Partition of label

Problem

Tc : O(nlogn) for sorting the intervals array + O(25) for other iterations
Sc :O(n) for using List

class Solution {
    public List<Integer> partitionLabels(String s) {
        int intervals[][] = new int[26][2];
        for(int i =0;i<intervals.length;i++){
            intervals[i][0] = 501;
            intervals[i][1] = -1;
        }
        for(int i = 0;i<s.length();i++){
            char  c = s.charAt(i);
            int index  = c-'a';
            intervals[index][0] = Math.min(intervals[index][0],i);
            intervals[index][1] = Math.max(intervals[index][1],i);
        }

        //sorting 
        Arrays.sort(intervals,(a,b)-> a[0]-b[0]);

        //merging
        int start = intervals[0][0];
        int end = intervals[0][1];
        List<Integer> list = new ArrayList<>();
        for(int i =1;i<intervals.length;i++){
            if(intervals[i][0]==501) break;// exit as no need to traverse any further
            if(end >= intervals[i][0]){
                start = Math.min(intervals[i][0],start);
                end = Math.max(intervals[i][1],end);
            }
            else{
               list.add(end-start+1); // instead of keeping the range [start, end] in the list we are addting their sizes
               start = intervals[i][0];
               end = intervals[i][1];
            }
        }
        list.add(end-start+1);
        return list;
    }
}