I just took DRW OA after today\'s contest ended. Hopefully I will finally become a guardian after ranking gets updated. I wonder how the top 500 Motorola Mobility Referral works...\n</br>\n__________\n#### Outline\nThere are in total of 3 questions and we are given 150 minutes, which is wayyyyyyyyyyyy more than enough. \nThe test is conducted on Codility and the DRW encourges that we use C++, Java, Python even though they do support other languages.\nI decided to code some problems in Java and some in C++. \nNote that Codility does **not** have any library included by default. You have to memorize what library to include.\nThe highest Java version they have is 11, and C++14.\n</br>\n__________________________\n#### Question 1\nGiven `A` and `B`, find the number of integer `X` such that `X*(X+1)` falls between `[A, B]`, both inclusive. \n\n**Constraint:**\n`1 <= A <= B <= 1e9`\n\n**My Thoughts**:\nUnclear problem statement! Could `X` be negative??\nI coded up the bruteforce `O(sqrt(N))` solution because I think that should be good enough, and I found out from the testcases that `X` can\'t be negative.\nWe can also give a `O(logN)` solution with binary search but I felt it unncessary during OA as that would require me to write a separate function and call `solve(B) - solve(A-1)`\n\nPassed all sample testcases in Java 11.\n\n</br>\n\n______________________________\n\n#### Question 2\nGiven `M` units of 1 by 1 square, and `N` units of 2 by 2 squares. Find the maximum size of square that you can make with them.\n\n**Constraints:**\n`0 <= M <= 1e9`\n`0 <= N <= 1e9`\n\n**My Solution**\n\nGreedily use as many as 2x2 as possible, and fill that gap with 1x1.\n- If we can fill the gap, then binary search for how much the remaining 1x1 can improve it.\n- If we can\'t fill the gap, backtrack the 2x2 length by 1 unit, and do binary search for all 1x1.\n\nPassed all sample testcases in C++14.\n\n</br>\n\n________________________________\n\n#### Question 3\nGiven an array `A` and `B` of the same length, we have to create an array `C`, `C[i]` can be either `A[i]` or `B[i]`, such that the MEX (Minimum Excluded Positive Integer) of `C` is minimized.\nReturn the MEX of `C`.\n\n**Constraints:**\n`1 <= length of A/B <= 100000`\n`1 <= A[i], B[i] <= 1e9`\n\n\n**My Solution**\nObserve that for any pair of `(A[i], B[i])`, we have an option to avoid a certain number unless `A[i] == B[i]`.\nLet `N = length of A`, create an `has` bool array of length `N+2` and mark value as present or not. \nIf `A[i] == B[i]`, then `has[A[i]] = true`.\nAfter we mark it all, use a `for (int ans = 1; true; ++ans)` and return the first element not present in `has`.\n\nPassed all sample testcases in C++14.

I just took DRW OA after today\'s contest ended. Hopefully I will finally become a guardian after ranking gets updated. I wonder how the top 500 Motorola Mobili