Check whether a number has consecutive 0's in the given base or not
Last Updated :
05 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 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 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# 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
<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>
<?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
?>
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 <iostream>
#include <string>
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;
}
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");
}
}
}
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")
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");
}
}
}
/**
* 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.
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