Partition Labels

/*
 * Problem : Cut into the most pieces.
 * Input   : One string.
 * Output  : The piece lengths.
 * Approach: Last occurrence greedy
 * Time    : O(n)
 * Space   : O(1)
 * Java    : 8 or newer.
 */
import java.util.Scanner;
import java.util.Arrays;
import java.util.stream.Collectors;

public class Main {
    // This method solves the task. It receives the input and returns the answer.
    static int[] solve(String s) {
        // Step 1: Remember the last position of every letter.
        int[] last = new int[128];
        for (int i = 0; i < s.length(); i++) {
            last[s.charAt(i)] = i;
        }
        java.util.List<Integer> sizes = new java.util.ArrayList<Integer>();
        int start = 0;
        int end = 0;
        for (int i = 0; i < s.length(); i++) {
            // Step 2: The piece must reach the last position of every letter that it contains.
            end = Math.max(end, last[s.charAt(i)]);
            // Step 3: When we reach that end, no letter of this piece appears later. Cut here.
            if (i == end) {
                sizes.add(i - start + 1);
                start = i + 1;
            }
        }
        int[] out = new int[sizes.size()];
        for (int i = 0; i < out.length; i++) {
            out[i] = sizes.get(i);
        }
        return out;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String s = scanner.hasNextLine() ? scanner.nextLine() : "";
        System.out.println(Arrays.stream(solve(s)).mapToObj(String::valueOf).collect(Collectors.joining(" ")));
    }
}