Smallest odd number with even sum of digits from the given number N - GeeksforGeeks

30 Jan, 2022
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++
Java
Python
C#
Javascript
// 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 stringint 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 codeint main(){    cout << (smallest("15470"));} // This code is contributed by Surendra_Gangwar
// Java program to find the smallest// odd number with even sum of digits// from the given number Nimport java.util.*;class GFG{     // Function to find the smallest// odd number whose sum of digits// is even from the given stringpublic 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 codepublic static void main(String[] args){    System.out.print(smallest("15470"));}} // This code is contributed by divyeshrabadiya07
# 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 stringdef 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 codeif __name__ == "__main__":    print(smallest("15470"))
// C# program to find the smallest// odd number with even sum of digits// from the given number Nusing System;using System.Collections; class GFG{      // Function to find the smallest// odd number whose sum of digits// is even from the given stringpublic 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 codepublic static void Main(string[] args){    Console.Write(smallest("15470"));}} // This code is contributed by rutvik_56
<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 stringfunction 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 codedocument.write(smallest("15470")); // This code is contributed by importantly </script>
15
Time Complexity: O(N) where N = length of string.
Auxiliary Space: O(N) 
SURENDRA_GANGWAR
divyeshrabadiya07
rutvik_56
importantly
subhammahato348
number-digits
Technical Scripter 2019
Mathematical
Strings
Technical Scripter
Strings
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