📌 Facebook | E5 | Bay Area | Phone Screen

I had my Facebook E5 phone screen interview, which lasted for 45 minutes. We started with a brief 3-5 minute introduction. The interviewer then clarified that the majority of the time would be dedicated to problem-solving, reserving the last 5 minutes for my questions.

The first question I encountered was Add Binary. Having practiced this problem before, I immediately explained the optimal approach, achieving a time complexity of O(max(l1, l2)). The interviewer was satisfied with my algorithm, and I quickly proceeded to code the solution in Java. After coding, I walked through my solution with a few examples. This entire segment, including discussion and coding, was completed within 10-15 minutes.

/**
 * Time Complexity: O(max(l1, l2))
 * Space Complexity: O(max(l1, l2))
 * l1 = Length of string A, l2 = Length of string B.
 */
public String addBinary(String a, String b) {
    if (a == null && b == null) {
        return "0";
    }
StringBuilder result = new StringBuilder();

int i = a.length() - 1;
int j = b.length() - 1;
int carry = 0;

while (i >= 0 || j >= 0) {
    int sum = carry;
    if (i >= 0) {
        sum += a.charAt(i) - '0';
        i--;
    }
    if (j >= 0) {
        sum += b.charAt(j) - '0';
        j--;
    }
    result.append(sum % 2);
    carry = sum / 2;
}

if (carry == 1) {
    result.append('1');
}

return result.reverse().toString();

}

Following that, the second question was Merge Intervals. Again, I was familiar with this problem and presented the optimal O(NlogN) approach. Once the interviewer approved the algorithm, I implemented the solution in Java and demonstrated its functionality with test cases. This question took approximately 20-25 minutes to complete.

/**
 * Time Complexity O(N log N)
 * Space Complexity: O(N) (Including the List<int[]> result)
 * N = Length of input array.
 */
public int[][] merge(int[][] intervals) {
    if (intervals == null || intervals.length <= 1) {
        return intervals;
    }
Arrays.sort(intervals, (a, b) -&gt; (a[0] - b[0]));

List&lt;int[]&gt; result = new ArrayList&lt;&gt;();
int start = intervals[0][0];
int end = intervals[0][1];

for (int i = 1; i &lt; intervals.length; i++) {
    if (intervals[i][0] &lt;= end) {
        end = Math.max(end, intervals[i][1]);
    } else {
        result.add(new int[] { start, end });
        start = intervals[i][0];
        end = intervals[i][1];
    }
}

result.add(new int[] { start, end });
return result.toArray(new int[result.size()][2]);

}

With a few minutes remaining, I took the opportunity to ask the interviewer two questions about their current work at Facebook. I was thrilled to later receive a call from the recruiter informing me of positive feedback from the hiring committee and that an offer would be extended.