2138. Divide a String Into Groups of Size k

Description

A string s can be partitioned into groups of size k using the following procedure:

• The first group consists of the first k characters of the string, the second group consists of the next k characters of the string, and so on. Each element can be a part of exactly one group.
• For the last group, if the string does not have k characters remaining, a character fill is used to complete the group.

Note that the partition is done so that after removing the fill character from the last group (if it exists) and concatenating all the groups in order, the resultant string should be s.

Given the string s, the size of each group k and the character fill, return a string array denoting the composition of every group s has been divided into, using the above procedure.

Example 1:

Input: s = "abcdefghi", k = 3, fill = "x"
Output: ["abc","def","ghi"]
Explanation:
The first 3 characters "abc" form the first group.
The next 3 characters "def" form the second group.
The last 3 characters "ghi" form the third group.
Since all groups can be completely filled by characters from the string, we do not need to use fill.
Thus, the groups formed are "abc", "def", and "ghi".

Example 2:

Input: s = "abcdefghij", k = 3, fill = "x"
Output: ["abc","def","ghi","jxx"]
Explanation:
Similar to the previous example, we are forming the first three groups "abc", "def", and "ghi".
For the last group, we can only use the character 'j' from the string. To complete this group, we add 'x' twice.
Thus, the 4 groups formed are "abc", "def", "ghi", and "jxx".

Constraints:

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

Solutions

Solution 1: Simulation

We can directly simulate the process described in the problem statement, dividing the string \(s\) into groups of length \(k\). For the last group, if it contains fewer than \(k\) characters, we use the character \(\textit{fill}\) to pad it.

The time complexity is \(O(n)\), and the space complexity is \(O(n)\), where \(n\) is the length of the string \(s\).

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class Solution:
    def divideString(self, s: str, k: int, fill: str) -> List[str]:
        return [s[i : i + k].ljust(k, fill) for i in range(0, len(s), k)]

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class Solution {
    public String[] divideString(String s, int k, char fill) {
        int n = s.length();
        String[] ans = new String[(n + k - 1) / k];
        if (n % k != 0) {
            s += String.valueOf(fill).repeat(k - n % k);
        }
        for (int i = 0; i < ans.length; ++i) {
            ans[i] = s.substring(i * k, (i + 1) * k);
        }
        return ans;
    }
}

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class Solution {
public:
    vector<string> divideString(string s, int k, char fill) {
        int n = s.size();
        if (n % k) {
            s += string(k - n % k, fill);
        }
        vector<string> ans;
        for (int i = 0; i < s.size() / k; ++i) {
            ans.push_back(s.substr(i * k, k));
        }
        return ans;
    }
};

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func divideString(s string, k int, fill byte) (ans []string) {
    n := len(s)
    if n%k != 0 {
        s += strings.Repeat(string(fill), k-n%k)
    }
    for i := 0; i < len(s)/k; i++ {
        ans = append(ans, s[i*k:(i+1)*k])
    }
    return
}

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function divideString(s: string, k: number, fill: string): string[] {
    const ans: string[] = [];
    for (let i = 0; i < s.length; i += k) {
        ans.push(s.slice(i, i + k).padEnd(k, fill));
    }
    return ans;
}

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