My approach was Union-Find: group indices that must share the same character (from both the palindrome constraint and the periodicity constraint), then for each group just count how many characters aren't the majority character.
Model the constraints as a graph where each position must equal its mirror and positions congruent modulo k must be equal. Use union-find to group positions that must share the same character, then for each group choose the character that minimizes changes. Sum the minimum changes across all groups to get the answer.
Pro tip: Mention that the solution runs in O(n α(n)) time and O(n) space, and that it handles edge cases like k=1 or k≥n gracefully. This shows you consider efficiency and robustness.
Clarify that the string must be a palindrome (s[i] = s[n-1-i]) and k-periodic (s[i] = s[i+k] for all valid i). These are equality constraints on positions.
Create a graph where each position is a node, and add edges between positions that must be equal: (i, n-1-i) for palindrome and (i, i+k) for periodicity.
Use union-find (or BFS/DFS) to group positions that must all have the same character. Each component represents a set of positions that must be identical.
For each component, count the frequency of each character. The minimum changes for that component is the total size minus the maximum frequency. Sum these over all components.
The sum from step 4 is the minimum number of character changes needed to satisfy both constraints simultaneously.
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