Run-Length Decoded String Iterator - Google Top Interview Questions


Problem Statement :


Given a run-length encoded lowercase alphabet string s, implement an iterator which is the decoded version of s:

next() polls the next element in the iterator

hasnext() which returns whether the next element exists

Constraints

n ≤ 100,000 where n is the number of calls to next and hasnext

Example 1

Input

methods = ["constructor", "next", "hasnext", "next", "next", "hasnext"]

arguments = [["2a1b"], [], [], [], [], []]`

Output

[None, "a", True, "a", "b", False]


Solution :



title-img 




                        Solution in C++ :

class RunLengthDecodedIterator {
    string s = "";
    int i = -1;
    int count = 0;

    public:
    RunLengthDecodedIterator(string s) : s(std::move(s)) {
    }

    string next() {
        if (!count) {
            i++;
            while (s[i] >= '0' && s[i] <= '9') {
                count *= 10;
                count += s[i] - '0';
                i++;
            }
        }
        count--;
        return string(1, s[i]);
    }

    bool hasnext() {
        return count || i < int(s.size() - 1);
    }
};
                    


                        Solution in Java :

import java.util.*;

class RunLengthDecodedIterator {
    class Alphabet {
        char c;
        long count;

        public Alphabet(char c, long count) {
            this.c = c;
            this.count = count;
        }
    }

    LinkedList<Alphabet> list;
    int index = 0;

    public RunLengthDecodedIterator(String s) {
        list = new LinkedList<>();
        int idx = 0;
        StringBuilder sb = new StringBuilder();
        while (idx < s.length()) {
            int j = idx;
            StringBuilder num = new StringBuilder();
            while (j < s.length()) {
                if (Character.isDigit(s.charAt(j))) {
                    num.append(s.charAt(j));
                } else {
                    break;
                }
                j++;
            }
            int limit = Integer.parseInt(num.toString());
            char c = s.charAt(j);
            list.add(new Alphabet(c, limit));
            idx = j + 1;
        }
    }

    public String next() {
        if (list.size() == 0)
            return "";
        Alphabet a = list.get(index);
        if (a.count == 1) {
            list.remove(index);
            return "" + a.c;
        }
        a.count -= 1;
        return "" + a.c;
    }

    public boolean hasnext() {
        return list.size() > 0;
    }
}
                    


                        Solution in Python : 
                            
class RunLengthDecodedIterator:
    def __init__(self, s):
        self.c, self.cnt = "", 0
        self.i, self.N = 0, len(s)
        self.s = s

    def move(self):
        char = False
        n = 0
        while not char:
            if self.s[self.i].isdigit():
                n = 10 * n + int(self.s[self.i])
            else:
                self.c = self.s[self.i]
                self.cnt = n
                char = True
            self.i += 1

    def next(self):
        if not self.cnt:
            self.move()
        self.cnt -= 1
        return self.c

    def hasnext(self):
        return self.i < self.N or self.cnt > 0


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