← SIG (Susquehanna) Interview Insights
I stared at the example for longer than I should have.
Use a prefix parity approach: track the parity of the number of zeros seen so far and count how many previous prefixes have the opposite parity. This yields an O(n) time, O(1) space solution by maintaining counts of even and odd parity prefixes.
Pro tip: Start by clarifying that 'odd number of zeros' means the count of zeros in the subarray is odd, not the sum of elements. Then mention that the same technique works for any condition based on parity of a count.
Confirm that we need to count contiguous subarrays where the number of zeros is odd. Ensure the array can contain any integers, and zeros are the only elements that affect the count.
Let P[i] be the parity (0 for even, 1 for odd) of the number of zeros in the first i elements. A subarray from j+1 to i has odd zeros iff P[i] != P[j].
Initialize counts of even and odd parity prefixes. Start with even count = 1 (empty prefix). Iterate through the array, update parity when encountering a zero, and add the count of opposite parity to the answer.
After processing each element, increment the count for the current parity. This ensures future subarrays can use this prefix.
Time complexity is O(n) because we traverse the array once. Space complexity is O(1) since we only store two counters and the current parity.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.