# Maximize the number by flipping at most K bits

Given an integer N, the task is to find the greatest number that can be obtained by flipping at most K bits in the binary representation of N.

Examples:

Input: N = 4, K = 1
Output: 6
The binary equivalent of 4 is 100.
On flipping the 1st 0, we get 110
which is equivalent to 6.

Input: N = 5, K = 2
Output: 7

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

Approach:

• If the number of 0s in the binary representation of N is less than K then flip all the 0s.
• If the number of 0s is greater than or equal to K then flip the most significant K 0s.
• Finally, print the maximized integer.

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; ` `    ``} ` `} ` ` `  `// Funtion 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 maximized ` `// number by flipping atmost k bits ` `int` `maxNum(``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); ` ` `  `    ``// To count the number of 0s flipped ` `    ``int` `cn = 0; ` `    ``for` `(``int` `i = 0; i < l; i++) { ` `        ``if` `(a[i] == 0 && cn < k) { ` `            ``a[i] = 1; ` `            ``cn++; ` `        ``} ` `    ``} ` ` `  `    ``// Return the decimal equivalent ` `    ``// of the maximized number ` `    ``return` `binaryDec(a, l); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 4, k = 1; ` ` `  `    ``cout << maxNum(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``;  ` `        ``}  ` `    ``}  ` `     `  `    ``// Funtion 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 maximized  ` `    ``// number by flipping atmost k bits  ` `    ``static` `int` `maxNum(``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);  ` `     `  `        ``// To count the number of 0s flipped  ` `        ``int` `cn = ``0``;  ` `        ``for` `(``int` `i = ``0``; i < l; i++) ` `        ``{  ` `            ``if` `(a[i] == ``0` `&& cn < k) ` `            ``{  ` `                ``a[i] = ``1``;  ` `                ``cn++;  ` `            ``}  ` `        ``}  ` `     `  `        ``// Return the decimal equivalent  ` `        ``// of the maximized number  ` `        ``return` `binaryDec(a, l);  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{  ` `        ``int` `n = ``4``, k = ``1``;  ` `     `  `        ``System.out.println(maxNum(n, k));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# 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` ` `  `# Funtion to convert the number  ` `# represented as a binary array  ` `# arr[] into 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 return the maximized  ` `# number by flipping atmost k bits  ` `def` `maxNum(n, k): ` `     `  `    ``# Number of bits in n  ` `    ``l ``=` `int``(math.log2(n)) ``+` `1` ` `  `    ``# Find the binary representation of n  ` `    ``a ``=` `[``0` `for` `i ``in` `range``(``0``, l)] ` `    ``decBinary(a, n)  ` ` `  `    ``# To count the number of 0s flipped  ` `    ``cn ``=` `0` `    ``for` `i ``in` `range``(``0``, l): ` `        ``if` `(a[i] ``=``=` `0` `and` `cn < k): ` `            ``a[i] ``=` `1` `            ``cn ``=` `cn ``+` `1` `             `  `    ``# Return the decimal equivalent  ` `    ``# of the maximized number  ` `    ``return` `binaryDec(a, l) ` ` `  `# Driver code  ` `n ``=` `4` `k ``=` `1` ` `  `print``(maxNum(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;  ` `        ``}  ` `    ``}  ` `     `  `    ``// Funtion 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 maximized  ` `    ``// number by flipping atmost k bits  ` `    ``static` `int` `maxNum(``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);  ` `     `  `        ``// To count the number of 0s flipped  ` `        ``int` `cn = 0;  ` `        ``for` `(``int` `i = 0; i < l; i++) ` `        ``{  ` `            ``if` `(a[i] == 0 && cn < k) ` `            ``{  ` `                ``a[i] = 1;  ` `                ``cn++;  ` `            ``}  ` `        ``}  ` `     `  `        ``// Return the decimal equivalent  ` `        ``// of the maximized number  ` `        ``return` `binaryDec(a, l);  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main(String[] args)  ` `    ``{  ` `        ``int` `n = 4, k = 1;  ` `     `  `        ``Console.WriteLine(maxNum(n, k));  ` `    ``}  ` `} ` ` `  `// This code is contributed by Rajput-Ji `

Output:

```6
```

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