Largest Odd Number
Largest Odd Number
PROBLEM STATEMENT :
The program must accept a string S and print the largest odd number L present in S. If there is no odd number in S, the program must print 0 as the output.
Boundary Condition(s):
1 <= Length of S <= 100
Input Format:
The first line contains S.
Output Format:
The first line contains L or 0 as per the given condition.
Example Input/Output 1:
Input: ()
123456
Output:
12345
Explanation:
1
123
12345
23
3
2345
345
45
5
The largest odd integer in the string is 12345. So 12345 is printed as the output.
Example Input/Output 2:
Input: ()
4466
Output:
0
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3) APPLY ONCE
SOLUTION :
C (Programming Language)
#include<stdio.h>
#include<stdlib.h>
int main()
{
char s[101];
int ind=-1;
scanf("%s",s);
for(int i=0;i<strlen(s);i++)
{
if((s[i]-'0')%2==1)
ind=i;
}
if(ind==-1)
printf("0");
else
for(int i=0;i<=ind;i++)
printf("%c",s[i]);
}
C++ (CPP)
#include <bits/stdc++.h>
using namespace std;
int main(int argc, char** argv)
{
string s ;
cin >> s;
int ind = s.size()-1;
while(ind>=0 && (s[ind]-'0')%2==0)
ind--;
if(ind<0)
cout << 0;
else
cout << s.substr(0,ind+1);
}
JAVA
import java.util.*;
import java.util.Scanner;
public class Hello {
public static void main(String[] args) {
Scanner sc = new Scanner (System.in);
String S;
int flag=0;
S=sc.next();
for(int i=S.length()-1;i>-1;i--){
if(Character.getNumericValue(S.charAt(i))%2!=0){
for(int j=0;j<=i;j++){
System.out.print(S.charAt(j));
}
flag=1;
break;
}
}
if(flag==0){
System.out.print("0");
}
}
}
PYTHON
s=input().strip()
l=len(s)
lst=[]
for i in range(l):
for j in range(i,l):
num=int(s[i:j+1])
if num%2!=0:
lst.append(num)
print(0 if lst==[] else max(lst))

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