string-compression.py

# Time:  O(n)
# Space: O(1)

# 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.

# V0
import collections 
class Solution(object):
    def compress(self, chars):
        count = collections.Counter(chars)
        key_count = len(count.keys())
        val_count = len([ x for x in list(count.values()) if x > 1 ])
        return key_count + val_count

# V1 
# http://bookshadow.com/weblog/2017/10/29/leetcode-string-compression/
# IDEA : QUEUE 
class Solution(object):
    def compress(self, chars):
        """
        :type chars: List[str]
        :rtype: int
        """
        last, n, y = chars[0], 1, 0
        for x in range(1, len(chars)):
            c = chars[x]
            if c == last: n += 1
            else:
                for ch in last + str(n > 1 and n or ''):
                    chars[y] = ch
                    y += 1
                last, n = c, 1
        for ch in last + str(n > 1 and n or ''):
            chars[y] = ch
            y += 1
        while len(chars) > y: chars.pop()
        return y

# V1'
# https://blog.csdn.net/androidchanhao/article/details/81408345
# IDEA : 2 POINTERS 
class Solution(object):
    def compress(self, chars):
        """
        :type chars: List[str]
        :rtype: int
        """
        n = len(chars) 
        cur = 0 # index of current element, for compress the list 
        i = 0
        while i < n:
            j = i
            while j < n - 1 and chars[j] == chars[j+1]:# find the count of sequential exist of an elment 
                j += 1
            chars[cur] = chars[i] # record current element 
            cur += 1
            if i != j:
                times = str(j-i+1) # write the count into original list 
                tLen = len(times)
                for k in range(tLen):
                    chars[cur+k] = times[k]
                cur += tLen
            i = j + 1 # move to next element 
        return cur

# V2 
# Time:  O(n)
# Space: O(1)
class Solution(object):
    def compress(self, chars):
        """
        :type chars: List[str]
        :rtype: int
        """
        anchor, write = 0, 0
        for read, c in enumerate(chars):
            if read+1 == len(chars) or chars[read+1] != c:
                chars[write] = chars[anchor]
                write += 1
                if read > anchor:
                    n, left = read-anchor+1, write
                    while n > 0:
                        chars[write] = chr(n%10+ord('0'))
                        write += 1
                        n /= 10
                    right = write-1
                    while left < right:
                        chars[left], chars[right] = chars[right], chars[left]
                        left += 1
                        right -= 1
                anchor = read+1
        return write