#1737

Change Minimum Characters to Satisfy One of Three Conditions

Medium
Hash TableStringCountingPrefix SumHash MapArray
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Approaches

Brute ForceOptimal
Complexity Comparison
Brute ForceOptimal Solution
Time
O(n²)
O(n)
Space
O(1)
O(1)
💡

Intuition

Time O(n)Space O(1)

The optimal approach leverages the properties of characters in the strings to minimize operations by calculating the necessary changes based on the smallest and largest characters in both strings. This avoids redundant checks and significantly reduces the number of operations needed.

⚙️

Algorithm

5 steps
  1. 1Step 1: Find the minimum and maximum characters in both strings 'a' and 'b'.
  2. 2Step 2: Calculate the number of changes needed to make all characters in 'a' less than the minimum character of 'b'.
  3. 3Step 3: Calculate the number of changes needed to make all characters in 'b' less than the minimum character of 'a'.
  4. 4Step 4: Calculate the number of changes needed to make both strings consist of the same character, which is the minimum character of either string.
  5. 5Step 5: Return the minimum of the three calculated operations.
solution.py7 lines
1def min_operations(a, b):
2    min_a, max_a = min(a), max(a)
3    min_b, max_b = min(b), max(b)
4    ops1 = sum(1 for x in a if x >= min_b) + len(b)
5    ops2 = sum(1 for y in b if y >= min_a) + len(a)
6    ops3 = len(a) + len(b) - (a.count(min_a) + b.count(min_a))
7    return min(ops1, ops2, ops3)

Complexity note: The time complexity is O(n) because we only need to traverse each string a constant number of times (to find min/max and calculate operations). The space complexity is O(1) since we are using a constant amount of extra space.

  • 1Understanding character relationships is crucial for minimizing operations.
  • 2Iterating through the alphabet can help identify optimal character changes.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.