Collections in action: interview questions & scenarios

Model answer

For the first prompt, I hear frequency counting plus order. Count each product ID in a Map<String, Integer>, and keep first-seen order with a LinkedHashMap or a separate ordered list. After the scan, return the first key whose count is 1.

For the second prompt, I hear Top-K. Keep a size-20 min-heap with PriorityQueue<Request> ordered by latency. The root is the smallest latency among the current winners, so a slower request can evict it in O(log k).

For the third prompt, I hear range query over intervals. Store bookings in a TreeMap<Integer, Integer> from start to end. Check floorEntry(start) for the previous interval and ceilingEntry(start) for the next interval. Only those two neighbors can overlap the new interval.

Model answer

For lowercase English letters, an int[26] is the cleanest tool. Count in one pass, then scan again and return the first character whose count is 1.

final class FirstNonRepeatingCharDemo {
    static char firstNonRepeatingChar(String s) {
        int[] counts = new int[26];
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            counts[ch - 'a']++;
        }

        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            if (counts[ch - 'a'] == 1) {
                return ch;
            }
        }
        return '\0';
    }

    public static void main(String[] args) {
        System.out.println(firstNonRepeatingChar("swiss")); // w
    }
}

This is O(n) time and O(1) extra space because the alphabet size is fixed at 26. If returning a sentinel feels unclear, use OptionalInt or return the index instead.