In this Leetcode String Compression problem solution we have 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:

  1. If the group's length is 1, append the character to s.
  2. 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.

Leetcode String Compression problem solution


Problem solution in Python.

class Solution:
    def compress(self, array: List[str]) -> int:
        ans=[]
        i=0
        while i < len(array):
            temp=array[i]
            count=i+1
            while count<len(array)and array[count]==temp:
                count+=1
                if count==len(array):
                    break
            ans.append(temp)
            if count-i>1 and count-i<10:
                ans.append(str(count-i))
            if count-i>9:
                num=count-i
                dup=[]
                while num:
                    dup.append(str(num%10))
                    num//=10
                for j in range(len(dup)-1,-1,-1):
                    ans.append(dup[j])
            i=count
        array.clear()
        for i in range(len(ans)):
            array.append(ans[i])
        return len(ans)



Problem solution in Java.

class Solution {
    public int compress(char[] chars) {
     if (chars.length == 1) return chars.length;
    int index = 0, newChar_start_at = 0,start = 0;
    while (start< chars.length) {
        int count = 0;
        char val = chars[newChar_start_at];
        for (;start < chars.length && chars[start] == val;count++,start++); 
        chars[index] = val;
        index++;
        if (count > 1) {
            for (char c : Integer.toString(count).toCharArray()) {
                chars[index] = c;
                index++;
            }
        }
        newChar_start_at = start;
    }
    return index;
    }
}


Problem solution in C++.

class Solution {
public:
    int compress(vector<char>& chars) {
        int n=(int)chars.size(), write_idx=0, cnt=1;
        if (n<=1) return n;
        for (int i=1; i<n; ++i){
            if (chars[i-1]==chars[i]){
                ++cnt;
            } else {
                write(chars, write_idx, chars[i-1], cnt);
                cnt=1;
            }
        }
        write(chars, write_idx, chars.back(), cnt);
        chars.resize(write_idx);
        return write_idx;
    }

    void write(vector<char>& chars, int& write_idx, char c, int cnt){
        chars[write_idx++]=c;
        if (cnt==1) return;
        string s=to_string(cnt);
        for (int i=0; i<s.size(); ++i){
            chars[write_idx++]=s[i];
        }
    }
};


Problem solution in C.

int compress(char* chars, int charsSize){

int char_count = 1;
int cnt = 0;
int count_index = 1; 
char test = chars[0];
char count_buf[20]; 

for(int i = 1;i<charsSize;i++)
{
    if(chars[i]==test)
    {
        char_count++;
    }
    else
    {
        if(char_count>1)
        {
            cnt = sprintf(count_buf,"%d",char_count); 
            memcpy(chars+count_index, count_buf, cnt);
            
            count_index = count_index + cnt;
            char_count = 1;
        }
       
        test = chars[i];
        chars[count_index] = test;
        count_index++;    
    }
}

if(char_count > 1)
{
    cnt = sprintf(count_buf,"%d",char_count);
    memcpy(chars+count_index, count_buf, cnt);
    count_index = count_index + cnt;
}

return count_index;
}