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 Kint 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 0bool 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 notvoid hasConsecutiveZeroes(int N, int K){ int z = toK(N, K); if (check(z)) cout<<"Yes"<<endl; else cout<<"No"<<endl;} // Driver codeint main(){int N = 15;int K = 8;hasConsecutiveZeroes(N, K);}// This code is contributed by// Surendra_Gangwar |
Java
// Java implementation of the above approachimport java.util.*;class GFG { // Function to convert N into base Kstatic 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 0static 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 notstatic void hasConsecutiveZeroes(int N, int K){ int z = toK(N, K); if (check(z)) System.out.println("Yes"); else System.out.println("No");}// Driver codepublic 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 notdef hasConsecutiveZeroes(N, K): z = toK(N, K) if (check(z)): print("Yes") else: print("No")# Function to convert N into base Kdef 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 0def 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 codeN, K = 15, 8hasConsecutiveZeroes(N, K) |
C#
// C# implementation of the above approachusing System;class GFG { // Function to convert N into base Kstatic 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 0static 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 notstatic void hasConsecutiveZeroes(int N, int K){ int z = toK(N, K); if (check(z)) Console.WriteLine("Yes"); else Console.WriteLine("No");}// Driver codepublic static void Main(String[] args){ int N = 15; int K = 8; hasConsecutiveZeroes(N, K);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the above approach// Function to convert N into base Kfunction 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 0function 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 notfunction hasConsecutiveZeroes(N, K){ let z = toK(N, K); if (check(z)) document.write("Yes"); else document.write("No");}// Driver codelet N = 15;let K = 8;hasConsecutiveZeroes(N, K);// This code is contributed by souravmahato348</script> |
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 notfunction hasConsecutiveZeroes($N, $K){ $z = toK($N, $K); if (check($z)) print("Yes"); else print("No");}// Function to convert N into base Kfunction 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 $s;}// Function to check for consecutive 0function check($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?> |
Output
Yes
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;} |
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"); }}} |
Python3
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 TrueN = 15K = 8if 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"); } |
Output
Yes
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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