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Two pointers ended up being the move here.
Use a two-pointer technique: one pointer to read through the array and identify runs, and another to write the compressed result in place. For each run, write the character and then the digits of the count (if >1) as separate characters. Return the write pointer's final position as the new length.
Pro tip: Clarify that the input array is mutable and that we must modify it in place with O(1) extra space. Also, handle edge cases like empty array and runs of length 1, and mention that the count can be multi-digit, so you need to convert the integer to individual digit characters.
Confirm that the array is mutable, compression is in place, and return the new length. Discuss edge cases: empty array, single character, runs longer than 9 (multi-digit counts).
Set a read pointer to scan the array and a write pointer to overwrite with compressed data. Start both at index 0.
While read < length, record the current character, then advance read while the same character repeats, counting the run length.
Write the character at the write pointer, increment write. If count > 1, convert count to string and write each digit character, incrementing write for each.
After processing all runs, return the write pointer as the new length of the compressed array.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Looks like a simple greedy but I kept doubting myself.
Start by clarifying the problem and edge cases, then propose a greedy strategy based on the binary representation of n. Explain that for odd n, the optimal choice depends on n mod 4, and prove why this greedy approach yields the minimum number of operations.
Pro tip: Mention that this problem is equivalent to finding the shortest path in a graph where each number connects to n/2, n+1, or n-1, and that the greedy solution is optimal due to the structure of binary numbers. This shows depth and can lead to a discussion on BFS vs greedy trade-offs.
Restate the problem in your own words and confirm the rules: for even n, only divide by 2; for odd n, you can add or subtract 1. Ask about constraints (e.g., n up to 2^31) and whether n=0 is possible.
Manually compute the minimum operations for small n (e.g., 1, 2, 3, 4, 5, 6, 7, 8) to identify patterns. Notice that for odd n, the choice depends on n mod 4.
For even n, always divide by 2. For odd n > 1, if n % 4 == 1, subtract 1; if n % 4 == 3, add 1 (except when n == 3, subtract 1). For n == 1, subtract 1 to reach 0.
Argue that the greedy choice is optimal by considering the binary representation: adding 1 to an odd number ending in '11' causes a carry that reduces the number of 1s, while subtracting 1 from an odd number ending in '01' does the same. This minimizes future operations.
The algorithm runs in O(log n) time and O(1) space. Provide pseudocode or code, and test with edge cases like n=1, n=3, and large powers of 2.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.