Divide a number into two parts such that sum of digits is maximum
Given a number N. The task is to find the maximum possible value of SumOfDigits(A) + SumOfDigits(B) such that A + B = n (0<=A, B<=n).
Examples:
Input: N = 35 Output: 17 35 = 9 + 26 SumOfDigits(26) = 8, SumOfDigits(9) = 9 So, 17 is the answer. Input: N = 7 Output: 7
Approach: The idea is to divide the number into two parts A and B such that A is in terms of 9 i.e. closest number to N and B = N-A. For example, N = 35, Smallest Number that is closest to 35 is 29.
C++
// C++ implementation of above approach #include <bits/stdc++.h> using namespace std; // Returns sum of digits of x int sumOfDigitsSingle(int x) { int ans = 0; while (x) { ans += x % 10; x /= 10; } return ans; } // Returns closest number to x in terms of 9's. int closest(int x) { int ans = 0; while (ans * 10 + 9 <= x) ans = ans * 10 + 9; return ans; } int sumOfDigitsTwoParts(int N) { int A = closest(N); return sumOfDigitsSingle(A) + sumOfDigitsSingle(N - A); } // Driver code int main() { int N = 35; cout << sumOfDigitsTwoParts(N); return 0; } |
Java
// Java implementation of above approach // Returns sum of digits of x import java.util.*; import java.lang.*; import java.io.*; class GFG { static int sumOfDigitsSingle(int x) { int ans = 0; while (x != 0) { ans += x % 10; x /= 10; } return ans; } // Returns closest number to x // in terms of 9's. static int closest(int x) { int ans = 0; while (ans * 10 + 9 <= x) ans = ans * 10 + 9; return ans; } static int sumOfDigitsTwoParts(int N) { int A = closest(N); return sumOfDigitsSingle(A) + sumOfDigitsSingle(N - A); } // Driver code public static void main(String args[]) { int N = 35; System.out.print(sumOfDigitsTwoParts(N)); } } // This code is contributed by // Subhadeep Gupta |
Python 3
# Python 3 implementation of above approach # Returns sum of digits of x def sumOfDigitsSingle(x) : ans = 0 while x : ans += x % 10 x //= 10 return ans # Returns closest number to x in terms of 9's def closest(x) : ans = 0 while (ans * 10 + 9 <= x) : ans = ans * 10 + 9 return ans def sumOfDigitsTwoParts(N) : A = closest(N) return sumOfDigitsSingle(A) + sumOfDigitsSingle(N - A) # Driver Code if __name__ == "__main__" : N = 35 print(sumOfDigitsTwoParts(N)) # This code is contributed by ANKITRAI1 |
C#
// C# implementation of above approach // Returns sum of digits of x using System; class GFG { static int sumOfDigitsSingle(int x) { int ans = 0; while (x != 0) { ans += x % 10; x /= 10; } return ans; } // Returns closest number to x // in terms of 9's. static int closest(int x) { int ans = 0; while (ans * 10 + 9 <= x) ans = ans * 10 + 9; return ans; } static int sumOfDigitsTwoParts(int N) { int A = closest(N); return sumOfDigitsSingle(A) + sumOfDigitsSingle(N - A); } // Driver code public static void Main() { int N = 35; Console.Write(sumOfDigitsTwoParts(N)); } } |
PHP
<?php // PHP implementation of above approach // Returns sum of digits of x function sumOfDigitsSingle($x) { $ans = 0; while ($x) { $ans += $x % 10; $x /= 10; } return $ans; } // Returns closest number // to x in terms of 9's. function closest($x) { $ans = 0; while ($ans * 10 + 9 <= $x) $ans = $ans * 10 + 9; return $ans; } function sumOfDigitsTwoParts($N) { $A = closest($N); return sumOfDigitsSingle($A) + sumOfDigitsSingle($N - $A); } // Driver code $N = 35; echo sumOfDigitsTwoParts($N); // This code is contributed // by inder_verma ?> |
Output:
17
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