# Find next greater element with no consecutive 1 in it’s binary representation

Given Q queries where each query consists of an integer N and the task is to find the smallest integer greater than N such that there are no consecutive 1s in its binary representation.

Examples:

Input: Q[] = {4, 6}
Output:
5
8

Input: Q[] = {50, 23, 456}
Output:
64
32
512

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

Approach: Store all the numbers in a list whose binary representation does not contain consecutive 1s upto a fixed limit. Now for every given N, find the next greater element in the list generated previously using binary search.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `const` `int` `MAX = 100000; ` ` `  `// To store the pre-computed integers ` `vector<``int``> v; ` ` `  `// Function that returns true if the ` `// binary representation of x contains ` `// consecutive 1s ` `int` `consecutiveOnes(``int` `x) ` `{ ` ` `  `    ``// To store the previous bit ` `    ``int` `p = 0; ` `    ``while` `(x > 0) { ` ` `  `        ``// Check whether the previous bit ` `        ``// and the current bit are both 1 ` `        ``if` `(x % 2 == 1 and p == 1) ` `            ``return` `true``; ` ` `  `        ``// Update previous bit ` `        ``p = x % 2; ` ` `  `        ``// Go to the next bit ` `        ``x /= 2; ` `    ``} ` `    ``return` `false``; ` `} ` ` `  `// Function to pre-compute the ` `// valid numbers from 0 to MAX ` `void` `preCompute() ` `{ ` `    ``// Store all the numbers which do ` `    ``// not have consecutive 1s ` `    ``for` `(``int` `i = 0; i <= MAX; i++) { ` `        ``if` `(!consecutiveOnes(i)) ` `            ``v.push_back(i); ` `    ``} ` `} ` ` `  `// Function to return the minimum ` `// number greater than n which does ` `// not contain consecutive 1s ` `int` `nextValid(``int` `n) ` `{ ` `    ``// Search for the next greater element ` `    ``// with no consecutive 1s ` `    ``int` `it = upper_bound(v.begin(), ` `                         ``v.end(), n) ` `             ``- v.begin(); ` `    ``int` `val = v[it]; ` `    ``return` `val; ` `} ` ` `  `// Function to perform the queries ` `void` `performQueries(``int` `queries[], ``int` `q) ` `{ ` `    ``for` `(``int` `i = 0; i < q; i++) ` `        ``cout << nextValid(queries[i]) << ``"\n"``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `queries[] = { 4, 6 }; ` `    ``int` `q = ``sizeof``(queries) / ``sizeof``(``int``); ` ` `  `    ``// Pre-compute the numbers ` `    ``preCompute(); ` ` `  `    ``// Perform the queries ` `    ``performQueries(queries, q); ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 implementation of the approach ` `from` `bisect ``import` `bisect_right as upper_bound ` ` `  `MAX` `=` `100000` ` `  `# To store the pre-computed integers ` `v ``=` `[] ` ` `  `# Function that returns true if the ` `# binary representation of x contains ` `# consecutive 1s ` `def` `consecutiveOnes(x): ` ` `  `    ``# To store the previous bit ` `    ``p ``=` `0` `    ``while` `(x > ``0``): ` ` `  `        ``# Check whether the previous bit ` `        ``# and the current bit are both 1 ` `        ``if` `(x ``%` `2` `=``=` `1` `and` `p ``=``=` `1``): ` `            ``return` `True` ` `  `        ``# Update previous bit ` `        ``p ``=` `x ``%` `2` ` `  `        ``# Go to the next bit ` `        ``x ``/``/``=` `2` ` `  `    ``return` `False` ` `  `# Function to pre-compute the ` `# valid numbers from 0 to MAX ` `def` `preCompute(): ` `     `  `    ``# Store all the numbers which do ` `    ``# not have consecutive 1s ` `    ``for` `i ``in` `range``(``MAX` `+` `1``): ` `        ``if` `(consecutiveOnes(i) ``=``=` `0``): ` `            ``v.append(i) ` ` `  `# Function to return the minimum ` `# number greater than n which does ` `# not contain consecutive 1s ` `def` `nextValid(n): ` `     `  `    ``# Search for the next greater element ` `    ``# with no consecutive 1s ` `    ``it ``=` `upper_bound(v, n) ` `    ``val ``=` `v[it] ` `    ``return` `val ` ` `  `# Function to perform the queries ` `def` `performQueries(queries, q): ` `    ``for` `i ``in` `range``(q): ` `        ``print``(nextValid(queries[i])) ` ` `  `# Driver code ` `queries ``=` `[``4``, ``6``] ` `q ``=` `len``(queries) ` ` `  `# Pre-compute the numbers ` `preCompute() ` ` `  `# Perform the queries ` `performQueries(queries, q) ` ` `  `# This code is contributed by Mohit Kumar `

Output:

```5
8
```

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