Number of Lines To Write String - LeetCode solutions

Problem

You are given a string s of lowercase English letters and an array widths denoting how many pixels wide each lowercase English letter is. Specifically, widths[0] is the width of 'a', widths[1] is the width of 'b', and so on.

You are trying to write s across several lines, where **each line is no longer than *100* pixels**. Starting at the beginning of s, write as many letters on the first line such that the total width does not exceed 100 pixels. Then, from where you stopped in s, continue writing as many letters as you can on the second line. Continue this process until you have written all of s.

Return **an array *result* of length 2 where:**

result[0]** is the total number of lines.**
result[1]** is the width of the last line in pixels.**

Example 1:

Input: widths = [10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10], s = "abcdefghijklmnopqrstuvwxyz"
Output: [3,60]
Explanation: You can write s as follows:
abcdefghij  // 100 pixels wide
klmnopqrst  // 100 pixels wide
uvwxyz      // 60 pixels wide
There are a total of 3 lines, and the last line is 60 pixels wide.

Example 2:

Input: widths = [4,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10], s = "bbbcccdddaaa"
Output: [2,4]
Explanation: You can write s as follows:
bbbcccdddaa  // 98 pixels wide
a            // 4 pixels wide
There are a total of 2 lines, and the last line is 4 pixels wide.

Constraints:

widths.length == 26
2 <= widths[i] <= 10
1 <= s.length <= 1000
s contains only lowercase English letters.

Solution (Java)

class Solution {
    public int[] numberOfLines(int[] widths, String s) {
        int count = 0;
        int line = 0;
        int i = 0;
        while (i < s.length()) {
            count += widths[s.charAt(i) - 'a'];
            if (count == 100) {
                line++;
                count = 0;
            }
            if (count > 100) {
                line++;
                i--;
                count = 0;
            }
            i++;
        }
        if (count > 0 && count < 100) {
            line++;
        }
        if (count == 0) {
            count = 100;
        }
        return new int[] {line, count};
    }
}

Explain:

nope.

Complexity:

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