Short Problem Definition:
Shashank loves to play with arrays a lot. Today, he has an array A consisting of N positive integers. At first, Shashank listed all the subarrays of his array A on a paper and later replaced all the subarrays on the paper with the maximum element present in the respective subarray.
Link
Complexity:
time complexity is O(N)
space complexity is O(1)
Execution:
You do not actually need to construct all the subarrays, as they reduce to only one element. You also can ignore all subarrays, that do not contain elements E > K. I also observed that there are x*y subarrays that match the above specified criteria for each element E > K. X is the distance from E to any previous e > K. Y is the distance from E to the end of the array. This way you crate all the subarrays that contain E and are not part of another e.
Solution:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

#!/usr/bin/py def numberList(a ,k): result = 0 last_biggest =  1 a_len = len (a) for idx in xrange (a_len): if a[idx] > k: result + = (idx  last_biggest) * (a_len  idx) last_biggest = idx return result if __name__ = = '__main__' : t = int ( raw_input ()) for _ in xrange (t): n,k = map ( int , raw_input ().split()) a = map ( int , raw_input ().split()) print numberList(a ,k) 