String Compression

LeetCode 443 | Difficulty: Medium

Medium

Problem Description

Given an array of characters chars, compress it using the following algorithm:

Begin with an empty string s. For each group of consecutive repeating characters in chars:

- If the group's length is `1`, append the character to `s`.

- Otherwise, append the character followed by the group's length.

The compressed string s should not be returned separately, but instead, be stored in the input character array chars. Note that group lengths that are 10 or longer will be split into multiple characters in chars.

After you are done modifying the input array, return the new length of the array.

You must write an algorithm that uses only constant extra space.

Note: The characters in the array beyond the returned length do not matter and should be ignored.

Example 1:

Input: chars = ["a","a","b","b","c","c","c"]
Output: Return 6, and the first 6 characters of the input array should be: ["a","2","b","2","c","3"]
Explanation: The groups are "aa", "bb", and "ccc". This compresses to "a2b2c3".

Example 2:

Input: chars = ["a"]
Output: Return 1, and the first character of the input array should be: ["a"]
Explanation: The only group is "a", which remains uncompressed since it's a single character.

Example 3:

Input: chars = ["a","b","b","b","b","b","b","b","b","b","b","b","b"]
Output: Return 4, and the first 4 characters of the input array should be: ["a","b","1","2"].
Explanation: The groups are "a" and "bbbbbbbbbbbb". This compresses to "ab12".

Constraints:

- `1 <= chars.length <= 2000`

- `chars[i]` is a lowercase English letter, uppercase English letter, digit, or symbol.

Topics: Two Pointers, String

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Solutions

Solution 1: C# (Best: 132 ms)

Metric	Value
Runtime	132 ms
Memory	46.5 MB
Date	2022-01-05

Solution

public class Solution {
    public int Compress(char[] chars) {
    int n = chars.Length;
    
    int idx = 0, i=0;
    while(i<n)
    {
        int count = 0;
        int j = i;
        while(j<n && chars[j]==chars[i])
        {
            count++;
            j++;
        }
        chars[idx++] = chars[i];
        if(count > 1)
        {
            string s = Convert.ToString(count);
            for (int k = 0; k < s.Length; k++)
            {
                chars[idx++] = s[k];
            }           
        }
        i= j;
    }
    return idx;
    }
}

📜 3 more C# submission(s)

Submission (2022-01-13) — 152 ms, 46.4 MB

public class Solution {
    public int Compress(char[] chars) {
    int n = chars.Length;
    
    int idx = 0, i=0;
    while(i<n)
    {
        int count = 0;
        int j = i;
        while(j<n && chars[j]==chars[i])
        {
            count++;
            j++;
        }
        chars[idx++] = chars[i];
        if(count > 1)
        {
            //int digits = Convert.ToInt32(Math.Floor(Math.Log10(count)+1));
            int digits = 0;
            int num = count;
            while(num>0)
            {
                digits++;
                num = num/10;
            }
            int lastIndex = idx+digits-1;
            num = count;
            while(num>0)
            {
                chars[lastIndex--] = (char)(num%10 + '0');
                num = num/10;
            }
            idx = idx+digits;
        }
        i= j;
    }
    return idx;
    }
}

Submission (2022-01-13) — 254 ms, 44.8 MB

public class Solution {
    public int Compress(char[] chars) {
    int n = chars.Length;
    
    int idx = 0, i=0;
    while(i<n)
    {
        int count = 0;
        int j = i;
        while(j<n && chars[j]==chars[i])
        {
            count++;
            j++;
        }
        chars[idx++] = chars[i];
        if(count > 1)
        {
            int digits = Convert.ToInt32(Math.Floor(Math.Log10(count)+1));
            int lastIndex = idx+digits-1;
            int num = count;
            while(num>0)
            {
                chars[lastIndex--] = (char)(num%10 + '0');
                num = num/10;
            }
            idx = idx+digits;
        }
        i= j;
    }
    return idx;
    }
}

Submission (2022-01-05) — 269 ms, 44.1 MB

public class Solution {
    public int Compress(char[] chars) {
        int n = chars.Length;
    
    int idx = 0;
    int count = 0, i = 0;;
    while(i<n)
    {
        count = 0;
        char c = chars[i];
        int j = i;
        while(j<n && chars[j]==chars[i])
        {
            count++;
            j++;
        }
        chars[idx++] = c;
        if(count > 1)
        {
            string s = Convert.ToString(count);
            for (int k = 0; k < s.Length; k++)
            {
                chars[idx++] = s[k];
            }           
        }
        i= j;
    }
    return idx;
    }
}

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Complexity Analysis

Approach	Time	Space
Two Pointers	$O(n)$	$O(1)$

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Interview Tips

Key Points

• Discuss the brute force approach first, then optimize. Explain your thought process.
• LeetCode provides 1 hint(s) for this problem — try solving without them first.

💡 Hints

Hint 1: How do you know if you are at the end of a consecutive group of characters?