# Invert the Kth most significant bit of N

Given two non-negative integers N and K, the task is to invert the Kth most significant bit of N and print the number obtained after inverting the bit.

Examples:

Input: N = 10, K = 1
Output: 2
The binary representation of 10 is 1010.
After inverting the first bit it becomes 0010
whose decimal equivalent is 2.

Input: N = 56, K = 2
Output: 40

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

Approach: Find the number of bits in N, if the number of bits is less than K then N itself is the required answer else flip the Kth most significant bit of N and print the number obtained after flipping it.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to convert decimal number n ` `// to its binary representation ` `// stored as an array arr[] ` `void` `decBinary(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `k = log2(n); ` `    ``while` `(n > 0) { ` `        ``arr[k--] = n % 2; ` `        ``n /= 2; ` `    ``} ` `} ` ` `  `// Function to convert the number ` `// represented as a binary array ` `// arr[] into its decimal equivalent ` `int` `binaryDec(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `ans = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``ans += arr[i] << (n - i - 1); ` `    ``return` `ans; ` `} ` ` `  `// Function to return the updated integer ` `// after flipping the kth bit ` `int` `getNum(``int` `n, ``int` `k) ` `{ ` ` `  `    ``// Number of bits in n ` `    ``int` `l = log2(n) + 1; ` ` `  `    ``// Find the binary ` `    ``// representation of n ` `    ``int` `a[l] = { 0 }; ` `    ``decBinary(a, n); ` ` `  `    ``// The number of bits in n ` `    ``// are less than k ` `    ``if` `(k > l) ` `        ``return` `n; ` ` `  `    ``// Flip the kth bit ` `    ``a[k - 1] = (a[k - 1] == 0) ? 1 : 0; ` ` `  `    ``// Return the decimal equivalent ` `    ``// of the number ` `    ``return` `binaryDec(a, l); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 56, k = 2; ` ` `  `    ``cout << getNum(n, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `class` `GFG  ` `{ ` `     `  `    ``// Function to convert decimal number n  ` `    ``// to its binary representation  ` `    ``// stored as an array arr[]  ` `    ``static` `void` `decBinary(``int` `arr[], ``int` `n)  ` `    ``{  ` `        ``int` `k = (``int``)(Math.log(n) /  ` `                      ``Math.log(``2``));  ` `         `  `        ``while` `(n > ``0``) ` `        ``{  ` `            ``arr[k--] = n % ``2``;  ` `            ``n /= ``2``;  ` `        ``}  ` `    ``}  ` `     `  `    ``// Function to convert the number  ` `    ``// represented as a binary array  ` `    ``// arr[] into its decimal equivalent  ` `    ``static` `int` `binaryDec(``int` `arr[], ``int` `n)  ` `    ``{  ` `        ``int` `ans = ``0``;  ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `            ``ans += arr[i] << (n - i - ``1``);  ` `        ``return` `ans;  ` `    ``}  ` `     `  `    ``// Function to return the updated integer  ` `    ``// after flipping the kth bit  ` `    ``static` `int` `getNum(``int` `n, ``int` `k)  ` `    ``{  ` `     `  `        ``// Number of bits in n  ` `        ``int` `l = (``int``)(Math.log(n) /  ` `                      ``Math.log(``2``)) + ``1``;  ` `     `  `        ``// Find the binary  ` `        ``// representation of n  ` `        ``int` `a[] = ``new` `int``[l];  ` `        ``decBinary(a, n);  ` `     `  `        ``// The number of bits in n  ` `        ``// are less than k  ` `        ``if` `(k > l)  ` `            ``return` `n;  ` `     `  `        ``// Flip the kth bit  ` `        ``a[k - ``1``] = (a[k - ``1``] == ``0``) ? ``1` `: ``0``;  ` `     `  `        ``// Return the decimal equivalent  ` `        ``// of the number  ` `        ``return` `binaryDec(a, l);  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args) ` `    ``{  ` `        ``int` `n = ``56``; ` `        ``int` `k = ``2``;  ` `     `  `        ``System.out.println(getNum(n, k));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python

