Minimum Character Changes to Make Every k-Length Block a Palindrome
Company: Ramp
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: medium
Interview Round: Take-home Project
## Minimum Character Changes to Make Every k-Length Block a Palindrome
You are given a `password` string and an integer `k`. The length of `password` is guaranteed to be a multiple of `k`.
Partition `password` into consecutive, non-overlapping blocks of exactly `k` characters: the first block is characters `0..k-1`, the second block is characters `k..2k-1`, and so on. The new password is **valid** when **every block is a palindrome**.
In one operation you may replace a single character of `password` with any lowercase English letter. Return the **minimum number of replacements** needed so that every `k`-length block becomes a palindrome.
### Input
- A string `password` consisting of lowercase English letters (`'a'`–`'z'`).
- An integer `k` — the block size. `len(password)` is divisible by `k`.
### Output
- A single integer: the minimum number of character replacements required.
### Constraints
- `1 <= k <= len(password) <= 200000`
- `len(password) % k == 0`
- `password` contains only lowercase English letters.
- Test data includes large inputs, so an efficient near-linear, $O(\text{len}(password))$ solution is expected; per-block string rebuilding or quadratic scanning will time out.
### Definitions
- A string is a **palindrome** if it reads the same forwards and backwards. For a block `b` of length `k`, this means `b[j] == b[k-1-j]` for every `j` in `0 <= j < k/2`. The middle character (when `k` is odd) is unconstrained.
### Example 1
```
password = "abca"
k = 2
Output: 2
```
Explanation: Blocks are `"ab"` and `"ca"`. `"ab"` needs `a == b` — one mismatched pair, so 1 change. `"ca"` needs `c == a` — one mismatched pair, so 1 change. Total `2`.
### Example 2
```
password = "aabb"
k = 2
Output: 0
```
Explanation: Blocks are `"aa"` and `"bb"`, both already palindromes. No changes needed.
### Example 3
```
password = "abxyba"
k = 6
Output: 1
```
Explanation: There is a single block `"abxyba"`. The mirror pairs are `(a, a)`, `(b, b)`, and `(x, y)`. Only `(x, y)` mismatches, so exactly `1` replacement is needed (e.g., change `y` to `x`).
Quick Answer: This question evaluates a candidate's ability to combine greedy pairing logic with efficient string traversal to minimize edits across multiple palindrome-constrained segments. It tests string manipulation and algorithmic optimization skills, since a naive per-block approach fails to scale on large inputs. This type of coding problem is commonly used to assess practical application of linear-time algorithm design under strict performance constraints.
You are given a string `password` and an integer `k`, where `len(password)` is a multiple of `k`. Partition `password` into consecutive, non-overlapping blocks of exactly `k` characters (characters `0..k-1`, `k..2k-1`, and so on). In one operation you may replace a single character with any lowercase English letter. Return the minimum number of replacements needed so that every `k`-length block becomes a palindrome.
A block `b` of length `k` is a palindrome when `b[j] == b[k-1-j]` for every `j` in `0 <= j < k/2`; the middle character (when `k` is odd) is unconstrained. For each block, count one required change per mismatched mirror pair, then sum over all blocks.
Examples:
- `password = "abca", k = 2` -> `2` (blocks "ab" and "ca", each needs 1 change).
- `password = "aabb", k = 2` -> `0` (blocks "aa" and "bb" already palindromes).
- `password = "abxyba", k = 6` -> `1` (only the (x, y) mirror pair mismatches).
Constraints
- 1 <= k <= len(password) <= 200000
- len(password) % k == 0
- password contains only lowercase English letters ('a'-'z')
- An efficient near-linear O(len(password)) solution is expected; per-block string rebuilding or quadratic scanning will time out.
Examples
Input: ("abca", 2)
Expected Output: 2
Explanation: Blocks "ab" and "ca"; each has one mismatched mirror pair, so 1 + 1 = 2.
Input: ("aabb", 2)
Expected Output: 0
Explanation: Blocks "aa" and "bb" are already palindromes, so no changes are needed.
Hints
- Process the string block by block using a step of k over the start index; you never need to build a separate substring.
- Within each block, only mirror pairs (i, k-1-i) matter. Walk two pointers from the block's ends toward the middle.
- Each mismatched mirror pair costs exactly one replacement (change one side to match the other). The middle character of an odd-length block is free.
- Sum the mismatch counts across all blocks. This is a single linear pass, so it handles the largest inputs comfortably.