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  1. Prepare
  2. Algorithms
  3. Greedy
  4. Reverse Shuffle Merge
  5. Discussions

Reverse Shuffle Merge

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215 Discussions

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  • daideneg4
     + 0 comments

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  • 1337er
     + 0 comments

    O(n) time complexity because even with nested while loop an element of s is only pushed and popped a maximum of 1 times

    from collections import Counter

def reverseShuffleMerge(s):
    char_counts = Counter(s)
    required = Counter()
    for key, value in char_counts.items():
        required[key] = value // 2
    last_char = s[-1]
    char_counts[last_char] -= 1
    required[last_char] -= 1
    result = [last_char]
    for i in range(len(s)-2, -1, -1):
        char = s[i]
        char_counts[char] -= 1
        if not required[char]:
            pass
        elif (not result) or (char >= result[-1]):
            result.append(char)
            required[char] -= 1
        else:
            while result and char < result[-1] and char_counts[result[-1]] >= required[result[-1]] +1:
                old_char = result.pop(-1)
                required[old_char] += 1
            result.append(char)
            required[char] -= 1
    return "".join(result)

  • 1337er
     + 0 comments

    I hate problems that require you to reread 50 times to just understand what they are asking you to do.

  • dangminhhoangdz
     + 0 comments

    Worst problem statement ever. Unclear

  • shreyavishwalin1
     + 0 comments
    def reverseShuffleMerge(s):
    from collections import Counter

    # Calculate the frequency of each character in s
    freq = Counter(s)
    
    # Calculate the required frequency for A
    required = {char: freq[char] // 2 for char in freq}
    
    # Initialize counts of used characters
    used = {char: 0 for char in freq}
    
    result = []
    # Iterate over the string in reverse order
    for char in reversed(s):
        if used[char] < required[char]:
            # Ensure that we are making the lexicographically smallest result
            while result and result[-1] > char and used[result[-1]] + freq[result[-1]] > required[result[-1]]:
                removed_char = result.pop()
                used[removed_char] -= 1

            result.append(char)
            used[char] += 1
        
        # Decrease the frequency count as we have processed this character
        freq[char] -= 1
    
    # Convert result list to a string and return
    return ''.join(result)