# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ program for the ` `// above approach ` `#include ` `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; ` `} `

## Python3

 `# 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 `

Output:

```10
```

Time Complexity: O(log(n))

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Improved By : himanshu77