# 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 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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Improved By : mohit kumar 29