One Important Thing Missed Out The complexity of the code is O(2n) and not O(n2) because the inner while loop executes at most n times in all loop iterations combined we are not starting from 0 everytime again and again we are continuing from where we ended in last operation hence the loop executes at most n times making total complexity = O(2n) Sub Link : codeforces.com/contest/1982/submission/267418199

Very nice explanation sir, Thank you!

Thank you for explaining I was waiting for your solution !

#include using namespace std; int main(){ long long int t; cin>>t; for(long long int j=0;j>n>>l>>r; long long int a[n]; for(long long int i=0;i>a[i]; } long long int s=0; long long int m=0; for(long long int i=0;ir){s=0; if(a[i]

@formidablechief27

2 күн бұрын

The problem is in s > r condition There you are not taking care of the situation u can get a suitable subarray if u trim your starting elements. Example counter case n = 5 l = 10 r = 10 a = [1, 1, 1, 1 10] In this case the suitable ans is just [10] but ur code will work right upto [1 1 1 1] and then it doesnt trim the start, instead u are directly making sum = 0 and hence not getting correct ans Another mistake i encountered, u are performing i--, wont that add the same element twice ?

@madhursharma7200

2 күн бұрын

@@formidablechief27 yeah after [1111] ,s become 4 and when i encounter 10, s become 14 and after it s>r that's why s=0 and i again add 10 in next loop so s become 10 and ans will come 1. My code also gives output 1.

@formidablechief27

2 күн бұрын

@madhursharma7200 Ok then this case might just do it n = 4, l = 10, r = 10 a = [1, 1, 1, 8] Ur code gives what for this case ?