Start by computing the character frequency difference between S and T to identify which characters are in surplus and which are in deficit. Then categorize the surplus characters into vowels and consonants, and greedily match cross-category replacements (cost 1) before handling same-category replacements (cost 2). This greedy approach minimizes total cost by always preferring cheaper operations first.
Pro tip: Explicitly mention that cross-category swaps (vowel ↔ consonant) should always be prioritized over same-category swaps because they resolve two imbalances at once for a lower cost — this demonstrates you understand the greedy optimality argument, which interviewers at Infosys look for in algorithm design questions.
Build frequency maps for both S and T, then compute the net surplus/deficit for each character (freq[S][c] - freq[T][c]). Positive values indicate characters S has in excess; negative values indicate characters S needs more of.
Partition the surplus characters (those with positive difference) into two buckets: surplus vowels and surplus consonants. Similarly, partition the deficit characters into deficit vowels and deficit consonants.
Match surplus vowels against deficit consonants, and surplus consonants against deficit vowels. Each such pairing costs 1 and resolves one unit of imbalance on each side — apply as many of these as possible first.
Handle any remaining surplus vowels paired with deficit vowels, and remaining surplus consonants paired with deficit consonants. Each such replacement costs 2 since it stays within the same character category.
Accumulate the costs from both phases and return the total. Verify your solution with edge cases such as strings that are already anagrams (cost 0) or strings composed entirely of vowels or consonants.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Classic house robber DP breaks down here because the extra constraint makes transitions O(n²) and that's too slow.
Treat this as a dynamic programming problem where the state must track both the current house index and the last robbed house index (or its profit value) to enforce both the adjacency and profit-difference constraints. Define dp[i][j] as the maximum profit when considering house i with house j being the most recently robbed house, then build transitions carefully. Clearly articulate the state space, transition logic, and how the two constraints interact before coding.
Pro tip: Explicitly call out the state space complexity trade-off: tracking the last robbed house index gives O(n²) states which is manageable, but naively tracking profit values could explode — showing you reason about scalability signals strong engineering maturity to interviewers.
Confirm input size, whether profits can be negative, and what D represents. Walk through a small example (e.g., [2, 7, 9, 3, 1], D=3) to validate your understanding of both constraints before designing a solution.
Define dp[i][j] as the maximum profit achievable up to house i where house j was the last robbed house (j < i). This two-dimensional state captures both which house we are at and what the last robbed profit was, enabling enforcement of the |profit[i] - profit[j]| >= D constraint.
For each house i and each valid previous house j (where i - j >= 2 and |profit[i] - profit[j]| >= D), set dp[i][j] = max(dp[i][j], dp[j][k] + profit[i]) for all valid k. Also handle the base case where house i is the first house robbed: dp[i][-1] = profit[i].
Identify that the naive approach is O(n³) time and O(n²) space, then discuss whether sorting or pruning can reduce constant factors. Mention that for large n, segment trees or sorted structures could optimize the inner lookup, demonstrating awareness of scalability.
Implement the solution cleanly, then test against edge cases: all houses skipped (answer is 0 or best single house), D=0 (reduces to classic house robber), negative profits, and arrays of size 1 or 2. Verbally trace through your example to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Recognize this as a dynamic programming problem where you build up an optimal solution by deciding where to place partition boundaries. Define dp[i] as the maximum profit achievable from the first i elements, then for each position i, try all valid group endings (groups of size >= 3) and use the third-smallest element of that group as the profit contribution. Sort each candidate group or use a selection algorithm to efficiently find the third-smallest value.
Pro tip: Mention that pre-sorting subarrays naively leads to O(n³ log n) complexity, but you can optimize by maintaining a running sorted structure or noting that for a fixed right boundary you can incrementally track the third-smallest as you extend the group leftward, bringing it closer to O(n²) — this shows you think beyond brute force.
Ask about array size, value ranges, and confirm what 'third-smallest' means for groups of exactly 3 (it's the maximum of the group). Identify the base case: if the array has fewer than 3 elements, return -1 immediately.
Let dp[i] represent the maximum total profit using the first i elements with a valid partition, initialized to -infinity (invalid). dp[0] = 0 as the base case (empty array, zero profit).
For each index i (1-indexed), iterate over all starting points j such that the group [j..i] has at least 3 elements. If dp[j-1] is valid, compute the third-smallest of the subarray and update dp[i] = max(dp[i], dp[j-1] + third_smallest(j, i)).
As you extend the left boundary j leftward from i, maintain a sorted structure (e.g., insertion into a sorted list or a max-heap of size 3) to track the third-smallest incrementally, avoiding a full sort for each subarray.
Return dp[n] if it's valid (not -infinity), otherwise return -1. Discuss time complexity: O(n²) for the DP transitions with O(n log n) for sorting per transition in the naive case, and how incremental tracking improves this.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model this as a modified Dijkstra's shortest path problem where the state includes not just the current city but also whether the VIP pass has been used and whether the next city's cost should be doubled. Extend the state space to (city, vip_used, next_doubled) and run Dijkstra over this augmented graph to find the globally optimal cost. This cleanly handles the pass's side effect without requiring exhaustive search.
Pro tip: Explicitly call out that the VIP pass introduces a 'state dependency' between consecutive nodes — this is the crux of the problem and shows you recognize it's not a vanilla shortest path problem. Mentioning that naive greedy application of the pass on the most expensive city can be suboptimal (due to the doubling penalty) will impress the interviewer.
Confirm key details: is the graph weighted by node visit costs only (no edge weights), can the VIP pass be used on the start or end city, and is the graph connected? Clarifying these prevents wasted effort on wrong assumptions.
Represent each state as a tuple (current_city, vip_used: bool, next_cost_doubled: bool) to capture all relevant information at any point in the traversal. This transforms the problem into a standard shortest path on a larger but well-defined state graph.
For each state, enumerate transitions: moving to a neighbor without using the pass, or using the pass on the current city (making it free but flagging the next city for double cost). Carefully compute the cost of entering each neighbor based on the 'next_cost_doubled' flag.
Apply Dijkstra's algorithm using a min-heap over the augmented states, initializing with the start city's cost (considering pass usage options). Track visited states to avoid reprocessing and extract the minimum cost when the destination is reached.
State space is O(N × 2 × 2) = O(N), so overall complexity is O(N log N) with a binary heap, which is efficient. Discuss edge cases: using the pass on the last city (no doubling penalty applies), using it on the start city, and graphs with a single node.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.