Given string str consisting of lowercase characters, the task is to check whether the string is divisible by 6 after changing it according to the given rules:
- ‘a’ gets changed to 1.
- ‘b’ gets changed to 2 …
- and similarly, ‘z’ gets changed to 26.
For example, the string “abz” will be changed to 1226.
Example:
Input: str = “ab”
Output: Yes
“ab” is equivalent to 12 which is divisible by 6.Input: str = “abc”
Output: No
123 is not divisible by 6.
Approach: It can be solved by using a simple math trick that a number is divisible by 6 only if the sum of all of its digits is divisible by 3 and the last digit of the number is divisible by 2. Find the sum of the digits of the formed number and store it in a variable sum. Also, find the last digit of the number and store it in lastDigit.
Now, if the sum is divisible by 3 and the lastDigit is divisible by 2 then print “Yes” else print “No”.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the sum// of the digits of nint sumDigits(int n){ int sum = 0; while (n > 0) { int digit = n % 10; sum += digit; n /= 10; } return sum;}// Function that return true if the// decoded string is divisible by 6bool isDivBySix(string str, int n){ // To store the sum of the digits int sum = 0; // For each character, get the // sum of the digits for (int i = 0; i < n; i++) { sum += (int)(str[i] - 'a' + 1); } // If the sum of digits is // not divisible by 3 if (sum % 3 != 0) return false; // Get the last digit of // the number formed int lastDigit = ((int)(str[n - 1] - 'a' + 1)) % 10; // If the last digit is // not divisible by 2 if (lastDigit % 2 != 0) return false; return true;}// Driver codeint main(){ string str = "ab"; int n = str.length(); if (isDivBySix(str, n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation of the approachclass GFG{// Function to return the sum// of the digits of nstatic int sumDigits(int n){ int sum = 0; while (n > 0) { int digit = n % 10; sum += digit; n /= 10; } return sum;}// Function that return true if the// decoded string is divisible by 6static boolean isDivBySix(String str, int n){ // To store the sum of the digits int sum = 0; // For each character, get the // sum of the digits for (int i = 0; i < n; i++) { sum += (int)(str.charAt(i) - 'a' + 1); } // If the sum of digits is // not divisible by 3 if (sum % 3 != 0) return false; // Get the last digit of // the number formed int lastDigit = ((int)(str.charAt(n - 1) - 'a' + 1)) % 10; // If the last digit is // not divisible by 2 if (lastDigit % 2 != 0) return false; return true;}// Driver codepublic static void main(String []args){ String str = "ab"; int n = str.length(); if (isDivBySix(str, n)) System.out.println("Yes"); else System.out.println("No");}}// This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach # Function to return the sum # of the digits of n def sumDigits(n) : sum = 0; while (n > 0) : digit = n % 10; sum += digit; n //= 10; return sum; # Function that return true if the # decoded string is divisible by 6 def isDivBySix(string , n) : # To store the sum of the digits sum = 0; # For each character, get the # sum of the digits for i in range(n) : sum += (ord(string[i]) - ord('a') + 1); # If the sum of digits is # not divisible by 3 if (sum % 3 != 0) : return False; # Get the last digit of # the number formed lastDigit = (ord(string[n - 1]) - ord('a') + 1) % 10; # If the last digit is # not divisible by 2 if (lastDigit % 2 != 0) : return False; return True; # Driver code if __name__ == "__main__" : string = "ab"; n = len(string); if (isDivBySix(string, n)) : print("Yes"); else : print("No"); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System;class GFG{// Function to return the sum// of the digits of nstatic int sumDigits(int n){ int sum = 0; while (n > 0) { int digit = n % 10; sum += digit; n /= 10; } return sum;}// Function that return true if the// decoded string is divisible by 6static bool isDivBySix(String str, int n){ // To store the sum of the digits int sum = 0; // For each character, get the // sum of the digits for (int i = 0; i < n; i++) { sum += (int)(str[i] - 'a' + 1); } // If the sum of digits is // not divisible by 3 if (sum % 3 != 0) return false; // Get the last digit of // the number formed int lastDigit = ((int)(str[n - 1] - 'a' + 1)) % 10; // If the last digit is // not divisible by 2 if (lastDigit % 2 != 0) return false; return true;}// Driver codepublic static void Main(String []args){ String str = "ab"; int n = str.Length; if (isDivBySix(str, n)) Console.WriteLine("Yes"); else Console.WriteLine("No");}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the approach// Function to return the sum// of the digits of nfunction sumDigits(n){ var sum = 0; while (n > 0) { var digit = n % 10; sum += digit; n = parseInt(n/10); } return sum;}// Function that return true if the// decoded string is divisible by 6function isDivBySix(str, n){ // To store the sum of the digits var sum = 0; // For each character, get the // sum of the digits for (var i = 0; i < n; i++) { sum += (str[i].charCodeAt(0) - 'a'.charCodeAt(0) + 1); } // If the sum of digits is // not divisible by 3 if (sum % 3 != 0) return false; // Get the last digit of // the number formed var lastDigit = ((str[n - 1].charCodeAt(0) - 'a'.charCodeAt(0) + 1)) % 10; // If the last digit is // not divisible by 2 if (lastDigit % 2 != 0) return false; return true;}// Driver codevar str = "ab";var n = str.length;if (isDivBySix(str, n)) document.write( "Yes");else document.write( "No");</script> |
Output
Yes
Time Complexity: O(N)
Auxiliary Space: O(1)
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