##### 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 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.

##### Solution:

#!/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)

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