Merge to Palindrome

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Unit 4 Session 2 (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 happens if the list is already a palindrome?
  • If the list is already a palindrome, the function should return 0 as no merges are needed.
• How should the function handle very small lists, like those with one or two elements?
  • For a single element, no operations are needed as a single element is trivially a palindrome. For two elements, if they are not equal, one merge operation will be needed.

P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Use two pointers starting at both ends of the list, moving towards the center. Merge adjacent numbers as needed to match the ends and make progress towards forming a palindrome.

1) Initialize two pointers, one at the start (`left`) and one at the end (`right`) of the list.
2) Keep track of the number of merge operations needed with a counter (`merge_count`).
3) While the left pointer is less than the right pointer:
  a) If the numbers at both pointers are equal, move both pointers towards the center.
  b) If the number at the left pointer is less, merge it with the next element, increment the merge count, and adjust the left pointer.
  c) If the number at the right pointer is less, merge it with its previous element, increment the merge count, and adjust the right pointer.
4) Return the total number of merge operations needed to make the list a palindrome.

⚠️ Common Mistakes

• Forgetting to handle the cases where merging affects subsequent elements, potentially leading to incorrect next steps.
• Not correctly adjusting the pointers after a merge, which could cause infinite loops or missed operations.

I-mplement

def min_merge_operations(nums):
    left, right = 0, len(nums) - 1
    merge_count = 0
    
    while left < right:
        # If elements at current pointers are equal, move towards center
        if nums[left] == nums[right]:
            left += 1
            right -= 1
        # If left element is smaller, merge it with the next element
        elif nums[left] < nums[right]:
            nums[left + 1] += nums[left]  # Merge
            left += 1  # Move pointer
            merge_count += 1  # Increment merge operation count
        # If right element is smaller, merge it with the previous element
        else:
            nums[right - 1] += nums[right]  # Merge
            right -= 1  # Move pointer
            merge_count += 1  # Increment merge operation count
            
    return merge_count