The naive approach is obviously too slow so you need to count occurrences of each string first, then walk through every possible split of accesscode and multiply the prefix count by the suffix count.
Use a hash map to store the frequency of each string in the array. For each string, compute its complement (the part needed to complete the accesscode) and look it up in the map, handling the case where the complement equals the current string to avoid counting the same index twice.
Pro tip: Clarify whether pairs are ordered (i, j) with i ≠ j and whether concatenation order matters (i.e., s[i] + s[j] vs s[j] + s[i]). This shows attention to detail and avoids off-by-one errors in counting.
Confirm if pairs are ordered (i, j) with i ≠ j, and whether concatenation order matters (i.e., s[i] + s[j] vs s[j] + s[i]). Also check if strings can be empty or if there are duplicate strings.
Create a hash map mapping each string to its frequency in the array. This allows O(1) lookups for complements.
For each string s, compute the complement needed to form the accesscode. If the complement exists in the map, add its frequency to the count. If the complement equals s, subtract 1 to avoid using the same index twice.
Consider cases where the accesscode length is less than 2, or when strings are longer than the accesscode. Also handle duplicates correctly by using frequencies.
State that the time complexity is O(n * L) where n is the number of strings and L is the average string length (due to substring operations), and space complexity is O(n) for the hash map.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.