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

Examples:

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[2] is 5 and left[4] 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[0] is 1 and left[3] 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[0] is 1 which is value 3 and left[3] 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.

## C/C++

// C++ program to check whether a subarray is in // mountain form or not #include <bits/stdc++.h> using namespace std; // Utility method to construct left and right array int preprocess(int arr[], int N, int left[], int right[]) { // initialize first left index as that index only left[0] = 0; int lastIncr = 0; for (int i = 1; i < N; i++) { // if current value is greater than previous, // update last increasing if (arr[i] > arr[i - 1]) lastIncr = i; left[i] = lastIncr; } // initialize last right index as that index only right[N - 1] = N - 1; int firstDecr = N - 1; for (int i = N - 2; i >= 0; i--) { // if current value is greater than next, // update first decreasing if (arr[i] > arr[i + 1]) firstDecr = i; right[i] = firstDecr; } } // method returns true if arr[L..R] is in mountain form bool isSubarrayMountainForm(int arr[], int left[], int right[], int L, int R) { // return true only if right at starting range is // greater than left at ending range return (right[L] >= left[R]); } // Driver code to test above methods int main() { int arr[] = {2, 3, 2, 4, 4, 6, 3, 2}; int N = sizeof(arr) / sizeof(int); int left[N], right[N]; preprocess(arr, N, left, right); int L = 0; int R = 2; if (isSubarrayMountainForm(arr, left, right, L, R)) cout << "Subarray is in mountain form\n"; else cout << "Subarray is not in mountain form\n"; L = 1; R = 3; if (isSubarrayMountainForm(arr, left, right, L, R)) cout << "Subarray is in mountain form\n"; else cout << "Subarray is not in mountain form\n"; return 0; }

## Java

// Java program to check whether a subarray is in // mountain form or not class SubArray { // Utility method to construct left and right array static void preprocess(int arr[], int N, int left[], int right[]) { // initialize first left index as that index only left[0] = 0; int lastIncr = 0; for (int i = 1; i < N; i++) { // if current value is greater than previous, // update last increasing if (arr[i] > arr[i - 1]) lastIncr = i; left[i] = lastIncr; } // initialize last right index as that index only right[N - 1] = N - 1; int firstDecr = N - 1; for (int i = N - 2; i >= 0; i--) { // if current value is greater than next, // update first decreasing if (arr[i] > arr[i + 1]) firstDecr = i; right[i] = firstDecr; } } // method returns true if arr[L..R] is in mountain form static boolean isSubarrayMountainForm(int arr[], int left[], int right[], int L, int R) { // return true only if right at starting range is // greater than left at ending range return (right[L] >= left[R]); } public static void main(String[] args) { int arr[] = {2, 3, 2, 4, 4, 6, 3, 2}; int N = arr.length; int left[] = new int[N]; int right[] = new int[N]; preprocess(arr, N, left, right); int L = 0; int R = 2; if (isSubarrayMountainForm(arr, left, right, L, R)) System.out.println("Subarray is in mountain form"); else System.out.println("Subarray is not in mountain form"); L = 1; R = 3; if (isSubarrayMountainForm(arr, left, right, L, R)) System.out.println("Subarray is in mountain form"); else System.out.println("Subarray is not in mountain form"); } } // This Code is Contributed by Saket Kumar

Output:

Subarray is in mountain form Subarray is not in mountain form

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