Find N distinct numbers whose bitwise Or is equal to K
Given two integers N and K, the task is to find N distinct integers whose bit-wise OR is equal to K. If there does not exist any possible answer then print -1.
Examples:
Input: N = 3, K = 5
Output: 5 0 1
5 OR 0 OR 1 = 5Input: N = 10, K = 5
Output: -1
It is not possible to find any solution.
Approach:
- We know that if bit-wise OR of a sequence of numbers is K then all the bit positions which are 0 in K must also be zero in all the numbers.
- So, we only have those positions to change where bit is 1 in K. Say that count is Bit_K.
- Now, we can form pow(2, Bit_K) distinct numbers with Bit_K bits. So if, we set one number to be K itself, then rest N – 1 numbers can be built by setting 0 all the bits in every number which are 0 in K and for other bit positions any permutation of Bit_K bits other than number K.
- If pow(2, Bit_K) < N then we cannot find any possible answer.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; #define ll long long int #define MAX 32 ll pow2[MAX]; bool visited[MAX]; vector<int> ans; // Function to pre-calculate // all the powers of 2 upto MAX void power_2() { ll ans = 1; for (int i = 0; i < MAX; i++) { pow2[i] = ans; ans *= 2; } } // Function to return the // count of set bits in x int countSetBits(ll x) { // To store the count // of set bits int setBits = 0; while (x != 0) { x = x & (x - 1); setBits++; } return setBits; } // Function to add num to the answer // by setting all bit positions as 0 // which are also 0 in K void add(ll num) { int point = 0; ll value = 0; for (ll i = 0; i < MAX; i++) { // Bit i is 0 in K if (visited[i]) continue; else { if (num & 1) { value += (1 << i); } num /= 2; } } ans.push_back(value); } // Function to find and print N distinct // numbers whose bitwise OR is K void solve(ll n, ll k) { // Choosing K itself as one number ans.push_back(k); // Find the count of set bits in K int countk = countSetBits(k); // Impossible to get N // distinct integers if (pow2[countk] < n) { cout << -1; return; } int count = 0; for (ll i = 0; i < pow2[countk] - 1; i++) { // Add i to the answer after // setting all the bits as 0 // which are 0 in K add(i); count++; // If N distinct numbers are generated if (count == n) break; } // Print the generated numbers for (int i = 0; i < n; i++) { cout << ans[i] << " "; } } // Driver code int main() { ll n = 3, k = 5; // Pre-calculate all // the powers of 2 power_2(); solve(n, k); return 0; } |
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Java
// Java implementation of the approach import java.util.*; class GFG { static final int MAX = 32; static long []pow2 = new long[MAX]; static boolean []visited = new boolean[MAX]; static Vector<Long> ans = new Vector<>(); // Function to pre-calculate // all the powers of 2 upto MAX static void power_2() { long ans = 1; for (int i = 0; i < MAX; i++) { pow2[i] = ans; ans *= 2; } } // Function to return the // count of set bits in x static int countSetBits(long x) { // To store the count // of set bits int setBits = 0; while (x != 0) { x = x & (x - 1); setBits++; } return setBits; } // Function to add num to the answer // by setting all bit positions as 0 // which are also 0 in K static void add(long num) { int point = 0; long value = 0; for (int i = 0; i < MAX; i++) { // Bit i is 0 in K if (visited[i]) continue; else { if (num %2== 1) { value += (1 << i); } num /= 2; } } ans.add(value); } // Function to find and print N distinct // numbers whose bitwise OR is K static void solve(long n, long k) { // Choosing K itself as one number ans.add(k); // Find the count of set bits in K int countk = countSetBits(k); // Impossible to get N // distinct integers if (pow2[countk] < n) { System.out.print(-1); return; } int count = 0; for (long i = 0; i < pow2[countk] - 1; i++) { // Add i to the answer after // setting all the bits as 0 // which are 0 in K add(i); count++; // If N distinct numbers are generated if (count == n) break; } // Print the generated numbers for (int i = 0; i < n; i++) { System.out.print(ans.get(i)+" "); } } // Driver code public static void main(String[] args) { long n = 3, k = 5; // Pre-calculate all // the powers of 2 power_2(); solve(n, k); } } // This code has been contributed by 29AjayKumar |
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Python3
# Pyhton 3 implementation of the approach MAX = 32 pow2 = [0 for i in range(MAX)] visited = [False for i in range(MAX)] ans = [] # Function to pre-calculate # all the powers of 2 upto MAX def power_2(): an = 1 for i in range(MAX): pow2[i] = an an *= 2 # Function to return the # count of set bits in x def countSetBits(x): # To store the count # of set bits setBits = 0 while (x != 0): x = x & (x - 1) setBits += 1 return setBits # Function to add num to the answer # by setting all bit positions as 0 # which are also 0 in K def add(num): point = 0 value = 0 for i in range(MAX): # Bit i is 0 in K if (visited[i]): continue else: if (num & 1): value += (1 << i) num = num//2 ans.append(value) # Function to find and print N distinct # numbers whose bitwise OR is K def solve(n, k): # Choosing K itself as one number ans.append(k) # Find the count of set bits in K countk = countSetBits(k) # Impossible to get N # distinct integers if (pow2[countk] < n): print(-1) return count = 0 for i in range(pow2[countk] - 1): # Add i to the answer after # setting all the bits as 0 # which are 0 in K add(i) count += 1 # If N distinct numbers are generated if (count == n): break # Print the generated numbers for i in range(n): print(ans[i],end = " ") # Driver code if __name__ == '__main__': n = 3 k = 5 # Pre-calculate all # the powers of 2 power_2() solve(n, k) # This code is contributed by # Surendra_Gangwar |
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C#
// C# implementation of the approach using System; using System.Collections.Generic; class GFG { static int MAX = 32; static long [] pow2 = new long[MAX]; static bool [] visited = new bool[MAX]; static List<long> ans = new List<long>(); // Function to pre-calculate // all the powers of 2 upto MAX static void power_2() { long ans = 1; for (int i = 0; i < MAX; i++) { pow2[i] = ans; ans *= 2; } } // Function to return the // count of set bits in x static int countSetBits(long x) { // To store the count // of set bits int setBits = 0; while (x != 0) { x = x & (x - 1); setBits++; } return setBits; } // Function to add num to the answer // by setting all bit positions as 0 // which are also 0 in K static void add(long num) { long value = 0; for (int i = 0; i < MAX; i++) { // Bit i is 0 in K if (visited[i]) continue; else { if (num %2== 1) { value += (1 << i); } num /= 2; } } ans.Add(value); } // Function to find and print N distinct // numbers whose bitwise OR is K static void solve(long n, long k) { // Choosing K itself as one number ans.Add(k); // Find the count of set bits in K int countk = countSetBits(k); // Impossible to get N // distinct integers if (pow2[countk] < n) { Console.WriteLine(-1); return; } int count = 0; for (long i = 0; i < pow2[countk] - 1; i++) { // Add i to the answer after // setting all the bits as 0 // which are 0 in K add(i); count++; // If N distinct numbers are generated if (count == n) break; } // Print the generated numbers for (int i = 0; i < n; i++) { Console.Write(ans[i]+ " "); } } // Driver code public static void Main() { long n = 3, k = 5; // Pre-calculate all // the powers of 2 power_2(); solve(n, k); } } // This code is contributed by ihritik |
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Output:
5 0 1
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