Given a number N, the task is to find the sum of digits of a number at even and odd places.
Examples:
Input: N = 54873
Output:
Sum odd = 16
Sum even = 11Input: N = 457892
Output:
Sum odd = 20
Sum even = 15
Approach:
- First, calculate the reverse of the given number.
- To the reverse number we apply modulus operator and extract its last digit which is actually the first digit of a number so it is odd positioned digit.
- The next digit will be even positioned digit, and we can take the sum in alternating turns.
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 reverse of a number int reverse(int n) { int rev = 0; while (n != 0) { rev = (rev * 10) + (n % 10); n /= 10; } return rev; } // Function to find the sum of the odd // and even positioned digits in a number void getSum(int n) { n = reverse(n); int sumOdd = 0, sumEven = 0, c = 1; while (n != 0) { // If c is even number then it means // digit extracted is at even place if (c % 2 == 0) sumEven += n % 10; else sumOdd += n % 10; n /= 10; c++; } cout << "Sum odd = " << sumOdd << "\n"; cout << "Sum even = " << sumEven; } // Driver code int main() { int n = 457892; getSum(n); return 0; } |
Java
// Java implementation of the approach import java.util.*; class GFG { // Function to return the reverse of a number static int reverse(int n) { int rev = 0; while (n != 0) { rev = (rev * 10) + (n % 10); n /= 10; } return rev; } // Function to find the sum of the odd // and even positioned digits in a number static void getSum(int n) { n = reverse(n); int sumOdd = 0, sumEven = 0, c = 1; while (n != 0) { // If c is even number then it means // digit extracted is at even place if (c % 2 == 0) sumEven += n % 10; else sumOdd += n % 10; n /= 10; c++; } System.out.println("Sum odd = " + sumOdd); System.out.println("Sum even = " + sumEven); } // Driver code public static void main(String args[]) { int n = 457892; getSum(n); } } // This code is contributed by // Surendra_Gangwar |
Python3
# Python3 implementation of the approach # Function to return the # reverse of a number def reverse(n): rev = 0 while (n != 0): rev = (rev * 10) + (n % 10) n //= 10 return rev # Function to find the sum of the odd # and even positioned digits in a number def getSum(n): n = reverse(n) sumOdd = 0 sumEven = 0 c = 1 while (n != 0): # If c is even number then it means # digit extracted is at even place if (c % 2 == 0): sumEven += n % 10 else: sumOdd += n % 10 n //= 10 c += 1 print("Sum odd =", sumOdd) print("Sum even =", sumEven) # Driver code n = 457892getSum(n) # This code is contributed # by mohit kumar |
C#
// C# implementation of the approach using System; class GFG { // Function to return the reverse of a number static int reverse(int n) { int rev = 0; while (n != 0) { rev = (rev * 10) + (n % 10); n /= 10; } return rev; } // Function to find the sum of the odd // and even positioned digits in a number static void getSum(int n) { n = reverse(n); int sumOdd = 0, sumEven = 0, c = 1; while (n != 0) { // If c is even number then it means // digit extracted is at even place if (c % 2 == 0) sumEven += n % 10; else sumOdd += n % 10; n /= 10; c++; } Console.WriteLine("Sum odd = " + sumOdd); Console.WriteLine("Sum even = " + sumEven); } // Driver code public static void Main() { int n = 457892; getSum(n); } } // This code is contributed by // Akanksha Rai |
PHP
<?php // PHP implementation of the above approach // Function to return the reverse of a number function reverse($n) { $rev = 0; while ($n != 0) { $rev = ($rev * 10) + ($n % 10); $n = floor($n / 10); } return $rev; } // Function to find the sum of the odd // and even positioned digits in a number function getSum($n) { $n = reverse($n); $sumOdd = 0; $sumEven = 0; $c = 1; while ($n != 0) { // If c is even number then it means // digit extracted is at even place if ($c % 2 == 0) $sumEven += $n % 10; else $sumOdd += $n % 10; $n = floor($n / 10); $c++; } echo "Sum odd = ", $sumOdd, "\n"; echo "Sum even = ", $sumEven; } // Driver code $n = 457892; getSum($n); // This code is contributed by Ryuga ?> |
Javascript
<script>//JavaScript implementation of the approach // Function to return the // reverse of a number function reverse(n) { let rev = 0; while (n != 0) { rev = (rev * 10) + (n % 10); n = Math.floor(n / 10); } return rev; } // Function to find the sum of the odd // and even positioned digits in a number function getSum(n) { n = reverse(n); let sumOdd = 0, sumEven = 0, c = 1; while (n != 0) { // If c is even number then it means // digit extracted is at even place if (c % 2 == 0) sumEven += n % 10; else sumOdd += n % 10; n = Math.floor(n / 10); c++; } document.write("Sum odd = " + sumOdd); document.write("<br>"); document.write("Sum even = " + sumEven);} let n = 457892; // function call getSum(n);// This code is contributed by Surbhi Tyagi</script> |
