We are given an array of integers and a range, we need to find whether the subarray which falls in this range has values in form of a mountain or not. All values of the subarray are said to be in form of a mountain if either all values are increasing or decreasing or first increasing and then decreasing.
More formally a subarray [a1, a2, a3 … aN] is said to be in form of a mountain if there exist an integer K, 1 <= K <= N such that,
a1 <= a2 <= a3 .. <= aK >= a(K+1) >= a(K+2) …. >= aN
Arr = [2 3 2 4 4 6 3 2] Range = [0, 2] Output yes because [2 3 2] subarray first increases and then decreases Range = [2, 7] Output yes because [2 4 4 6 3 2] subarray first increases and then decreases Range = [2, 3] Output yes because [2 4] subarray increases Range = [1, 3] Output no because [3 2 4] is not in the form above stated
We can solve this problem by first some preprocessing then we can answer for each subarray in the constant amount of time. We maintain two arrays left and right where left[i] stores the last index on left side which is increasing i.e. greater than its previous element and right[i] will store the first index on the right side which is decreasing i.e. greater than its next element. Once we maintained these arrays we can answer each subarray in constant time. Suppose range [l, r] is given then only if right[l] >= left[r], the subarray will be in form of a mountain otherwise not because the first index in decreasing form (i.e. right[l]) should come later than last index in increasing form (i.e. left[r]).
This procedure is demonstrated for above example array.
Arr = [2 3 2 4 4 6 3 2] Using above procedure building left and right array, left = [0 1 1 3 3 5 5 5] right = [1 1 5 5 5 5 6 7] Range = [2, 4] Now right is 5 and left is 3 that means at index 2 first decreasing element is right to the last increasing element at index 4, so they should have a mountain form. Range = [0, 3] Now right is 1 and left is 3 that means at index 0 first decreasing element is left to the last increasing element at index 3, so the subarray corresponding to this range does not have mountain form. We can see this in the array itself, right is 1 which is value 3 and left is 3 which is value 4 so 4 which is in increasing form (due to previous value 2) comes later to 3 which is in decreasing form (due to next value 2), mountain form was not possible here, same information is carried out with the help of left and right array.
Auxiliary space for this solution is O(N) and preprocessing takes O(N) time, after that each subarray can be handled in constant time.
Subarray is in mountain form Subarray is not in mountain form
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