35. Partition Labels

Topic :

hashing

Difficulty :

medium

Problem Link :

problem statement

You are given a string s. We want to partition the string into as many parts as possible so that each letter appears in at most one part.

Note that the partition is done so that after concatenating all the parts in order, the resultant string should be s.

Return a list of integers representing the size of these parts.

Example 1:

Input: s = "ababcbacadefegdehijhklij"
Output: [9,7,8]
Explanation:
The partition is "ababcbaca", "defegde", "hijhklij".
This is a partition so that each letter appears in at most one part.
A partition like "ababcbacadefegde", "hijhklij" is incorrect,
because it splits s into less parts.

Example 2:

Input: s = "eccbbbbdec"
Output: [10]

Constraints:

1 <= s.length <= 500
s consists of lowercase English letters.

solution

import java.io.*;
import java.util.*;
class Partition_Labels
{
  public static void main(String args[]){
      String s="ababcbacadefegdehijhklij";
      System.out.println(partitionLabels(s));
    } 
   static List<Integer> partitionLabels(String s) {
        HashMap<Character,Integer> map=new HashMap<>();
        for(int i=0;i<s.length();i++)
        {
            char ch=s.charAt(i); 
            map.put(ch,i); //store the index upto which a charachter impacts most
        }
        
        List<Integer> ans= new ArrayList<>();
        int max=0;
        int prev=-1;
        
        for(int i=0;i<s.length();i++){
            
            char ch=s.charAt(i);
            
            max=Math.max(max,map.get(ch));
            
            if(i==max){
                //partition 
                ans.add(max-prev);
                prev=max;
            }
        }
        
        return ans;
    }  
}