Reduce a given number to form a key by the given operations
Given an integer N, the task is to reduce the number and form a key by following the operations given below:
- Extract the Most Significant Digit of the number:
- If the digit is even: Add the consecutive digits until the sum of digits is odd.
- If the digit is odd: Add the consecutive digits until the sum of digits is even.
- Repeat the process for all the remaining digits.
- Finally, concatenate the sum computed together to get the key.
Examples:
Input: N = 1667848270
Output: 20290
Explanation:
Step 1: First Digit(= 1) is odd. So, add up the next digits until the sum is even.
Therefore, digits 1, 6, 6, and 7 are added up to form 20.
Step 2: Next digit(= 8) is even. So, add up the next digits until the sum is odd.
Therefore, digits 8, 4, 8, 2, and 7 are added up to form 29.
Step 3: Last digit(= 0) is even.
Therefore, the final answer after concatenating the results will be: 20290Input: N = 7246262412
Output: 342
Explanation:
Step 1: First Digit(= 7) is odd. So, add up the next digits until the sum is even.
Therefore, digits 7, 2, 4, 6, 2, 6, 2, 4, and 1 are added up to form 34.
Step 2: Last digit(= 2) is even.
Therefore, the final answer after concatenating the results will be: 342.
Approach: The idea is to iterate the digits of the number and check the parity of the digit. If it is even, then proceed to the next digits until an odd digit is encountered. For odd digit, add consecutive digits until the sum of digits is even. Finally, concatenate the sum computed to get the desired key.
Below is the implementation of the above approach:
C++
// C++ program of the // above approach #include <bits/stdc++.h> using namespace std; // Function to find the key // of the given number int key( int N) { // Convert the integer // to String string num = "" + to_string(N); int ans = 0; int j = 0; // Iterate the num-string // to get the result for (j = 0; j < num.length(); j++) { // Check if digit is even or odd if ((num[j] - 48) % 2 == 0) { int add = 0; int i; // Iterate until odd sum // is obtained by adding // consecutive digits for (i = j; j < num.length(); j++) { add += num[j] - 48; // Check if sum becomes odd if (add % 2 == 1) break ; } if (add == 0) { ans *= 10; } else { int digit = ( int ) floor ( log10 (add) + 1); ans *= ( pow (10, digit)); // Add the result in ans ans += add; } // Assign the digit index // to num string i = j; } else { // If the number is odd int add = 0; int i; // Iterate until odd sum // is obtained by adding // consecutive digits for (i = j; j < num.length(); j++) { add += num[j] - 48; // Check if sum becomes even if (add % 2 == 0) { break ; } } if (add == 0) { ans *= 10; } else { int digit = ( int ) floor ( log10 (add) + 1); ans *= ( pow (10, digit)); // Add the result in ans ans += add; } // assign the digit index // to main numstring i = j; } } // Check if all digits // are visited or not if (j + 1 >= num.length()) { return ans; } else { return ans += num[num.length() - 1] - 48; } } // Driver code int main() { int N = 1667848271; cout << key(N); return 0; } // This code is contributed by divyeshrabadiya07 |
Java
// Java program of the // above approach import java.io.*; import java.util.*; import java.lang.*; public class Main { // Function to find the key // of the given number static int key( int N) { // Convert the integer // to String String num = "" + N; int ans = 0 ; int j = 0 ; // Iterate the num-string // to get the result for (j = 0 ; j < num.length(); j++) { // Check if digit is even or odd if ((num.charAt(j) - 48 ) % 2 == 0 ) { int add = 0 ; int i; // Iterate until odd sum // is obtained by adding // consecutive digits for (i = j; j < num.length(); j++) { add += num.charAt(j) - 48 ; // Check if sum becomes odd if (add % 2 == 1 ) break ; } if (add == 0 ) { ans *= 10 ; } else { int digit = ( int )Math.floor( Math.log10(add) + 1 ); ans *= (Math.pow( 10 , digit)); // Add the result in ans ans += add; } // Assign the digit index // to num string i = j; } else { // If the number is odd int add = 0 ; int i; // Iterate until odd sum // is obtained by adding // consecutive digits for (i = j; j < num.length(); j++) { add += num.charAt(j) - 48 ; // Check if sum becomes even if (add % 2 == 0 ) { break ; } } if (add == 0 ) { ans *= 10 ; } else { int digit = ( int )Math.floor( Math.log10(add) + 1 ); ans *= (Math.pow( 10 , digit)); // Add the result in ans ans += add; } // assign the digit index // to main numstring i = j; } } // Check if all digits // are visited or not if (j + 1 >= num.length()) { return ans; } else { return ans += num.charAt( num.length() - 1 ) - 48 ; } } // Driver Code public static void main(String[] args) { int N = 1667848271 ; System.out.print(key(N)); } } |
