# Smallest number greater or equals to N such that it has no odd positioned bit set

Given an integer N, the task is to find the smallest integer X such that it has no odd position set and X ≥ N.
Note: The positioning of bits is assumed from the right side and the first bit is assumed to be 0th bit.

Examples:

Input: N = 9
Output: 16
16’s binary representation is 10000, which has its 4th bit
set which is the smallest number possible satisfying the given condition.

Input: N = 5
Output: 5

Input: N = 19
Output: 20

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

Approach: The problem can be solved using a greedy approach and some bit properties. The property that if smaller powers of two are taken exactly once and added up they can never exceed a higher power of two (e.g., (1 + 2 + 4) < 8). The following greedy approach is used to solve the above problem:

• Initially count the number of bits.
• Get the leftmost index of the set bit.
• If the leftmost set bit is at an odd index, then answer will always be (1 << (leftmost_bit_index + 1)).
• Else, greedily form a number by setting all the even bits from 0 to leftmost_bit_index. Now greedily remove a power of two from right to check if we get a number which satisfies the given conditions.
• If the above condition does not give us our number, then we simply set the next leftmost even bit and return the answer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to count the total bits ` `int` `countBits(``int` `n) ` `{ ` `    ``int` `count = 0; ` ` `  `    ``// Iterate and find the ` `    ``// number of set bits ` `    ``while` `(n) { ` `        ``count++; ` ` `  `        ``// Right shift the number by 1 ` `        ``n >>= 1; ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Function to find the nearest number ` `int` `findNearestNumber(``int` `n) ` `{ ` ` `  `    ``// Count the total number of bits ` `    ``int` `cnt = countBits(n); ` ` `  `    ``// To get the position ` `    ``cnt -= 1; ` ` `  `    ``// If the last set bit is ` `    ``// at odd position then ` `    ``// answer will always be a number ` `    ``// with the left bit set ` `    ``if` `(cnt % 2) { ` `        ``return` `1 << (cnt + 1); ` `    ``} ` ` `  `    ``else` `{ ` ` `  `        ``int` `tempnum = 0; ` ` `  `        ``// Set all the even bits which ` `        ``// are possible ` `        ``for` `(``int` `i = 0; i <= cnt; i += 2) ` `            ``tempnum += 1 << i; ` ` `  `        ``// If the number still is less than N ` `        ``if` `(tempnum < n) { ` ` `  `            ``// Return the number by setting the ` `            ``// next even set bit ` `            ``return` `(1 << (cnt + 2)); ` `        ``} ` ` `  `        ``else` `if` `(tempnum == n) ` `            ``return` `n; ` ` `  `        ``// If we have reached this position ` `        ``// it means tempsum > n ` `        ``// hence turn off even bits to get the ` `        ``// first possible number ` `        ``for` `(``int` `i = 0; i <= cnt; i += 2) { ` ` `  `            ``// Turn off the bit ` `            ``tempnum -= (1 << i); ` ` `  `            ``// If it gets lower than N ` `            ``// then set it and return that number ` `            ``if` `(tempnum < n) ` `                ``return` `tempnum += (1 << i); ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 19; ` `    ``cout << findNearestNumber(n); ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// Function to count the total bits ` `    ``static` `int` `countBits(``int` `n)  ` `    ``{ ` `        ``int` `count = ``0``; ` ` `  `        ``// Iterate and find the ` `        ``// number of set bits ` `        ``while` `(n > ``0``) ` `        ``{ ` `            ``count++; ` ` `  `            ``// Right shift the number by 1 ` `            ``n >>= ``1``; ` `        ``} ` `        ``return` `count; ` `    ``} ` ` `  `    ``// Function to find the nearest number ` `    ``static` `int` `findNearestNumber(``int` `n)  ` `    ``{ ` ` `  `        ``// Count the total number of bits ` `        ``int` `cnt = countBits(n); ` ` `  `        ``// To get the position ` `        ``cnt -= ``1``; ` ` `  `        ``// If the last set bit is ` `        ``// at odd position then ` `        ``// answer will always be a number ` `        ``// with the left bit set ` `        ``if` `(cnt % ``2` `== ``1``)  ` `        ``{ ` `            ``return` `1` `<< (cnt + ``1``); ` `        ``}  ` `        ``else`  `        ``{ ` ` `  `            ``int` `tempnum = ``0``; ` ` `  `            ``// Set all the even bits which ` `            ``// are possible ` `            ``for` `(``int` `i = ``0``; i <= cnt; i += ``2``)  ` `            ``{ ` `                ``tempnum += ``1` `<< i; ` `            ``} ` ` `  `            ``// If the number still is less than N ` `            ``if` `(tempnum < n)  ` `            ``{ ` ` `  `                ``// Return the number by setting the ` `                ``// next even set bit ` `                ``return` `(``1` `<< (cnt + ``2``)); ` `            ``}  ` `            ``else` `            ``if` `(tempnum == n) ` `            ``{ ` `                ``return` `n; ` `            ``} ` ` `  `            ``// If we have reached this position ` `            ``// it means tempsum > n ` `            ``// hence turn off even bits to get the ` `            ``// first possible number ` `            ``for` `(``int` `i = ``0``; i <= cnt; i += ``2``) ` `            ``{ ` ` `  `                ``// Turn off the bit ` `                ``tempnum -= (``1` `<< i); ` ` `  `                ``// If it gets lower than N ` `                ``// then set it and return that number ` `                ``if` `(tempnum < n)  ` `                ``{ ` `                    ``return` `tempnum += (``1` `<< i); ` `                ``} ` `            ``} ` `        ``} ` `        ``return` `Integer.MIN_VALUE; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `n = ``19``; ` ` `  `        ``System.out.println(findNearestNumber(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// Function to count the total bits ` `    ``static` `int` `countBits(``int` `n)  ` `    ``{ ` `        ``int` `count = 0; ` ` `  `        ``// Iterate and find the ` `        ``// number of set bits ` `        ``while` `(n > 0) ` `        ``{ ` `            ``count++; ` ` `  `            ``// Right shift the number by 1 ` `            ``n >>= 1; ` `        ``} ` `        ``return` `count; ` `    ``} ` ` `  `    ``// Function to find the nearest number ` `    ``static` `int` `findNearestNumber(``int` `n)  ` `    ``{ ` ` `  `        ``// Count the total number of bits ` `        ``int` `cnt = countBits(n); ` ` `  `        ``// To get the position ` `        ``cnt -= 1; ` ` `  `        ``// If the last set bit is ` `        ``// at odd position then ` `        ``// answer will always be a number ` `        ``// with the left bit set ` `        ``if` `(cnt % 2 == 1)  ` `        ``{ ` `            ``return` `1 << (cnt + 1); ` `        ``}  ` `        ``else` `        ``{ ` ` `  `            ``int` `tempnum = 0; ` ` `  `            ``// Set all the even bits which ` `            ``// are possible ` `            ``for` `(``int` `i = 0; i <= cnt; i += 2)  ` `            ``{ ` `                ``tempnum += 1 << i; ` `            ``} ` ` `  `            ``// If the number still is less than N ` `            ``if` `(tempnum < n)  ` `            ``{ ` ` `  `                ``// Return the number by setting the ` `                ``// next even set bit ` `                ``return` `(1 << (cnt + 2)); ` `            ``}  ` `            ``else` `            ``if` `(tempnum == n) ` `            ``{ ` `                ``return` `n; ` `            ``} ` ` `  `            ``// If we have reached this position ` `            ``// it means tempsum > n ` `            ``// hence turn off even bits to get the ` `            ``// first possible number ` `            ``for` `(``int` `i = 0; i <= cnt; i += 2) ` `            ``{ ` ` `  `                ``// Turn off the bit ` `                ``tempnum -= (1 << i); ` ` `  `                ``// If it gets lower than N ` `                ``// then set it and return that number ` `                ``if` `(tempnum < n)  ` `                ``{ ` `                    ``return` `tempnum += (1 << i); ` `                ``} ` `            ``} ` `        ``} ` `        ``return` `int``.MinValue; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int` `n = 19; ` ` `  `        ``Console.WriteLine(findNearestNumber(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by anuj_67.. `

