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.
time complexity is
space complexity is
You do not actually need to construct all the sub-arrays, as they reduce to only one element. You also can ignore all sub-arrays, that do not contain elements E > K. I also observed that there are x*y sub-arrays 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 sub-arrays that contain E and are not part of another e.
#!/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)