Algorithm: given a list of strings, return all the strings with the nth longest length in that list for example: list – Yuri, Ron, Interview, Longest, List, Contain and nth = 1 will return just “Interview” whereas nth = 2 will return both “Longest” and “Contain”.
Though solution to “How to find the kth largest element in an unsorted array of length n in O(n)?” can be applied to string length, how to translate back to print all the strings with n length?
Solution
I have written a simple java program which is able to find “all Nth longest elements” from a list of strings. Program is as below:
package com.howtodoinjava.examples;
import java.util.ArrayList;
import java.util.List;
import java.util.TreeMap;
public class NthLogestStringAlgorithm
{
public static void main(String[] args)
{
int n = 0;
List<String> list = new ArrayList<String>();
list.add("Yuri");
list.add("Ron");
list.add("Interview");
list.add("Longest");
list.add("List");
list.add("Contain");
System.out.println( findNthLongestElement(list, n) );
}
private static List<String> findNthLongestElement(List<String> list, int n)
{
if(n < 1) {
return null; //Handle invalid case
}
TreeMap<Integer, List<String>> map = new TreeMap<>();
for(String str : list)
{
Integer length = str.length();
List<String> tempList = map.get(length) != null ? map.get(length) : new ArrayList<String>();
tempList.add(str);
map.put(length, tempList);
}
return map.get( map.descendingKeySet().toArray()[n-1] );
}
}========Output of program======= n = 0 => null n = 1 => [Interview] n = 2 => [Longest, Contain] n = 3 => [Yuri, List] n = 4 => [Ron]
Happy Learning !!
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Robert
similar solution
public static List getLongestNthString(List strings, int n) {
TreeMap<Integer, List> grouped = strings.stream().collect(
Collectors.groupingBy(
String::length,
TreeMap::new,
Collectors.toList()));
return grouped.get(grouped.descendingKeySet().toArray()[n – 1]);
}
Juan David Gómez
Using stream
return input.stream() .sorted((s1, s2) -> s2.length() - s1.length()) .limit(n) .collect(Collectors.toList());kd
if i remove interview it will return yuri and list
Lokesh Gupta
I am not sure how you are testing the program. I removed Interview from the list and these are new outputs:
n = 0 => nulln = 1 => [Longest, Contain]
n = 2 => [Yuri, List]
n = 3 => [Ron]
kd
it will always return elements from position n-1.
priyesh
Hi Lokesh,
1 problem though. It will always return null. The if condition will always be true.
Lokesh Gupta
Indeed, if you do not change the value of n in main() function.