Output:
Sum odd = 20 Sum even = 15
Time Complexity: O(log10n)
Auxiliary Space: O(1)
Another approach: The problem can be solved without reversing the number. We can extract all the digits from the number one by one from the end. If the original number was odd then the last digit must be odd positioned else it will be even positioned. After processing a digit, we can invert the state from odd to even and vice versa.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to find the sum of the odd // and even positioned digits in a number void getSum(int n) { // If n is odd then the last digit // will be odd positioned bool isOdd = (n % 2 == 1) ? true : false; // To store the respective sums int sumOdd = 0, sumEven = 0; // While there are digits left process while (n != 0) { // If current digit is odd positioned if (isOdd) sumOdd += n % 10; // Even positioned digit else sumEven += n % 10; // Invert state isOdd = !isOdd; // Remove last digit n /= 10; } cout << "Sum odd = " << sumOdd << "\n"; cout << "Sum even = " << sumEven; } // Driver code int main() { int n = 457892; getSum(n); return 0; } |
Java
// Java implementation of the above approach class GFG{ // Function to find the sum of the odd // and even positioned digits in a number static void getSum(int n) { // If n is odd then the last digit // will be odd positioned boolean isOdd = (n % 2 == 1) ? true : false; // To store the respective sums int sumOdd = 0, sumEven = 0; // While there are digits left process while (n != 0) { // If current digit is odd positioned if (isOdd) sumOdd += n % 10; // Even positioned digit else sumEven += n % 10; // Invert state isOdd = !isOdd; // Remove last digit n /= 10; } System.out.println("Sum odd = " + sumOdd); System.out.println("Sum even = " + sumEven); } // Driver code public static void main(String[] args) { int n = 457892; getSum(n); } } // This code is contributed by jrishabh99 |
Python3
# Python3 implementation of the approach # Function to find the sum of the odd # and even positioned digits in a number def getSum(n): # If n is odd then the last digit # will be odd positioned if (n % 2 == 1) : isOdd = True else: isOdd = False # To store the respective sums sumOdd = 0 sumEven = 0 # While there are digits left process while (n != 0) : # If current digit is odd positioned if (isOdd): sumOdd += n % 10 # Even positioned digit else: sumEven += n % 10 # Invert state isOdd = not isOdd # Remove last digit n //= 10 print( "Sum odd = " , sumOdd ) print("Sum even = " ,sumEven) # Driver code if __name__ =="__main__": n = 457892 getSum(n) # This code is contributed by chitranayal |
C#
// C# implementation of the above approach using System;class GFG{ // Function to find the sum of the odd// and even positioned digits in a numberstatic void getSum(int n){ // If n is odd then the last digit // will be odd positioned bool isOdd = (n % 2 == 1) ? true : false; // To store the respective sums int sumOdd = 0, sumEven = 0; // While there are digits left process while (n != 0) { // If current digit is odd positioned if (isOdd) sumOdd += n % 10; // Even positioned digit else sumEven += n % 10; // Invert state isOdd = !isOdd; // Remove last digit n /= 10; } Console.WriteLine("Sum odd = " + sumOdd); Console.Write("Sum even = " + sumEven);}// Driver code static public void Main (){ int n = 457892; getSum(n);}}// This code is contributed by offbeat |
Javascript
<script>// Javascript implementation of the approach // Function to find the sum of the odd // and even positioned digits in a number function getSum(n) { // If n is odd then the last digit // will be odd positioned let isOdd = (n % 2 == 1) ? true : false; // To store the respective sums let sumOdd = 0, sumEven = 0; // While there are digits left process while (n != 0) { // If current digit is odd positioned if (isOdd) sumOdd += n % 10; // Even positioned digit else sumEven += n % 10; // Invert state isOdd = !isOdd; // Remove last digit n = Math.floor(n/10); } document.write("Sum odd = " + sumOdd + "<br>"); document.write("Sum even = " + sumEven); } // Driver code let n = 457892; getSum(n); // This code is contributed by Mayank Tyagi</script> |
Output:
Sum odd = 20 Sum even = 15
Time Complexity: O(log10n) as while loop would run for log10n times
Auxiliary Space: O(1)
Method #3:Using string() method:
- Convert the integer to string. Traverse the string and store all even indices sum in one variable and all odd indices sum in another variable.
