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Change Minimum Characters to Satisfy One of Three Conditions

You are given two strings a and b that consist of lowercase letters. In one operation, you can change any character in a or b to any lowercase letter.

Your goal is to satisfy one of the following three conditions:

  • Every letter in a is strictly less than every letter in b in the alphabet.
  • Every letter in b is strictly less than every letter in a in the alphabet.
  • Both a and b consist of only one distinct letter.

Return the minimum number of operations needed to achieve your goal.

 

Example 1:

Input: a = "aba", b = "caa"
Output: 2
Explanation: Consider the best way to make each condition true:
1) Change b to "ccc" in 2 operations, then every letter in a is less than every letter in b.
2) Change a to "bbb" and b to "aaa" in 3 operations, then every letter in b is less than every letter in a.
3) Change a to "aaa" and b to "aaa" in 2 operations, then a and b consist of one distinct letter.
The best way was done in 2 operations (either condition 1 or condition 3).

Example 2:

Input: a = "dabadd", b = "cda"
Output: 3
Explanation: The best way is to make condition 1 true by changing b to "eee".

 

Constraints:

  • 1 <= a.length, b.length <= 105
  • a and b consist only of lowercase letters.

Solution Explanation for LeetCode 1737: Change Minimum Characters to Satisfy One of Three Conditions

This problem asks us to find the minimum number of character changes needed to make two strings, a and b, satisfy one of three conditions:

  1. Every character in a is strictly less than every character in b alphabetically.
  2. Every character in b is strictly less than every character in a alphabetically.
  3. Both a and b contain only one distinct character.

Approach: Counting and Enumeration

The optimal approach involves counting character frequencies and then enumerating possibilities to find the minimum operations.

1. Character Frequency Counting:

We first count the occurrences of each character (a-z) in both strings a and b. This is efficiently done using arrays cnt1 and cnt2 (of size 26).

2. Condition 3: Single Distinct Character:

For condition 3, we iterate through each character ('a' to 'z'). For each character c, we calculate the total number of characters in a and b that are not c. This represents the minimum operations needed to transform both strings to contain only c. We keep track of the minimum operations found across all characters.

3. Conditions 1 & 2: Alphabetical Ordering:

For conditions 1 and 2, we need to find the optimal "split point" in the alphabet. Let's consider condition 1 (a < b). We iterate through possible split points i (from 1 to 25). For each i, we calculate the cost:

  • The number of characters in a that are greater than or equal to the i-th character.
  • The number of characters in b that are less than the i-th character.

The sum of these costs represents the minimum operations needed to satisfy condition 1 for that split point. We repeat this for condition 2 (b < a), reversing the roles of a and b.

4. Minimum Operations:

Finally, we return the minimum of the operation counts obtained from the three conditions.

Time Complexity Analysis

  • Character counting: O(m + n), where m and n are the lengths of strings a and b.
  • Condition 3: O(26) (iterating through the alphabet).
  • Conditions 1 & 2: O(26) (iterating through possible split points).

Therefore, the overall time complexity is O(m + n + 26), which simplifies to O(m + n) since the constant term 26 is insignificant compared to the input string lengths.

Space Complexity Analysis

The space complexity is O(1) because we use constant-size arrays (cnt1, cnt2) to store character counts, regardless of the input string lengths.

Code Examples (Python)

def minCharacters(a: str, b: str) -> int:
    cnt1 = [0] * 26
    cnt2 = [0] * 26
    for c in a:
        cnt1[ord(c) - ord('a')] += 1
    for c in b:
        cnt2[ord(c) - ord('a')] += 1
    ans = len(a) + len(b)
    for i in range(26):
        ans = min(ans, len(a) + len(b) - cnt1[i] - cnt2[i])
    f = lambda cnt1, cnt2: min(sum(cnt1[i:]) + sum(cnt2[:i]) for i in range(1, 26))
    ans = min(ans, f(cnt1, cnt2), f(cnt2, cnt1))
    return ans

This Python code directly implements the algorithm described above. The f function uses a lambda expression for brevity to calculate the minimum operations for conditions 1 and 2. Other languages (Java, C++, Go, TypeScript) would follow a similar structure, adjusting syntax as needed.