Minimum Priority Conflicts When Interleaving Two Lowercase Strings
Company: Amazon
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: hard
Interview Round: Online Assessment
You are given two strings, `primary` (length `m`) and `secondary` (length `n`), made of lowercase English letters. Merge them into one string of length `m + n` that uses every character of both strings and keeps the original relative order of the characters within each string. In other words, the merged string is an interleaving of `primary` and `secondary`.
Letters have priorities: a smaller letter has a higher priority, so `'a'` has the highest priority and `'z'` the lowest. A conflict is a pair of positions in the merged string where a lower-priority letter appears before a higher-priority letter. For example, the merged string `"zab"` has 2 conflicts: `'z'` before `'a'`, and `'z'` before `'b'`.
Return the minimum number of conflicts over all valid merged strings.
### Function Signature
```python
def min_merge_conflicts(primary: str, secondary: str) -> int:
```
### Rules
- The merged string must use every character of `primary` and every character of `secondary` exactly once, and the characters of each input must appear in the same relative order as in that input.
- A conflict is a pair of indices `x < y` in the merged string with `merged[x] > merged[y]` in alphabetical order. Each such pair counts once.
- Two equal letters never form a conflict.
- Conflicts between two characters that come from the same input string count too, even though no merge can change their relative order.
- Return only the minimum count, not the merged string.
### Constraints
- `1 <= m <= 1000` and `1 <= n <= 1000`, where `m = len(primary)` and `n = len(secondary)`
- Both strings contain only the letters `'a'` to `'z'`.
- The answer is at most `(m + n) * (m + n - 1) / 2`, which is below 2,000,000, so it fits in a 32-bit signed integer.
### Examples
**Example 1**
```text
Input: primary = "ca", secondary = "b"
Output: 2
```
The three possible merges are `"bca"` (2 conflicts), `"cba"` (3 conflicts) and `"cab"` (2 conflicts).
**Example 2**
```text
Input: primary = "db", secondary = "ca"
Output: 3
```
The merge `"cadb"` has 3 conflicts (`'c'` before `'a'`, `'c'` before `'b'`, and `'d'` before `'b'`), and no valid merge has fewer.
**Example 3**
```text
Input: primary = "a", secondary = "a"
Output: 0
```
The only merge is `"aa"`, and equal letters do not conflict.
Overview: From an Amazon online assessment: merge two lowercase strings into one while keeping each string's internal order, where smaller letters have higher priority and every lower-priority letter placed before a higher-priority one counts as a conflict. Return the minimum possible number of conflicts over all valid merges.
Read the full Amazon Software Engineer interview experience this question came from
You are given two strings, `primary` (length `m`) and `secondary` (length `n`), made of lowercase English letters. Merge them into one string of length `m + n` that uses every character of both strings exactly once and keeps the original relative order of the characters within each string. In other words, the merged string is an interleaving of `primary` and `secondary`.
Letters have priorities: a smaller letter has a higher priority, so `'a'` has the highest priority and `'z'` the lowest. A conflict is a pair of indices `x < y` in the merged string with `merged[x] > merged[y]` in alphabetical order, that is, a lower-priority letter appearing before a higher-priority letter. Each such pair counts once. Two equal letters never form a conflict. Conflicts between two characters that come from the same input string count too, even though no merge can change their relative order. For example, the merged string `"zab"` has 2 conflicts: `'z'` before `'a'`, and `'z'` before `'b'`.
Return the minimum number of conflicts over all valid merged strings. Return only this count, not the merged string. The count is at most `(m + n) * (m + n - 1) / 2`, which is below 2,000,000, so it never exceeds 2^31 - 1 and fits in a 32-bit signed integer (`int` in Java and C++).
**Example 1**
Input: primary = "ca", secondary = "b"
Output: 2
The three possible merges are "bca" (2 conflicts), "cba" (3 conflicts) and "cab" (2 conflicts).
**Example 2**
Input: primary = "db", secondary = "ca"
Output: 3
The merge "cadb" has 3 conflicts ('c' before 'a', 'c' before 'b', and 'd' before 'b'), and no valid merge has fewer.
**Constraints**
- `1 <= m <= 1000` and `1 <= n <= 1000`, where `m = len(primary)` and `n = len(secondary)`
- Both strings contain only the letters `'a'` to `'z'`.
- The answer is at most `(m + n) * (m + n - 1) / 2`, which is below 2,000,000, so it fits in a 32-bit signed integer.
Constraints
- 1 <= m <= 1000 and 1 <= n <= 1000, where m = len(primary) and n = len(secondary)
- primary and secondary contain only the lowercase letters 'a' to 'z'
- The answer is at most (m + n) * (m + n - 1) / 2, which is below 2,000,000, so it fits in a 32-bit signed integer
Examples
Input: ('ca', 'b')
Expected Output: 2
Explanation: Source Example 1: merges 'bca' and 'cab' have 2 conflicts, 'cba' has 3.
Input: ('db', 'ca')
Expected Output: 3
Explanation: Source Example 2: 'cadb' has 3 conflicts and no merge has fewer.
Hints
- As the rules note, no merge can change the order of two characters from the same input string, so their conflicts are the same in every merge; only pairs with one character from each string depend on how you interleave.
- If you build the merged string left to right, appending a character creates exactly one conflict for each already-placed character that is strictly greater than it.
- When both strings offer the same letter next, the two choices can still lead to different totals later, even though equal letters never conflict with each other.