I'm looking at the second test case

5 1

5 3 4 5 6

And don't understand can it be solved in a day. We have 2 problems with the same difficulty level - 5.

Solving two problems with same difficulty in a day will be against the rules, right ?

it's a weird problem. Hard to understand.

Its really hard to understand the question. Can anyone please tell me what actually i have to do?

testcase #1 [113 , 128]

Expected output is 5, where as my solution solves in 3 dayz. And seems correct to me..

Day 1 :: 383 694 335 335 160 986 355 762 973 542 717 853 663 483 218 16 507 852 365 791 264 492 173 38 538 860 281 988 857 591 342 971 353 666 512 70 518 362 84 352 113 301 507 639 365 33 876 680 142 413 970 637 171 957 761 466 315 887 184 40 970 536 153 622 394 791 290 110 632 265 736 549 296 878 314 834 199 950 356 156 794 469 157 961 824 287 678 141

Day 2 :: 422 422 958 763 400 155 829 668 490 155 539 845 650 815 327 674 934 172 359

Day 3 :: 440 851 174 872 105 246

Am I missing any constraint??

include

include

include

include

include

using namespace std;

int t, n, v[300], k, match[300], visited[300];

bool dfs(int s) { if(visited[s]) return false; visited[s] = true; for(int i = s + 1; i < n; i ++) if(abs(v[i] - v[s]) >= k) if(match[i] == -1 || dfs(match[i])) { match[i] = s; return true; } return false; }

int main() { for(cin >> t; t--; ) { cin >> n >> k; for(int i = 0; i < n; i ++) cin >> v[i]; memset(match, -1, sizeof match); int result = n; for(int i = 0; i < n; i ++) { memset(visited, 0, sizeof visited); if(dfs(i)) result -= 1; } cout << result << endl; } return 0; }