Python -> stack based solution with detailed explanation

  • 1

    This is a stack based solution.
    1> First I go over all the letters and add it to dictionary along with their counts of appearance.
    2> I use a list as stack to store all the lexicographically arranged result set until i'th index where 0 <= i < len(input_array).
    3> I initialize a visited set to keep track of all elements that are already there in stack. Whenever I come across the same element which is already there in stack, I do not bother to do any operation except for decreasing its counter.
    4> If the stack is empty, I just add the element as is.
    5> if the letter under inspection is lesser than the elements on the top of the stack, then I continue to pop them out of stack but on one condition:
    -> If the stack letter counter is 0, I do not dare to remove it.
    If the stack letter counter is ZERO, I know that the element has no chance to appear in future. Hence I keep it as is in the stack and append our new letter on to the stack.
    6> Also, whenever I remove any element from the stack, I have to ensure they are removed from the visited set as well so in the future I can add them back to stack.

        def removeDuplicateLetters(self, s):
            :type s: str
            :rtype: str
            ldict = {}
            for letter in s:
                ldict[letter] = ldict.get(letter, 0) + 1
            stack = []
            visited = set()
            for letter in s:
                if not stack:
                    ldict[letter] -= 1
                if letter not in visited:
                    while stack and letter < stack[-1] and ldict[stack[-1]] > 0:
                        oldLetter = stack[-1]
                ldict[letter] -= 1
            return "".join(stack)```

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