << bit))): arr[1] |= (1 << bit) aa = sys.maxsize bb = 0 cc = 0 # Calculate AND, OR # and XOR of array for i in range (n): aa &= arr[i] bb |= arr[i] cc ^= arr[i] # Check if values are equal or not if (a == aa and b == bb and c == cc): for i in range (n): print (arr[i], end = " ") # If not, then array # is not possible else: print ("-1") # Driver Code if __name__ =="__main__": # Given Bitwise AND, OR, and XOR n = 3 a = 4 b = 6 c = 6 # Function Call findArray(n, a, b, c) # This code is contributed by Chitranayal C# // C# program for the above approach using System; class GFG{ // Function to find the array static void findArray(int n, int a, int b, int c) { int []arr = new int[n + 1]; // Loop through all bits in number for(int bit = 30; bit >= 0; bit--) { // If bit is set in AND // then set it in every element // of the array int set = a & (1 << bit); if (set != 0) { for(int i = 0; i < n; i++) arr[i] |= set; } // If bit is not set in AND else { // But set in b(OR) if ((b & (1 << bit)) != 0) { // Set bit position // in first element arr[0] |= (1 << bit); // If bit is not set in c // then set it in second // element to keep xor as // zero for bit position if ((c & (1 << bit)) == 0) { arr[1] |= (1 << bit); } } } } int aa = int.MaxValue, bb = 0, cc = 0; // Calculate AND, OR // and XOR of array for(int i = 0; i < n; i++) { aa &= arr[i]; bb |= arr[i]; cc ^= arr[i]; } // Check if values are equal or not if (a == aa && b == bb && c == cc) { for(int i = 0; i < n; i++) Console.Write(arr[i] + " "); } // If not, then array // is not possible else Console.WriteLine("-1"); } // Driver code public static void Main(String[] args) { // Given Bitwise AND, OR, and XOR int n = 3, a = 4, b = 6, c = 6; // Function Call findArray(n, a, b, c); } } // This code is contributed by gauravrajput1 JavaScript <script> // Javascript program for the above approach // Function to find the array function findArray(n, a, b, c) { let arr = new Array(n + 1); // Loop through all bits in number for (let bit = 30; bit >= 0; bit--) { // If bit is set in AND // then set it in every element // of the array let set = a & (1 << bit); if (set) { for (let i = 0; i < n; i++) arr[i] |= set; } // If bit is not set in AND else { // But set in b(OR) if (b & (1 << bit)) { // Set bit position // in first element arr[0] |= (1 << bit); // If bit is not set in c // then set it in second // element to keep xor as // zero for bit position if (!(c & (1 << bit))) { arr[1] |= (1 << bit); } } } } let aa = Number.MAX_SAFE_INTEGER, bb = 0, cc = 0; // Calculate AND, OR // and XOR of array for (let i = 0; i < n; i++) { aa &= arr[i]; bb |= arr[i]; cc ^= arr[i]; } // Check if values are equal or not if (a == aa && b == bb && c == cc) { for (let i = 0; i < n; i++) document.write(arr[i] + " "); } // If not, then array // is not possible else document.write("-1"); } // Driver Code // Given Bitwise AND, OR, and XOR let n = 3, a = 4, b = 6, c = 6; // Function Call findArray(n, a, b, c); // This code is contributed by gfgking </script> Output: 6 4 4 Time Complexity: O(31*N) Auxiliary Space: O(N) Comment More infoAdvertise with us Next Article Construct the Array using given bitwise AND, OR and XOR sarthak_eddy Follow Improve Article Tags : Bit Magic Algorithms Competitive Programming DSA Arrays Bitwise-XOR Bitwise-OR Bitwise-AND +4 More Practice Tags : AlgorithmsArraysBit Magic Similar Reads DSA Tutorial - Learn Data Structures and Algorithms DSA (Data Structures and Algorithms) is the study of organizing data efficiently using data structures like arrays, stacks, and trees, paired with step-by-step procedures (or algorithms) to solve problems effectively. 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