algorithm notes: leetcode#806 number of lines to write string

Problem

We are to write the letters of a given string S, from left to right into lines. Each line has maximum width 100 units, and if writing a letter would cause the width of the line to exceed 100 units, it is written on the next line. We are given an array widths, an array where widths[0] is the width of ‘a’, widths[1] is the width of ‘b’, …, and widths[25] is the width of ‘z’.

Now answer two questions: how many lines have at least one character from S, and what is the width used by the last such line? Return your answer as an integer list of length 2.

Example :
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:
All letters have the same length of 10. To write all 26 letters,
we need two full lines and one line with 60 units.

Example :
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:
All letters except ‘a’ have the same length of 10, and
"bbbcccdddaa” will cover 9 * 10 + 2 * 4 = 98 units.
For the last ‘a’, it is written on the second line because
there is only 2 units left in the first line.
So the answer is 2 lines, plus 4 units in the second line.

Note:

• The length of S will be in the range [1, 1000].
• S will only contain lowercase letters.
• widths is an array of length 26.
• widths[i] will be in the range of [2, 10].

Solution

Basic idea

Use two variables to help solve the problem, one to record the line number and one to store width units. Initialize line number with 1 and line width with 0, which means we start from the first line and we have not put anything in the line. Then iterate each character in the given string, find its width, and try to put it in the current line. If line width is over 100, the character will be put in the next line, and we will increase line number by 1 and set line width equal to width of the character.

Python implementation

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class :
    def numberOfLines(self, widths, S):
        """
        :type widths: List[int]
        :type S: str
        :rtype: List[int]
        """
        ans = [1, 0]
        for c in S:
            num = widths[ord(c) - ord("a")]
            if ans[1] + num > 100:
                ans[0] += 1
                ans[1] = num
            else:
                ans[1] += num
        return ans

Java implementation

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class  {
    public int[] numberOfLines(int[] widths, String S) {
        int[] ans = {1, 0};
        for(char c : S.toCharArray()){
            int length = widths[c - 'a'];
            if(ans[1] + length > 100){
                ans[0] += 1;
                ans[1] = length;
            }else{
                ans[1] += length;
            }
        }
        return ans;
    }
}

Time complexity analysis

O(N). N is the length of string S, because we need to iterate each character in the string.

Space complexity analysis

O(1). Use a list with two elements to store values.

Links

806. Number of Lines To Write String
806. 写字符串需要的行数
(中文版) 算法笔记: 力扣#806 写字符串需要的行数