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Treat each word independently and count the minimum substitutions needed to eliminate all adjacent identical characters. For each word, scan left to right and whenever two adjacent characters are the same, substitute the second one with a different character (e.g., a placeholder) and increment the count. Sum the counts across all words.
Pro tip: Clarify that substitutions can be any character, including ones not in the original alphabet, so a single substitution always suffices to break a pair. This shows you understand the problem's constraints and avoids overcomplicating the solution.
Confirm that substitutions can be any character (not limited to the original alphabet) and that each word is independent. Ask if the input is a list of strings and if the output is a single integer.
For each word, iterate through its characters from left to right. When you find two identical adjacent characters, replace the second one with a different character (e.g., a special symbol) and increment a counter.
Consider words with length 0 or 1 (no substitutions needed), and words with runs of identical characters (e.g., 'aaa') where each additional character in the run requires a substitution.
Accumulate the substitution counts for all words and return the total. Explain that the greedy left-to-right approach yields the minimum because each substitution can fix at most one adjacent pair.
State that the time complexity is O(N) where N is the total number of characters across all words, and space complexity is O(1) extra space (if modifying in place) or O(N) if creating a new string.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify the problem: you need to remove the minimum number of elements so that the resulting array can be made strictly ascending by deleting at most one more element. This is equivalent to finding the longest subsequence that is 'almost strictly increasing' (allowing one violation). Then, compute the minimum removals as the array length minus the length of this longest subsequence. Use dynamic programming to track the longest valid subsequence ending at each index with 0 or 1 deletion used.
Pro tip: Demonstrate strong problem-solving by discussing both the DP solution and potential optimizations (e.g., using patience sorting for LIS variants). Also, mention edge cases like arrays already strictly ascending or with only one element, and clarify the 'at most one more deletion' condition to ensure alignment with the interviewer.
Restate the problem in your own words and confirm with the interviewer. Ensure you understand that after removing elements, the resulting array must be such that deleting at most one more element makes it strictly ascending.
Recognize that this is equivalent to finding the longest subsequence that is strictly increasing except for at most one violation (i.e., one pair of adjacent elements where the order is not strictly increasing).
Define DP states: dp0[i] = length of longest valid subsequence ending at i with 0 deletions used; dp1[i] = length with 1 deletion used. Transition by considering previous elements and whether adding current element maintains the property.
After filling the DP table, the maximum value in dp0 and dp1 gives the length of the longest valid subsequence. The minimum removals is n minus this maximum length.
State the time complexity (O(n^2) for the DP) and space complexity (O(n)). Walk through a few test cases, including edge cases, to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.