Type an input and write what you think is the output . Each problem is converted to Web Assembly, so any possible input will show the corresponding output. Only correct predictions count.
Locked until you have predicted 4 outputs correctly.
| 1 | double solution(vector<int> a, vector<int> b) { |
| 2 | |
| 3 | if (a.size() > b.size()) { |
| 4 | swap(a, b); |
| 5 | } |
| 6 | |
| 7 | int m = a.size(); |
| 8 | int n = b.size(); |
| 9 | int half = (m + n + 1) / 2; |
| 10 | |
| 11 | int lo = 0; |
| 12 | int hi = m; |
| 13 | |
| 14 | while (lo <= hi) { |
| 15 | |
| 16 | int i = lo + (hi - lo) / 2; |
| 17 | int j = half - i; |
| 18 | |
| 19 | int aLeft = (i == 0) ? INT_MIN : a[i - 1]; |
| 20 | int aRight = (i == m) ? INT_MAX : a[i]; |
| 21 | int bLeft = (j == 0) ? INT_MIN : b[j - 1]; |
| 22 | int bRight = (j == n) ? INT_MAX : b[j]; |
| 23 | |
| 24 | if (aLeft <= bRight && bLeft <= aRight) { |
| 25 | |
| 26 | int leftTop = max(aLeft, bLeft); |
| 27 | |
| 28 | if ((m + n) % 2 == 1) { |
| 29 | return leftTop; |
| 30 | } |
| 31 | |
| 32 | int rightLow = min(aRight, bRight); |
| 33 | return (leftTop + rightLow) / 2.0; |
| 34 | } |
| 35 | |
| 36 | if (aLeft > bRight) { |
| 37 | hi = i - 1; |
| 38 | } else { |
| 39 | lo = i + 1; |
| 40 | } |
| 41 | } |
| 42 | |
| 43 | return 0.0; |
| 44 | } |
Names have been stripped. The signature is the only clue you get for free. Compiled as C++20 with the standard headers and using namespace std; already in scope.