Salesforce Coding: Left-Right Occurrence Binary Strings
Question Description
Problem summary
You are given an integer array of length n. Produce two binary strings (each length n) that mark whether each element appears elsewhere to its left or to its right.
For index i: the first string has '1' if arr[i] appears at any j < i; the second has '1' if arr[i] appears at any k > i. Return [left_mark_string, right_mark_string] preserving the original order.
What the interviewer expects
Start by clarifying edge cases (empty array, all-unique, all-equal). A brute-force approach checks every pair (O(n^2)). You should show how to optimize to O(n) time with O(n) extra space using hash-based counting and two efficient passes: a left-to-right pass to set left marks using a seen set, and a right-to-left pass (or counts) to set right marks. Discuss trade-offs between one-pass streaming variants and two-pass counting approaches.
Flow in an interview
- Clarify input/output and constraints
- Propose brute force and analyze complexity
- Move to the hash-table / two-pass solution and implement
- Discuss complexity (time O(n), space O(n)), memory limits, and follow-ups
Skills and signals to demonstrate
You should demonstrate correct use of hash tables (sets or frequency maps), two-pointer/scan thinking, careful indexing/ordering, and clear complexity analysis. Mention streaming or memory-constrained variants and test edge cases to show robustness.
Common Follow-up Questions
- •How would you modify the solution to produce the strings in a single pass (streaming) if you cannot store the entire array?
- •If allowed, how would you change the outputs to report counts (number of occurrences to the left/right) rather than binary flags? What changes in complexity?
- •How would you adapt the approach if the input is a stream and you must update left/right marks online as elements arrive?
- •How does the solution change if values are extremely large or not hashable? Consider using coordinate compression or sorting.
- •Describe how you would test the implementation and which edge cases (e.g., duplicates clustered, interleaved duplicates) are most likely to expose bugs.
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