CareerCup

List traversal, O(n) time, saving the largest you see.

// C#
public string FindLongest(string[] s)
{
  if (s == null || s.Length == 0)
  {
    return ""; // Or throw, depending on interviewer's preference.  Always error check
  }
  string maxSeen = string.Empty;
  for(int i = 0; i < s.Length; i++)
  {
    if(s[i].Length > maxSeen.Length)
    {
      maxSeen = s[i];
    }
  }
  return maxSeen;
}

make two empty stack. Push all the string in a stack
suppose the stack name is s in which you want to push

for(i=1;i<size+1;i++) {
b=s.pop();
int k=b.compareTo(max);
if(k>0) {
max=b;
}
t.push(b);
}
System.out.print(max+" ");
for(i=1;i<size+1;i++) {
b=t.pop();
s.push(b);
}

import java.util.Arrays;
public class Expand {
public static void main(String args[])
{
String[] arr={"careercu","career" ,"dewdeals" , "online","filewrit","filewrit"};
int [] arr1=new int[arr.length];
int [] arr2=new int[arr.length];
for(int i=0;i<arr.length;i++)
{
arr1[i]=arr[i].length(); //making two arrays to grab the length of one of the string
arr2[i]=arr[i].length();
}
Arrays.sort(arr2);//to sort the array lengths
int x=arr2[arr1.length-1];
for(int j=0;j<arr1.length;j++)
{
if(x==arr1[j])
System.out.println(arr[j]);
}
}
}

this code perfectly works for any no os the strings

#include<conio.h>
#include<iostream.h>
#include<string.h>

void main()
{

int i,num,max,max_pos;
clrscr();
cout<<"enter the number of words you want to compare";
cin>>num;
char ch[10][50];
for(i=0; i<num;i++)
{
cout<<"enter string"<<i;
cin>>ch[i];
}

if(ch==NULL)
{
 cout<<"error"; //or return();

}
else
{
 max= strlen(ch[0]);
 max_pos= 0;
 for (i=1;i<num;i++)
 {
  if (max< strlen(ch[i]))
   {
     max= strlen(ch[i]);
     max_pos=i;
   }

  }
    cout<< "found element"<<ch[max_pos]<<"\t"<<max;
 }
getch();
}

You can do it in sublinear time by using a trie.

scala> val xs = List("Google", "Microsoft", "Apple", "Amazon")
xs: List[java.lang.String] = List(Google, Microsoft, Apple, Amazon)

scala> xs.maxBy(_.length)
res1: java.lang.String = Microsoft

import java.util.HashMap;
import java.util.Map;

public class LargestString {
public static void main(String[] args) {
String[] stringTable = new String[] {"careercup", "career", "dewdeals", "onlineshopping", "onlineshoppin1"};

System.out.println("Result is:" + largestString(stringTable));

}

private static String largestString(String... stringTable) {
assert stringTable != null && stringTable.length > 0 : "Input cannot be null or empty table.";
Map<Integer, String> cache = new HashMap<Integer, String>(stringTable.length);
int l, maxLength = 0;
for(String s : stringTable) {
l = s.length();
cache.put(l, s);
maxLength = Math.max(maxLength, l);
}
return cache.get(maxLength);
}
}

I found this one simpler

public class LargestString {
    public static void main(String[] args) {
        String[] arr={"careercu","career" ,"dewdeals" , "online","filewrit","filewrit"};
        int maxLen=0;
        for(String str:arr){
            if(maxLen<str.length()){maxLen=str.length();}
        }

        for(String str:arr){
            if(str.length()==maxLen){System.out.println(str);}
        }

    }

}

char* str[] = {"Career_Cup","Vinay","India","No"};
int maxLen = 0;
int index = -1;
for(int i = 0; i < sizeof(*str); ++i)
{
int cur_len = strlen(str[i]);
if( cur_len > maxLen)
{
maxLen = cur_len;
index = i;
}
}
cout<<"Max Len string is:"<<str[index];
}