Number of Different Integers in a String

Problem

You are given a string word that consists of digits and lowercase English letters.

You will replace every non-digit character with a space. For example, "a123bc34d8ef34" will become " 123  34 8  34". Notice that you are left with some integers that are separated by at least one space: "123", "34", "8", and "34".

Return **the number of *different* integers after performing the replacement operations on **word.

Two integers are considered different if their decimal representations without any leading zeros are different.

Example 1:

Input: word = "a123bc34d8ef34"
Output: 3
Explanation: The three different integers are "123", "34", and "8". Notice that "34" is only counted once.

Example 2:

Input: word = "leet1234code234"
Output: 2

Example 3:

Input: word = "a1b01c001"
Output: 1
Explanation: The three integers "1", "01", and "001" all represent the same integer because
the leading zeros are ignored when comparing their decimal values.

Constraints:

1 <= word.length <= 1000
word consists of digits and lowercase English letters.

Solution (Java)

class Solution {
    public int numDifferentIntegers(String word) {
        Set<String> ints = new HashSet<>();
        char[] chars = word.toCharArray();
        int start = -1;
        int stop = 0;
        for (int i = 0; i < chars.length; i++) {
            if (chars[i] >= '0' && chars[i] <= '9') {
                if (start == -1) {
                    start = i;
                }
                stop = i;
            } else if (start != -1) {
                ints.add(extractInt(chars, start, stop));
                start = -1;
            }
        }
        if (start != -1) {
            ints.add(extractInt(chars, start, stop));
        }
        return ints.size();
    }

    private String extractInt(char[] chrs, int start, int stop) {
        StringBuilder stb = new StringBuilder();
        while (start <= stop && chrs[start] == '0') {
            start++;
        }
        if (start >= stop) {
            stb.append(chrs[stop]);
        } else {
            while (start <= stop) {
                stb.append(chrs[start++]);
            }
        }
        return stb.toString();
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).