Below is the implementation:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to find the sum of the odd// and even positioned digits in a numbervoid getSum(int n){ // To store the respective sums int sumOdd = 0, sumEven = 0; // Converting integer to string string num = to_string(n); // Traversing the string for(int i = 0; i < num.size(); i++) { if (i % 2 == 0) sumOdd = sumOdd + (int(num[i]) - 48); else sumEven = sumEven + (int(num[i]) - 48); } cout << "Sum odd = " << sumOdd << "\n"; cout << "Sum even = " << sumEven << "\n";}// Driver codeint main(){ int n = 457892; getSum(n); return 0;}// This code is contributed by souravmahato348 |
Java
// Java implementation of the approach import java.util.*; class GFG{ static void getSum(int n){ // To store the respective sum int sumOdd = 0; int sumEven = 0; // Converting integer to String String num = String.valueOf(n); // Traversing the String for(int i = 0; i < num.length(); i++) if (i % 2 == 0) sumOdd = sumOdd + (num.charAt(i) - '0'); else sumEven = sumEven + (num.charAt(i) - '0'); System.out.println("Sum odd = " + sumOdd); System.out.println("Sum even = " + sumEven);} // Driver codepublic static void main(String[] args){ int n = 457892; getSum(n);}}// Code contributed by swarnalii |
Python3
# Python3 implementation of the approach# Function to find the sum of the odd# and even positioned digits in a numberdef getSum(n): # To store the respective sums sumOdd = 0 sumEven = 0 # Converting integer to string num = str(n) # Traversing the string for i in range(len(num)): if(i % 2 == 0): sumOdd = sumOdd+int(num[i]) else: sumEven = sumEven+int(num[i]) print("Sum odd = ", sumOdd) print("Sum even = ", sumEven)# Driver codeif __name__ == "__main__": n = 457892 getSum(n)# This code is contributed by vikkycirus |
C#
// C# implementation of the approachusing System;class GFG{static void getSum(int n){ // To store the respective sum int sumOdd = 0; int sumEven = 0; // Converting integer to String String num = n.ToString(); // Traversing the String for(int i = 0; i < num.Length; i++) if (i % 2 == 0) sumOdd = sumOdd + (num[i] - '0'); else sumEven = sumEven + (num[i] - '0'); Console.WriteLine("Sum odd = " + sumOdd); Console.WriteLine("Sum even = " + sumEven);}// Driver codepublic static void Main(){ int n = 457892; getSum(n);}}// This code is contributed by subhammahato348 |
Javascript
<script>// Javascript implementation of the approachfunction getSum(n){ // To store the respective sum let sumOdd = 0; let sumEven = 0; // Converting integer to String let num = (n).toString(); // Traversing the String for(let i = 0; i < num.length; i++) if (i % 2 == 0) sumOdd = sumOdd + (num[i] - '0'); else sumEven = sumEven + (num[i] - '0'); document.write("Sum odd = " + sumOdd+"<br>"); document.write("Sum even = " + sumEven+"<br>");}// Driver codelet n = 457892;getSum(n);// This code is contributed by unknown2108</script> |
Output
Sum odd = 20 Sum even = 15
Time Complexity: O(log10n), as the length of the string will be log10n.
Auxiliary Space: O(log10n)
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