Check whether a number has consecutive 0's in the given base or not (original) (raw)

Last Updated : 5 May, 2025

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

Python

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

JavaScript

PHP

z=toK(z = toK(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=r = r=N % $K; N=(int)(N = (int)(N=(int)(N / $K); s=s = s=r * w+w + 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=r = r=N % 10; N=(int)(N = (int)(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 ?>

****Time Complexity:O(logkn + 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.

**Approach: 2

Here is another approach to solve the same problem:

Start with the given number N and initialize a variable called count to 0.
Convert N to base K by continuously dividing N by K and appending the remainder to the right of the result until N becomes 0. Let's call the result of this conversion s.
Traverse through the string s and check each character. If the character is '0', increment count by 1. If the character is not '0', reset count to 0.
If count becomes 2 or more during the traversal, then the number N has consecutive 0s in base K. Otherwise, it does not.

Here is the code of above approach:

C++ `

#include #include

using namespace std;

bool hasConsecutiveZeroes(int N, int K) { // Convert N to base K string s = ""; while (N > 0) { s = to_string(N % K) + s; N /= K; }

// Traverse through the converted string and check for consecutive 0s
int count = 0;
for (char c : s) {
    if (c == '0') {
        count++;
        if (count >= 2) {
            return false;
        }
    } else {
        count = 0;
    }
}

return true;

}

int main() { int N = 15; int K = 8; if (hasConsecutiveZeroes(N, K)) { cout << "Yes" << endl; } else { cout << "No" << endl; } return 0; }

Java

import java.util.*;

public class Main { static boolean hasConsecutiveZeroes(int N, int K) { // Convert N to base K String s = ""; while (N > 0) { s = Integer.toString(N % K) + s; N /= K; }

// Traverse through the converted string and check for consecutive 0s
int count = 0;
for (char c : s.toCharArray()) {
    if (c == '0') {
        count++;
        if (count >= 2) {
            return false;
        }
    } else {
        count = 0;
    }
}

return true;

}

public static void main(String[] args) { int N = 15; int K = 8; if (hasConsecutiveZeroes(N, K)) { System.out.println("Yes"); } else { System.out.println("No"); } } }

Python

def hasConsecutiveZeroes(N, K): # Convert N to base K s = "" while N > 0: s = str(N % K) + s N //= K

# Traverse through the converted string and check for consecutive 0s
count = 0
for c in s:
    if c == '0':
        count += 1
        if count >= 2:
            return False
    else:
        count = 0

return True

N = 15 K = 8 if hasConsecutiveZeroes(N, K): print("Yes") else: print("No")

C#

using System;

public class ConsecutiveZeroesCheck { public static bool HasConsecutiveZeroes(int N, int K) { // Convert N to base K string s = ""; while (N > 0) { s = (N % K).ToString() + s; N /= K; }

    // Traverse through the converted string and check for consecutive 0s
    int count = 0;
    foreach (char c in s)
    {
        if (c == '0')
        {
            count++;
            if (count >= 2)
            {
                return false;
            }
        }
        else
        {
            count = 0;
        }
    }

    return true;
}

public static void Main(string[] args)
{
    int N = 15;
    int K = 8;

    if (HasConsecutiveZeroes(N, K))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}

}

` JavaScript ``

/**

This function checks if the number N in base K has consecutive 0s.
@param n The number to check.
@param k The base to convert N to.
@returns True if N in base K has consecutive 0s, False otherwise. */ function hasConsecutiveZeroes(n, k) {

// Convert N to base K. // The variable s will store the converted string. let s = ""; while (n > 0) { // Append the remainder of N divided by K to the string s. s += String(n % k); // Get the quotient of N divided by K. n = Math.floor(n / k); }

// Traverse through the converted string and check for consecutive 0s. // The variable count will keep track of the number of consecutive 0s. let count = 0; for (const c of s) { // If the current character is a 0, increment count. if (c === "0") { count++; } else { // If the current character is not a 0, reset count to 0. count = 0; } // If count is greater than or equal to 2, return False. if (count >= 2) { return false; } }

// If count is less than 2, return True. return true;

}

/**

This is the main entry point of the program.
It takes two integers N and K as input and prints "Yes" if N in base K has consecutive 0s,
and "No" otherwise. */ const n = 15; const k = 8; if (hasConsecutiveZeroes(n, k)) { console.log("Yes"); } else { console.log("No"); }

**Time Complexity: O(logN + length of the string).
**Auxiliary Space: O(logN) , because it also requires storing the converted number as a string.