Interview Questions

Exponent - "This is a fun Fermi problem asked recently by Facebook — an estimation or approximation problem with limited information and back-of-the-envelope calculations. Recall the formula for Fermi problems: Ask clarifying questions Catalog what you know Make equation(s) Breakdown components of your equation Review and state assumptions Compute Sanity Check Summarize Ask clarifying questions It's wise here to slow down and get some clari"See full answer

Emma X. - "String commonStr(String str1, String str2) { int len1 = str1.length(); int len2 = str2.length(); if (len1 == 0 || len2 == 0) return ""; // let dpx reprsent the longest common str of 0...x int dp = new int len1 + 1; int maxLen = 0; int endIndex = 0; for (int i = 1; i <= len1; i++) { for (int j = 1; j <= len2; j++) { if (str1.charAt(i-1) == str2.charAt(j-1)) { dpi = dpi-1 + 1; "See full answer

Exponent - "You should identify this type of interview question as an Expansion problem, since we're asked to expand further into a market. This is similar to a growth problem, with a few additional components. This is the formula you should use when tackling these types of interview questions: Ask clarifying questions Perform user analysis Market risk analysis State goals Perform channel analysis Prioritize growth channels Strategy Summarize With"See full answer

Ashesh S. - "Steps: Validate K to be less than the number of elements in an array. Sort the array in ascending order. Get the K'th smallest element by array[k-1]."See full answer

Mubarak O. - "Tell a story that you do not need to sit with a laptop to be in career growing mood. "You can grow your career on the go." "Grow your career while you scroll." We will target our Messaging at every point where people go to read about job roles and descriptions."See full answer

Anonymous Firefly - "def findmaxone_index(A):    i = 0    n = len(A)    count_left = 0    count_right = 0    max_i = -1    last_i = -1    count_max = 0    while i count_max:                countmax = countleft + count_right + 1                maxi = lasti             last_i = i             countleft = countright             count_right = 0       i += 1    if last_i != -1: "See full answer

Fakhri A. - "static int countBinaryStrings(int n) { int a[] = new int[n]; int b[] = new int[n]; a[0] = b[0] = 1; for (int i = 1; i < n; i++) { a[i] = a[i - 1] + b[i - 1]; b[i] = a[i - 1]; } System.out.println(a[n - 1] + b[n - 1]); return a[n - 1] + b[n - 1]; }"See full answer