Check whether a number has consecutive 0’s in the given base or not

Given a decimal number N, the task is to check if a number has consecutive zeroes or not after converting the number to its K-based notation.

Examples:

Input: N = 4, K = 2 
Output: No 
4 in base 2 is 100, As there are consecutive 2 thus the answer is No.

Input: N = 15, K = 8 
Output: Yes 
15 in base 8 is 17, As there are no consecutive 0 so the answer is Yes. 

Approach: First convert the number N into base K and then simply check if the number has consecutive zeroes or not.

Below is the implementation of the above approach:  

C++

// C++ implementation of the above approach
#include<bits/stdc++.h>
using namespace std;



// Function to convert N into base K
int toK(int N, int K)
{

// Weight of each digit
    int w = 1;
    int s = 0;
    while (N != 0)
     {
        int r = N % K;
        N = N/K;
        s = r * w + s;
        w *= 10;
     }
    return s;

}

// Function to check for consecutive 0
bool check(int N)
{

// Flag to check if there are consecutive
    // zero or not
    bool fl = false;
    while (N != 0)
    {

        int r = N % 10;
        N = N/10;

        // If there are two consecutive zero
        // then returning False
        if (fl == true and r == 0)
            return false;
        if (r > 0)
            {
            fl = false;
            continue;
            }
        fl = true;

    }
     return true;

}

// We first convert to given base, then
// check if the converted number has two
// consecutive 0s or not
void hasConsecutiveZeroes(int N, int K)
{
    int z = toK(N, K);
    if (check(z))
       cout<<"Yes"<<endl;
    else
      cout<<"No"<<endl;
}



// Driver code
int main()
{
int N = 15;
int K = 8;
hasConsecutiveZeroes(N, K);

}
// This code is contributed by
// Surendra_Gangwar

Java

// Java implementation of the above approach
import java.util.*;

class GFG
{

// Function to convert N into base K
static int toK(int N, int K)
{

    // Weight of each digit
    int w = 1;
    int s = 0;
    while (N != 0)
    {
        int r = N % K;
        N = N / K;
        s = r * w + s;
        w *= 10;
    }
    return s;

}

// Function to check for consecutive 0
static boolean check(int N)
{

    // Flag to check if there are consecutive
    // zero or not
    boolean fl = false;
    while (N != 0)
    {

        int r = N % 10;
        N = N / 10;

        // If there are two consecutive zero
        // then returning False
        if (fl == true && r == 0)
            return false;
        if (r > 0)
        {
            fl = false;
            continue;
        }
        fl = true;
    }
    return true;
}

// We first convert to given base, then
// check if the converted number has two
// consecutive 0s or not
static void hasConsecutiveZeroes(int N, int K)
{
    int z = toK(N, K);
    if (check(z))
        System.out.println("Yes");
    else
        System.out.println("No");
}

// Driver code
public static void main(String[] args)
{
    int N = 15;
    int K = 8;
    hasConsecutiveZeroes(N, K);
}
}

// This code is contributed by Princi Singh

Python3

# Python implementation of the above approach

# We first convert to given base, then
# check if the converted number has two
# consecutive 0s or not
def hasConsecutiveZeroes(N, K):
    z = toK(N, K)
    if (check(z)):
        print("Yes")
    else:
        print("No")

# Function to convert N into base K
def toK(N, K):

    # Weight of each digit
    w = 1
    s = 0
    while (N != 0):
        r = N % K
        N = N//K
        s = r * w + s
        w* = 10
    return s

# Function to check for consecutive 0
def check(N):

    # Flag to check if there are consecutive
    # zero or not
    fl = False
    while (N != 0):
        r = N % 10
        N = N//10

        # If there are two consecutive zero
        # then returning False
        if (fl == True and r == 0):
            return False
        if (r > 0):
            fl = False
            continue
        fl = True
    return True

# Driver code
N, K = 15, 8
hasConsecutiveZeroes(N, K)

C#

// C# implementation of the above approach
using System;

class GFG
{

// Function to convert N into base K
static int toK(int N, int K)
{

    // Weight of each digit
    int w = 1;
    int s = 0;
    while (N != 0)
    {
        int r = N % K;
        N = N / K;
        s = r * w + s;
        w *= 10;
    }
    return s;
}

// Function to check for consecutive 0
static Boolean check(int N)
{

    // Flag to check if there are consecutive
    // zero or not
    Boolean fl = false;
    while (N != 0)
    {

        int r = N % 10;
        N = N / 10;

        // If there are two consecutive zero
        // then returning False
        if (fl == true && r == 0)
            return false;
        if (r > 0)
        {
            fl = false;
            continue;
        }
        fl = true;
    }
    return true;
}

// We first convert to given base, then
// check if the converted number has two
// consecutive 0s or not
static void hasConsecutiveZeroes(int N, int K)
{
    int z = toK(N, K);
    if (check(z))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}

// Driver code
public static void Main(String[] args)
{
    int N = 15;
    int K = 8;
    hasConsecutiveZeroes(N, K);
}
}

// This code is contributed by 29AjayKumar

PHP

<?php
// PHP implementation of the above approach

// We first convert to given base,
// then check if the converted number
// has two consecutive 0s or not
function hasConsecutiveZeroes($N, $K)
{
    $z = toK($N, $K);
    if (check($z))
        print("Yes");
    else
        print("No");
}

// Function to convert N into base K
function toK($N, $K)
{
    // Weight of each digit
    $w = 1;
    $s = 0;
    while ($N != 0)
    {
        $r = $N % $K;
        $N = (int)($N / $K);
        $s = $r * $w + $s;
        $w *= 10;
    }
    return $s;
}

// Function to check for consecutive 0
function check($N)
{
    // Flag to check if there are
    // consecutive zero or not
    $fl = false;
    while ($N != 0)
    {
        $r = $N % 10;
        $N = (int)($N / 10);

        // If there are two consecutive
        // zero then returning false
        if ($fl == true and $r == 0)
            return false;
        if ($r > 0)
        {
            $fl = false;
            continue;
        }
        $fl = true;
    }
    return true;
}

// Driver code
$N = 15;
$K = 8;
hasConsecutiveZeroes($N, $K);

// This code is contributed by mits
?>

Javascript

<script>

// Javascript implementation of the above approach

// Function to convert N into base K
function toK(N, K)
{

    // Weight of each digit
    let w = 1;
    let s = 0;

    while (N != 0)
    {
        let r = N % K;
        N = parseInt(N / K);
        s = r * w + s;
        w *= 10;
    }
    return s;
}

// Function to check for consecutive 0
function check(N)
{

    // Flag to check if there are consecutive
    // zero or not
    let fl = false;

    while (N != 0)
    {
        let r = N % 10;
        N = parseInt(N/10);

        // If there are two consecutive zero
        // then returning False
        if (fl == true && r == 0)
            return false;
        if (r > 0)
        {
            fl = false;
            continue;
        }
        fl = true;
    }
     return true;
}

// We first convert to given base, then
// check if the converted number has two
// consecutive 0s or not
function hasConsecutiveZeroes(N, K)
{
    let z = toK(N, K);
    if (check(z))
       document.write("Yes");
    else
      document.write("No");
}

// Driver code
let N = 15;
let K = 8;

hasConsecutiveZeroes(N, K);

// This code is contributed by souravmahato348

</script>

Output:

Yes

Time Complexity: O(log_kn + log₁₀n), where n and k represents the value of the given integers.
Auxiliary Space: O(1), no extra space is required, so it is a constant.