LeetCode 604: Design Compressed String Iterator Solution

Master LeetCode problem 604 (Design Compressed String Iterator), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

604. Design Compressed String Iterator

Easy

Array String Design Iterator

Problem Explanation

Explanation:

To design a compressed string iterator, we need to implement a class that takes a compressed string as input and allows iterating over each character in the uncompressed form. The compressed string consists of a sequence of characters followed by their frequencies.

We can use two pointers - one to keep track of the current character and another to keep track of the remaining count of that character. When iterating, we decrement the count until it becomes zero and move to the next character.

Algorithm:

Initialize class variables to store the current character, count of that character, and the compressed string.
Implement a constructor that takes the compressed string as input and initializes the class variables.
Implement a next() function that returns the next character in the uncompressed form.
Implement a hasNext() function that checks if there are more characters to iterate over.

Time Complexity:

Constructor: O(n), where n is the length of the compressed string.
next(): O(1)
hasNext(): O(1)

Space Complexity:

O(n) to store the compressed string.

: :

Solution Code

class StringIterator {
    private char ch;
    private int count;
    private int index;
    private String compressedString;

    public StringIterator(String compressedString) {
        this.compressedString = compressedString;
        this.index = 0;
        this.count = 0;
        this.ch = ' ';
    }

    public char next() {
        if (count == 0) {
            if (index >= compressedString.length()) {
                return ' ';
            }
            ch = compressedString.charAt(index++);
            while (index < compressedString.length() && Character.isDigit(compressedString.charAt(index))) {
                count = count * 10 + (compressedString.charAt(index++) - '0');
            }
        }
        count--;
        return ch;
    }

    public boolean hasNext() {
        return index < compressedString.length() || count > 0;
    }
}

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