Smallest odd number with even sum of digits from the given number N
Given a large number in form of string str. The task is to find the smallest odd number whose sum of digits is even by removing zero or more characters from the given string str, where the digits can be rearranged.
Examples
Input: str = “15470”
Output: 15
Explanation:
Two smallest odd digits are 1 & 5. Hence the required number is 15.Input: str = “124”
Output: -1
Explanation:
There is no smallest odd digit other than 1. Hence the required number can’t be form.
Approach:
On observing closely, by intuition, it can be understood that the number of digits in the smallest odd number possible is 2. And every digit in this number is odd because the sum of two odd digits is always even. Therefore, the idea to solve this problem is to iterate through the given string and store every odd number in an array. This array can be sorted and the first two digits together form the smallest odd number whose sum of its digits is even.
Below is the implementation of the above approach.
C++
// C++ program to find the smallest odd number // with even sum of digits from the given number N #include<bits/stdc++.h> using namespace std; // Function to find the smallest odd number // whose sum of digits is even from the given string int smallest(string s) { // Converting the given string // to a list of digits vector< int > a(s.length()); for ( int i = 0; i < s.length(); i++) a[i] = s[i]- '0' ; // An empty array to store the digits vector< int > b; // For loop to iterate through each digit for ( int i = 0; i < a.size(); i++) { // If the given digit is odd then // the digit is appended to the array b if ((a[i]) % 2 != 0) b.push_back(a[i]); } // Sorting the list of digits sort(b.begin(),b.end()); // If the size of the list is greater than 1 // then a 2 digit smallest odd number is returned // Since the sum of two odd digits is always even if (b.size() > 1) return (b[0])*10 + (b[1]); // Else, -1 is returned return -1; } // Driver code int main() { cout << (smallest( "15470" )); } // This code is contributed by Surendra_Gangwar |
Java
// Java program to find the smallest // odd number with even sum of digits // from the given number N import java.util.*; class GFG{ // Function to find the smallest // odd number whose sum of digits // is even from the given string public static int smallest(String s) { // Converting the given string // to a list of digits int [] a = new int [s.length()]; for ( int i = 0 ; i < s.length(); i++) a[i] = s.charAt(i) - '0' ; // An empty array to store the digits Vector<Integer> b = new Vector<Integer>(); // For loop to iterate through each digit for ( int i = 0 ; i < a.length; i++) { // If the given digit is odd // then the digit is appended // to the array b if (a[i] % 2 != 0 ) b.add(a[i]); } // Sorting the list of digits Collections.sort(b); // If the size of the list is greater // than 1 then a 2 digit smallest odd // number is returned. Since the sum // of two odd digits is always even if (b.size() > 1 ) return (b.get( 0 )) * 10 + (b.get( 1 )); // Else, -1 is returned return - 1 ; } // Driver code public static void main(String[] args) { System.out.print(smallest( "15470" )); } } // This code is contributed by divyeshrabadiya07 |
Python
# Python program to find the smallest odd number # with even sum of digits from the given number N # Function to find the smallest odd number # whose sum of digits is even from the given string def smallest(s): # Converting the given string # to a list of digits a = list (s) # An empty array to store the digits b = [] # For loop to iterate through each digit for i in range ( len (a)): # If the given digit is odd then # the digit is appended to the array b if ( int (a[i]) % 2 ! = 0 ): b.append(a[i]) # Sorting the list of digits b = sorted (b) # If the size of the list is greater than 1 # then a 2 digit smallest odd number is returned # Since the sum of two odd digits is always even if ( len (b)> 1 ): return int (b[ 0 ]) * 10 + int (b[ 1 ]) # Else, -1 is returned return - 1 # Driver code if __name__ = = "__main__" : print (smallest( "15470" )) |
C#
// C# program to find the smallest // odd number with even sum of digits // from the given number N using System; using System.Collections; class GFG{ // Function to find the smallest // odd number whose sum of digits // is even from the given string public static int smallest( string s) { // Converting the given string // to a list of digits int [] a = new int [s.Length]; for ( int i = 0; i < s.Length; i++) a[i] = ( int )(s[i] - '0' ); // An empty array to store the digits ArrayList b = new ArrayList(); // For loop to iterate through each digit for ( int i = 0; i < a.Length; i++) { // If the given digit is odd // then the digit is appended // to the array b if (a[i] % 2 != 0) b.Add(a[i]); } // Sorting the list of digits b.Sort(); // If the size of the list is greater // than 1 then a 2 digit smallest odd // number is returned. Since the sum // of two odd digits is always even if (b.Count > 1) return (( int )b[0] * 10 + ( int )b[1]); // Else, -1 is returned return -1; } // Driver code public static void Main( string [] args) { Console.Write(smallest( "15470" )); } } // This code is contributed by rutvik_56 |
Javascript
<script> // Javascript program to find the // smallest odd number with even // sum of digits from the given number N // Function to find the smallest odd // number whose sum of digits is even // from the given string function smallest(s) { // Converting the given string // to a list of digits var a = Array(s.length); for ( var i = 0; i < s.length; i++) a[i] = s[i].charCodeAt(0) - '0' .charCodeAt(0); // An empty array to store the digits var b = []; // For loop to iterate through each digit for ( var i = 0; i < a.length; i++) { // If the given digit is odd then // the digit is appended to the array b if ((a[i]) % 2 != 0) b.push(a[i]); } // Sorting the list of digits b.sort((a, b) => a - b); // If the size of the list is greater // than 1 then a 2 digit smallest odd // number is returned. Since the sum // of two odd digits is always even if (b.length > 1) return (b[0]) * 10 + (b[1]); // Else, -1 is returned return -1; } // Driver code document.write(smallest( "15470" )); // This code is contributed by importantly </script> |
15
Time Complexity: O(N) where N = length of string.
Auxiliary Space: O(N)
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