Partition String - Amazon Top Interview Questions

Problem Statement :

Given a lowercase alphabet string s, partition s into as many pieces as possible such that each letter appears in at most one piece and return the sizes of the partitions as a list.


0 ≤ n ≤ 100,000 where n is the length of s

Example 1


s = "cocoplaydae"


[4, 1, 1, 4, 1]


The string is split to["coco", "p", "l", "ayda", "e"].

Solution :


                        Solution in C++ :

vector<int> solve(string s) {
    // store the last index of all the chars
    vector<int> m(26);
    for (int i = 0; i < s.size(); i++) {
        m[s[i] - 'a'] = i;
    // integer which stores the index where we have to go(at least) for creating a piece
    int curr = 0;
    // interger which stores the last index, from where we had started
    int last = 0;
    vector<int> res;
    for (int i = 0; i < s.size(); i++) {
        // take the max index where we have to go for creating a piece
        curr = max(curr, m[s[i] - 'a']);
        if (curr == i) {
            // if the last index of current substring is the current index itself, create a piece
            res.push_back(curr - last + 1);
            // update the last index to next index
            last = curr + 1;
    return res;

                        Solution in Java :

import java.util.*;

class Solution {
    public int[] solve(String s) {
        int[] arr = new int[26];
        List<Integer> list = new ArrayList<>();
        int end;

        for (int i = 0; i < s.length(); i++) {
            arr[s.charAt(i) - 'a'] = i;

        for (int i = 0; i < s.length(); i++) {
            end = arr[s.charAt(i) - 'a'];

            for (int j = i; j < end; j++) {
                end = Math.max(end, arr[s.charAt(j) - 'a']);

            list.add(end - i + 1);
            i = end;

        int[] res = new int[list.size()];

        for (int i = 0; i < res.length; i++) res[i] = list.get(i);

        return res;

                        Solution in Python : 
class Solution:
    def solve(self, s):
        last = {c: i for i, c in enumerate(s)}
        lo = r = 0
        res = []

        for hi, c in enumerate(s):
            r = max(r, last[c])
            if hi == r:
                res.append(hi - lo + 1)
                lo = r = hi + 1

        return res

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