Given a number N, the task is to print the maximum between the sum and multiplication of the digits of the given number until the number is reduced to a single digit.
Note: Sum and multiplication of digits to be done until the number is reduced to a single digit.
Let’s take an example where N = 19,
19 breaks into 1+9=10 then 10 breaks into 1+0=1. 1 is a single digit sum.
Also, 19 breaks into 1*9 = 9. 9 is a single digit multiplication.
So, output is 9 i.e. maximum of 9 and 1.
Input: N = 631 Output: 8 Input: 110 Output: 2
Approach:
- Check if a number is less than 10 then the sum and product will be the same. So, return that number.
- Else,
- Find the sum of digits repeatedly using Method 2 of Finding sum of digits of a number until sum becomes single digit.
- And, Find the product of digits repeatedly using Method 1 of Finding sum of digits of a number until sum becomes single digit.
- Return the maximum of both.
Below is the implementation of above approach:
C++
// CPP implementation of above approach#include<bits/stdc++.h>using namespace std; // Function to sum the digits until it // becomes a single digit long repeatedSum(long n) { if (n == 0) return 0; return (n % 9 == 0) ? 9 : (n % 9); } // Function to product the digits until it // becomes a single digit long repeatedProduct(long n) { long prod = 1; // Loop to do sum while // sum is not less than // or equal to 9 while (n > 0 || prod > 9) { if (n == 0) { n = prod; prod = 1; } prod *= n % 10; n /= 10; } return prod; } // Function to find the maximum among // repeated sum and repeated product long maxSumProduct(long N) { if (N < 10) return N; return max(repeatedSum(N), repeatedProduct(N)); } // Driver code int main() { long n = 631; cout << maxSumProduct(n)<<endl; return 0; }// This code is contributed by mits |
Java
// Java implementation of above approachimport java.util.*;import java.lang.*;import java.io.*;class GFG { // Function to sum the digits until it // becomes a single digit public static long repeatedSum(long n) { if (n == 0) return 0; return (n % 9 == 0) ? 9 : (n % 9); } // Function to product the digits until it // becomes a single digit public static long repeatedProduct(long n) { long prod = 1; // Loop to do sum while // sum is not less than // or equal to 9 while (n > 0 || prod > 9) { if (n == 0) { n = prod; prod = 1; } prod *= n % 10; n /= 10; } return prod; } // Function to find the maximum among // repeated sum and repeated product public static long maxSumProduct(long N) { if (N < 10) return N; return Math.max(repeatedSum(N), repeatedProduct(N)); } // Driver code public static void main(String[] args) { long n = 631; System.out.println(maxSumProduct(n)); }} |
Python3
# Python 3 implementation of above approach# Function to sum the digits until # it becomes a single digitdef repeatedSum(n): if (n == 0): return 0 return 9 if(n % 9 == 0) else (n % 9)# Function to product the digits # until it becomes a single digitdef repeatedProduct(n): prod = 1 # Loop to do sum while # sum is not less than # or equal to 9 while (n > 0 or prod > 9) : if (n == 0) : n = prod prod = 1 prod *= n % 10 n //= 10 return prod# Function to find the maximum among# repeated sum and repeated productdef maxSumProduct(N): if (N < 10): return N return max(repeatedSum(N), repeatedProduct(N))# Driver codeif __name__ == "__main__": n = 631 print(maxSumProduct(n))# This code is contributed # by ChitraNayal |
C#
// C# implementation of // above approachusing System;class GFG {// Function to sum the digits // until it becomes a single digitpublic static long repeatedSum(long n){ if (n == 0) return 0; return (n % 9 == 0) ? 9 : (n % 9);}// Function to product the digits // until it becomes a single digitpublic static long repeatedProduct(long n){ long prod = 1; // Loop to do sum while // sum is not less than // or equal to 9 while (n > 0 || prod > 9) { if (n == 0) { n = prod; prod = 1; } prod *= n % 10; n /= 10; } return prod;}// Function to find the maximum among// repeated sum and repeated productpublic static long maxSumProduct(long N){ if (N < 10) return N; return Math.Max(repeatedSum(N), repeatedProduct(N));}// Driver codepublic static void Main(){ long n = 631; Console.WriteLine(maxSumProduct(n));}}// This code is contributed// by inder_verma |
Javascript
<script>// javascript implementation of above approach // Function to sum the digits until it // becomes a single digit function repeatedSum(n) { if (n == 0) return 0; return (n % 9 == 0) ? 9 : (n % 9); } // Function to product the digits until it // becomes a single digit function repeatedProduct(n) { var prod = 1; // Loop to do sum while // sum is not less than // or equal to 9 while (n > 0 || prod > 9) { if (n == 0) { n = prod; prod = 1; } prod *= n % 10; n = parseInt(n/10); } return prod; } // Function to find the maximum among // repeated sum and repeated product function maxSumProduct(N) { if (N < 10) return N; return Math.max(repeatedSum(N), repeatedProduct(N)); } // Driver code var n = 631; document.write(maxSumProduct(n));// This code contributed by aashish1995</script> |
Output:
8
Time Complexity: O(log10n)
Auxiliary Space: O(1)
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