Find K’th smallest number such that A + B = A | B

Given two number A and K, the task is to find K’th smallest positive integer B such that A + B = A | B, where | denotes the bitwise OR operator.

Examples:

Input: A = 10, K = 3
Output: 5
Explanation:
K = 1, 10 + 1 = 10 | 1 = 11
K = 2, 10 + 4 = 10 | 4 = 14
K = 3, 10 + 5 = 10 | 5 = 15

Input: A = 1, B = 1
Output: 2

Approach:

  • B is a solution of the given equation if and only if B has 0 in all positions where A has 1 (in binary notation).
  • So, we need to determine the bit of B for positions where A has 0. Let, if A = 10100001 then the last eight digits of B must be 0_0____0, where _ denotes either 0 or 1. Any replacement of all _ by 0 or 1 gives us a solution.
  • The k-th smallest number will be received by replacing all _ in y by digits of binary representation of number k.

Below is the implementation of the above approach:

C++

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// C++ program for the
// above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find k'th
// smallest number such that
// A + B = A | B
long long kthSmallest(long long a, long long k)
{
  
    // res will store
    // final answer
    long long res = 0;
    long long j = 0;
  
    for (long long i = 0; i < 32; i++) {
  
        // Skip when j'th position
        // has 1 in binary representation
        // as in res, j'th position will be 0.
        while (j < 32 && (a & (1 << j))) {
            // j'th bit is set
            j++;
        }
  
        // If i'th bit of k is 1
        // and i'th bit of j is 0
        // then set i'th bit in res.
        if (k & (1 << i)) {
            res |= (1LL << j);
        }
  
        // Proceed to next bit
        j++;
    }
  
    return res;
}
  
// Driver Code
int main()
{
  
    long long a = 5, k = 3;
    cout << kthSmallest(a, k) << "\n";
  
    return 0;
}

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Python3

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# Python3 program for the above approach
  
# Function to find k'th
# smallest number such that 
# A + B = A | B
def kthSmallest(a, k):
  
    # res will store
    # final answer
    res = 0
    j = 0
  
    for i in range(32):
  
        # Skip when j'th position 
        # has 1 in binary representation 
        # as in res, j'th position will be 0.
        while (j < 32 and (a & (1 << j))):
              
            # j'th bit is set
            j += 1
  
        # If i'th bit of k is 1 
        # and i'th bit of j is 0 
        # then set i'th bit in res. 
        if (k & (1 << i)):
            res |= (1 << j)
  
        # Proceed to next bit
        j += 1
  
    return res
  
# Driver Code
a = 5
k = 3
  
print(kthSmallest(a, k))
  
# This code is contributed by himanshu77

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Output:

10

Time Complexity: O(log(n))

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Improved By : himanshu77