 `# Python implementation of the approach  ` `import` `math ` ` `  `# Function to convert decimal number n  ` `# to its binary representation  ` `# stored as an array arr[]  ` `def` `decBinary(arr, n): ` `    ``k ``=` `int``(math.log2(n))  ` `    ``while` `(n > ``0``): ` `        ``arr[k] ``=` `n ``%` `2` `        ``k ``=` `k ``-` `1` `        ``n ``=` `n``/``/``2` ` `  `# Function to convert the number  ` `# represented as a binary array  ` `# arr[] its decimal equivalent  ` `def` `binaryDec(arr, n): ` `    ``ans ``=` `0` `    ``for` `i ``in` `range``(``0``, n): ` `        ``ans ``=` `ans ``+` `(arr[i] << (n ``-` `i ``-` `1``)) ` `    ``return` `ans ` ` `  `# Function to concatenate the binary  ` `# numbers and return the decimal result  ` `def` `getNum(n, k):  ` ` `  `    ``# Number of bits in both the numbers  ` `    ``l ``=` `int``(math.log2(n)) ``+` `1` ` `  `    ``# Convert the bits in both the gers  ` `    ``# to the arrays a[] and b[]  ` `    ``a ``=` `[``0` `for` `i ``in` `range``(``0``, l)] ` ` `  `    ``decBinary(a, n) ` `    ``# The number of bits in n  ` `    ``# are less than k  ` `    ``if``(k > l): ` `        ``return` `n  ` ` `  `    ``# Flip the kth bit ` `    ``if``(a[k ``-` `1``] ``=``=` `0``): ` `        ``a[k ``-` `1``] ``=` `1` `    ``else``: ` `        ``a[k ``-` `1``] ``=` `0` ` `  `    ``# Return the decimal equivalent  ` `    ``# of the number  ` `    ``return` `binaryDec(a, l) ` ` `  `# Driver code  ` `n ``=` `56` `k ``=` `2` ` `  `print``(getNum(n, k)) ` ` `  `# This code is contributed by Sanjit_Prasad `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `     `  `    ``// Function to convert decimal number n  ` `    ``// to its binary representation  ` `    ``// stored as an array []arr  ` `    ``static` `void` `decBinary(``int` `[]arr, ``int` `n)  ` `    ``{  ` `        ``int` `k = (``int``)(Math.Log(n) /  ` `                      ``Math.Log(2));  ` `         `  `        ``while` `(n > 0) ` `        ``{  ` `            ``arr[k--] = n % 2;  ` `            ``n /= 2;  ` `        ``}  ` `    ``}  ` `     `  `    ``// Function to convert the number  ` `    ``// represented as a binary array  ` `    ``// []arr into its decimal equivalent  ` `    ``static` `int` `binaryDec(``int` `[]arr, ``int` `n)  ` `    ``{  ` `        ``int` `ans = 0;  ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `            ``ans += arr[i] << (n - i - 1);  ` `        ``return` `ans;  ` `    ``}  ` `     `  `    ``// Function to return the updated integer  ` `    ``// after flipping the kth bit  ` `    ``static` `int` `getNum(``int` `n, ``int` `k)  ` `    ``{  ` `     `  `        ``// Number of bits in n  ` `        ``int` `l = (``int``)(Math.Log(n) /  ` `                      ``Math.Log(2)) + 1;  ` `     `  `        ``// Find the binary  ` `        ``// representation of n  ` `        ``int` `[]a = ``new` `int``[l];  ` `        ``decBinary(a, n);  ` `     `  `        ``// The number of bits in n  ` `        ``// are less than k  ` `        ``if` `(k > l)  ` `            ``return` `n;  ` `     `  `        ``// Flip the kth bit  ` `        ``a[k - 1] = (a[k - 1] == 0) ? 1 : 0;  ` `     `  `        ``// Return the decimal equivalent  ` `        ``// of the number  ` `        ``return` `binaryDec(a, l);  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{  ` `        ``int` `n = 56; ` `        ``int` `k = 2;  ` `     `  `        ``Console.WriteLine(getNum(n, k));  ` `    ``}  ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```40
```

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