leetcode

Divide a String Into Groups of Size k - Link

Question Description

Given a string s, divide the string into groups of size k (except for the last group, which may be shorter). If the last group is shorter than k, fill it with the character fill.

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Constraints

• 1 <= s.length <= 100
• 1 <= k <= 100
• s consists of lowercase and uppercase English letters.
• fill is a lowercase or uppercase English letter.

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Approach

• Calculate the number of resulting groups based on the string length and k.
• Iterate over the string in steps of k.
• For each group, build a substring of length k.
  • If characters run out, append the fill character.
• Store each group in the result array and return it.

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Dry Run

Example Input: s = “abcdef”, k = 3, fill = ‘x’

Step-by-step execution:

• sLen = 6, aLen = 2
• i=0: “abc”
• i=3: “def”
• Result: [“abc”, “def”]

Another example: s = “abc”, k = 3, fill = ‘x’

• i=0: “abc”
• Result: [“abc”]

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Solution

class Solution {
    public String[] divideString(String s, int k, char fill) {
        int sLen = s.length();
        int aLen = sLen % k == 0 ? sLen / k : sLen / k + 1;
        String[] str = new String[aLen];

        int index = 0;
        for (int i = 0; i < sLen; i += k) {
            StringBuilder sb = new StringBuilder();
            for (int j = i; j < k + i; j++) {
                if (j < sLen) {
                    sb.append(s.charAt(j));
                } else {
                    sb.append(fill);
                }
            }
            str[index++] = sb.toString();
        }
        return str;
    }
}

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Time and Space Complexity

• Time Complexity: O(n) where n is the length of string s.
• Space Complexity: O(n) for storing the resulting array of strings.

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