## Python3

 `# Python implementation of the above approach ` ` `  `  `  `# Function to count the total bits ` `def` `countBits(n): ` `    ``count ``=` `0``; ` `  `  `    ``# Iterate and find the ` `    ``# number of set bits ` `    ``while` `(n>``0``): ` `        ``count``+``=``1``; ` `  `  `        ``# Right shift the number by 1 ` `        ``n >>``=` `1``; ` `    ``return` `count; ` `  `  `# Function to find the nearest number ` `def` `findNearestNumber(n): ` `  `  `    ``# Count the total number of bits ` `    ``cnt ``=` `countBits(n); ` `  `  `    ``# To get the position ` `    ``cnt ``-``=` `1``; ` `  `  `    ``# If the last set bit is ` `    ``# at odd position then ` `    ``# answer will always be a number ` `    ``# with the left bit set ` `    ``if` `(cnt ``%` `2``): ` `        ``return` `1` `<< (cnt ``+` `1``); ` `  `  `    ``else``: ` `  `  `        ``tempnum ``=` `0``; ` `  `  `        ``# Set all the even bits which ` `        ``# are possible ` `        ``for` `i ``in` `range``(``0``,cnt``+``1``,``2``): ` `            ``tempnum ``+``=` `1` `<< i; ` `  `  `        ``# If the number still is less than N ` `        ``if` `(tempnum < n): ` `  `  `            ``# Return the number by setting the ` `            ``# next even set bit ` `            ``return` `(``1` `<< (cnt ``+` `2``)); ` `  `  `        ``elif` `(tempnum ``=``=` `n): ` `            ``return` `n; ` `  `  `        ``# If we have reached this position ` `        ``# it means tempsum > n ` `        ``# hence turn off even bits to get the ` `        ``# first possible number ` `        ``for` `i ``in` `range``(``0``,cnt``+``1``,``2``): ` `  `  `            ``# Turn off the bit ` `            ``tempnum ``-``=` `(``1` `<< i); ` `  `  `            ``# If it gets lower than N ` `            ``# then set it and return that number ` `            ``if` `(tempnum < n): ` `                ``tempnum ``+``=` `(``1` `<< i); ` `                ``return` `tempnum; ` `# Driver code ` `n ``=` `19``; ` `print``(findNearestNumber(n)); ` ` `  `# This code contributed by PrinciRaj1992  `

Output:

```20
```

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