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

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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}

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Python3

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# 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

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Output:

5
8


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Improved By : mohit kumar 29