Python3
# Python3 program of the # above approach import math # Function to find the key # of the given number def key(N) : # Convert the integer # to String num = "" + str (N) ans = 0 j = 0 # Iterate the num-string # to get the result while j < len (num) : # Check if digit is even or odd if (( ord (num[j]) - 48 ) % 2 = = 0 ) : add = 0 # Iterate until odd sum # is obtained by adding # consecutive digits i = j while j < len (num) : add + = ord (num[j]) - 48 # Check if sum becomes odd if (add % 2 = = 1 ) : break j + = 1 if (add = = 0 ) : ans * = 10 else : digit = int (math.floor(math.log10(add) + 1 )) ans * = ( pow ( 10 , digit)) # Add the result in ans ans + = add # Assign the digit index # to num string i = j else : # If the number is odd add = 0 # Iterate until odd sum # is obtained by adding # consecutive digits i = j while j < len (num) : add + = ord (num[j]) - 48 # Check if sum becomes even if (add % 2 = = 0 ) : break j + = 1 if (add = = 0 ) : ans * = 10 else : digit = int (math.floor(math.log10(add) + 1 )) ans * = ( pow ( 10 , digit)) # Add the result in ans ans + = add # assign the digit index # to main numstring i = j j + = 1 # Check if all digits # are visited or not if (j + 1 ) > = len (num) : return ans else : ans + = ord (num[ len (num) - 1 ]) - 48 return ans N = 1667848271 print (key(N)) # This code is contributed by divyesh072019 |
C#
// C# program of the // above approach using System; class GFG{ // Function to find the key // of the given number static int key( int N) { // Convert the integer // to String String num = "" + N; int ans = 0; int j = 0; // Iterate the num-string // to get the result for (j = 0; j < num.Length; j++) { // Check if digit is even or odd if ((num[j] - 48) % 2 == 0) { int add = 0; int i; // Iterate until odd sum // is obtained by adding // consecutive digits for (i = j; j < num.Length; j++) { add += num[j] - 48; // Check if sum becomes odd if (add % 2 == 1) break ; } if (add == 0) { ans *= 10; } else { int digit = ( int )Math.Floor( Math.Log10(add) + 1); ans *= ( int )(Math.Pow(10, digit)); // Add the result in ans ans += add; } // Assign the digit index // to num string i = j; } else { // If the number is odd int add = 0; int i; // Iterate until odd sum // is obtained by adding // consecutive digits for (i = j; j < num.Length; j++) { add += num[j] - 48; // Check if sum becomes even if (add % 2 == 0) { break ; } } if (add == 0) { ans *= 10; } else { int digit = ( int )Math.Floor( Math.Log10(add) + 1); ans *= ( int )(Math.Pow(10, digit)); // Add the result in ans ans += add; } // assign the digit index // to main numstring i = j; } } // Check if all digits // are visited or not if (j + 1 >= num.Length) { return ans; } else { return ans += num[num.Length - 1] - 48; } } // Driver Code public static void Main(String[] args) { int N = 1667848271; Console.Write(key(N)); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript program of the above approach // Function to find the key // of the given number function key(N) { // Convert the integer // to String let num = "" + N.toString(); let ans = 0; let j = 0; // Iterate the num-string // to get the result for (j = 0; j < num.length; j++) { // Check if digit is even or odd if ((num[j].charCodeAt() - 48) % 2 == 0) { let add = 0; let i; // Iterate until odd sum // is obtained by adding // consecutive digits for (i = j; j < num.length; j++) { add += num[j].charCodeAt() - 48; // Check if sum becomes odd if (add % 2 == 1) break ; } if (add == 0) { ans *= 10; } else { let digit = Math.floor(Math.log10(add) + 1); ans *= parseInt(Math.pow(10, digit), 10); // Add the result in ans ans += add; } // Assign the digit index // to num string i = j; } else { // If the number is odd let add = 0; let i; // Iterate until odd sum // is obtained by adding // consecutive digits for (i = j; j < num.length; j++) { add += num[j].charCodeAt() - 48; // Check if sum becomes even if (add % 2 == 0) { break ; } } if (add == 0) { ans *= 10; } else { let digit = Math.floor(Math.log10(add) + 1); ans *= parseInt(Math.pow(10, digit), 10); // Add the result in ans ans += add; } // assign the digit index // to main numstring i = j; } } // Check if all digits // are visited or not if (j + 1 >= num.length) { return ans; } else { return ans += num[num.length - 1].charCodeAt() - 48; } } let N = 1667848271; document.write(key(N)); // This code is contributed by mukesh07. </script> |
20291
Time Complexity: O(N)
Auxiliary Space: O(1)
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