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3163. String Compression III

Description

Given a string word, compress it using the following algorithm:

  • Begin with an empty string comp. While word is not empty, use the following operation:
    • Remove a maximum length prefix of word made of a single character c repeating at most 9 times.
    • Append the length of the prefix followed by c to comp.

Return the string comp.

 

Example 1:

Input: word = "abcde"

Output: "1a1b1c1d1e"

Explanation:

Initially, comp = "". Apply the operation 5 times, choosing "a", "b", "c", "d", and "e" as the prefix in each operation.

For each prefix, append "1" followed by the character to comp.

Example 2:

Input: word = "aaaaaaaaaaaaaabb"

Output: "9a5a2b"

Explanation:

Initially, comp = "". Apply the operation 3 times, choosing "aaaaaaaaa", "aaaaa", and "bb" as the prefix in each operation.

  • For prefix "aaaaaaaaa", append "9" followed by "a" to comp.
  • For prefix "aaaaa", append "5" followed by "a" to comp.
  • For prefix "bb", append "2" followed by "b" to comp.

 

Constraints:

  • 1 <= word.length <= 2 * 105
  • word consists only of lowercase English letters.

Solutions

Solution 1: Two Pointers

We can use two pointers to count the consecutive occurrences of each character. Suppose the current character $c$ appears consecutively $k$ times, then we divide $k$ into several $x$, each $x$ is at most $9$, then we concatenate $x$ and $c$, and append each $x$ and $c$ to the result.

Finally, return the result.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the

  • class Solution {
    public String compressedString(String word) {
        StringBuilder ans = new StringBuilder();
        int n = word.length();
        for (int i = 0; i < n;) {
            int j = i + 1;
            while (j < n && word.charAt(j) == word.charAt(i)) {
                ++j;
            }
            int k = j - i;
            while (k > 0) {
                int x = Math.min(9, k);
                ans.append(x).append(word.charAt(i));
                k -= x;
            }
            i = j;
        }
        return ans.toString();
    }
}

  • class Solution {
public:
    string compressedString(string word) {
        string ans;
        int n = word.length();
        for (int i = 0; i < n;) {
            int j = i + 1;
            while (j < n && word[j] == word[i]) {
                ++j;
            }
            int k = j - i;
            while (k > 0) {
                int x = min(9, k);
                ans.push_back('0' + x);
                ans.push_back(word[i]);
                k -= x;
            }
            i = j;
        }
        return ans;
    }
};

  • class Solution:
    def compressedString(self, word: str) -> str:
        g = groupby(word)
        ans = []
        for c, v in g:
            k = len(list(v))
            while k:
                x = min(9, k)
                ans.append(str(x) + c)
                k -= x
        return "".join(ans)


  • func compressedString(word string) string {
	ans := []byte{}
	n := len(word)
	for i := 0; i < n; {
		j := i + 1
		for j < n && word[j] == word[i] {
			j++
		}
		k := j - i
		for k > 0 {
			x := min(9, k)
			ans = append(ans, byte('0'+x))
			ans = append(ans, word[i])
			k -= x
		}
		i = j
	}
	return string(ans)
}

  • function compressedString(word: string): string {
    const ans: string[] = [];
    const n = word.length;
    for (let i = 0; i < n; ) {
        let j = i + 1;
        while (j < n && word[j] === word[i]) {
            ++j;
        }
        let k = j - i;
        while (k) {
            const x = Math.min(k, 9);
            ans.push(x + word[i]);
            k -= x;
        }
        i = j;
    }
    return ans.join('');
}

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