Reverse Prefix

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Unit 4 Session 1 (Click for link to problem statements)

U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• What should happen if the character ch does not exist in the string word?
  • The function should return the original string unchanged.
• Does the function need to handle cases where ch appears multiple times?
  • The function only considers the first occurrence of ch.

P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Locate the first occurrence of ch, then reverse the segment from the start of the string to this index.

1) Convert the string into a list to enable easy manipulation of characters.
2) Check if `ch` is in the string. If not, return the original string.
3) Find the index of the first occurrence of `ch` and store it as `right`.
4) Initialize a pointer `left` at the start of the string.
5) Use a two-pointer technique to reverse the substring from `left` to `right`:
  a) Swap the characters at `left` and `right`.
  b) Increment `left` and decrement `right` until they meet or cross.
6) Convert the list of characters back to a string and return it.

⚠️ Common Mistakes

• Not returning the original string when ch is not found.
• Incorrectly handling the indices or the two-pointer swap, leading to partial reversals or errors.

I-mplement

def reverse_prefix(word, ch):
    # Convert the string to a list to modify its characters
    chars = list(word)
    
    # Find the index of the first occurrence of ch
    if ch in chars:
	    right = chars.index(ch)
	  else:
		  right = -1 
    
    # If ch is not found, return the original string
    if right == -1:
        return word
    
    # Initialize the left pointer
    left = 0
    
    # Reverse the substring using the two-pointer approach
    while left < right:
        # Swap the characters at the left and right pointers
        chars[left], chars[right] = chars[right], chars[left]
        left += 1
        right -= 1
    
    # Convert the list back to a string and return it
    return ''.join(chars)