Formatted question description: https://leetcode.ca/all/443.html

443. String Compression (Easy)

Given an array of characters, compress it in-place.

The length after compression must always be smaller than or equal to the original array.

Every element of the array should be a character (not int) of length 1.

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

Follow up:
Could you solve it using only O(1) extra space?

Example 1:

Input:
["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:
"aa" is replaced by "a2". "bb" is replaced by "b2". "ccc" is replaced by "c3".

Example 2:

Input:
["a"]

Output:
Return 1, and the first 1 characters of the input array should be: ["a"]

Explanation:
Nothing is replaced.

Example 3:

Input:
["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:
Since the character "a" does not repeat, it is not compressed. "bbbbbbbbbbbb" is replaced by "b12".
Notice each digit has it's own entry in the array.

Note:

All characters have an ASCII value in [35, 126].
1 <= len(chars) <= 1000.

Companies:
Microsoft, Goldman Sachs, Bloomberg

Related Topics:
String

Similar Questions:

Solution 1.

class Solution {
    public int compress(char[] chars) {
        if (chars == null)
            return 0;
        int length = chars.length;
        if (length <= 1)
            return length;
        int index = 0;
        char prevChar = chars[0];
        int count = 1;
        for (int i = 1; i < length; i++) {
            char c = chars[i];
            if (c == prevChar)
                count++;
            else {
                chars[index] = prevChar;
                index++;
                if (count > 1) {
                    String countStr = String.valueOf(count);
                    int countLength = countStr.length();
                    for (int j = 0; j < countLength; j++) {
                        chars[index] = countStr.charAt(j);
                        index++;
                    }
                    count = 1;
                }
                prevChar = c;
            }
        }
        if (count > 0) {
            chars[index] = prevChar;
            index++;
            if (count > 1) {
                String countStr = String.valueOf(count);
                int countLength = countStr.length();
                for (int j = 0; j < countLength; j++) {
                    chars[index] = countStr.charAt(j);
                    index++;
                }
            }
        }
        return index;
    }
}

############

class Solution {
    public int compress(char[] chars) {
        int k = 0, n = chars.length;
        for (int i = 0, j = i + 1; i < n;) {
            while (j < n && chars[j] == chars[i]) {
                ++j;
            }
            chars[k++] = chars[i];
            if (j - i > 1) {
                String cnt = String.valueOf(j - i);
                for (char c : cnt.toCharArray()) {
                    chars[k++] = c;
                }
            }
            i = j;
        }
        return k;
    }
}

// OJ: https://leetcode.com/problems/string-compression/
// Time: O(N)
// Space: O(log_10^N)
class Solution {
public:
    int compress(vector<char>& A) {
        int j = 0;
        for (int i = 0; i < A.size(); ++i) {
            char c = A[i];
            int cnt = 1;
            while (i + 1 < A.size() && A[i + 1] == c) ++i, ++cnt;
            A[j++] = c;
            if (cnt == 1) continue;
            auto s = to_string(cnt);
            for (char d : s) A[j++] = d;
        }
        return j;
    }
};

class Solution:
    def compress(self, chars: List[str]) -> int:
        i, k, n = 0, 0, len(chars)
        while i < n:
            j = i + 1
            while j < n and chars[j] == chars[i]:
                j += 1
            chars[k] = chars[i]
            k += 1
            if j - i > 1:
                cnt = str(j - i)
                for c in cnt:
                    chars[k] = c
                    k += 1
            i = j
        return k

############

class Solution(object):
    def compress(self, chars):
        """
        :type chars: List[str]
        :rtype: int
        """
        marks = ""
        length = -1
        cur = chars[0]
        for i, value in enumerate(chars):
            length += 1
            if value != cur:
                count = str(length) if length != 1 else ''
                marks += cur + count
                cur = value
                length = 0
            if i == len(chars) - 1:
                length += 1
                count = str(length) if length != 1 else ''
                marks += cur + count
                cur = value
                length = 0
        print marks
        for i, mark in enumerate(marks):
            chars[i] = mark
        return len(marks)

func compress(chars []byte) int {
	i, k, n := 0, 0, len(chars)
	for i < n {
		j := i + 1
		for j < n && chars[j] == chars[i] {
			j++
		}
		chars[k] = chars[i]
		k++
		if j-i > 1 {
			cnt := strconv.Itoa(j - i)
			for _, c := range cnt {
				chars[k] = byte(c)
				k++
			}
		}
		i = j
	}
	return k
}

443 - String Compression

443. String Compression (Easy)

Solution 1.

All Problems

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443. String Compression (Easy)

Solution 1.

All Problems

All Solutions