Find whether a subarray is in form of a mountain or not

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 the form of a mountain or not. All values of the subarray are said to be in the 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:

Input : Arr[]  = [2 3 2 4 4 6 3 2], Range = [0, 2]
Output :  Yes

Explanation: The output is yes , subarray is [2 3 2], 
so subarray first increases and then decreases

Input:  Arr[] = [2 3 2 4 4 6 3 2], Range = [2, 7]
Output: Yes

Explanation: The output is yes , subarray is [2 4 4 6 3 2], 
so subarray first increases and then decreases


Input: Arr[]= [2 3 2 4 4 6 3 2], Range = [1, 3]
Output: no

Explanation: The output is no, subarray is [3 2 4], 
so subarray is not in the form above stated

Solution:

• Approach: The problem has multiple queries so for each query the solution should be calculated with least possible time complexity. So create two extra spaces of the length of the original array. For every element find the last index on the left side which is increasing i.e. greater than its previous element and find the element on the right side will store the first index on the right side which is decreasing i.e. greater than its next element. If these value can be calculated for every index in constant time then for every given range the answer can be given in constant time.
• Algorithm:
  1. Create two extra spaces of length n,left and right and a extra variable lastptr
  2. Initialize left[0] = 0 and lastptr = 0
  3. Traverse the original array from second index to the end
  4. For every index check if it is greater than the previous element, if yes then update the lastptr with the current index.
  5. For every index store the lastptr in left[i]
  6. initialize right[N-1] = N-1 and lastptr = N-1
  7. Traverse the original array from second last index to the start
  8. For every index check if it is greater than the next element, if yes then update the lastptr with the current index.
  9. For every index store the lastptr in right[i]
  10. Now process the queries
  11. for every query l, r, if right[l] >= left[r] then print yes else no
• Implementation:

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

Python3

# Python 3 program to check whether a subarray is in
# mountain form or not

# Utility method to construct left and right array
def preprocess(arr, N, left, right):
    # initialize first left index as that index only
    left[0] = 0
    lastIncr = 0

    for i in range(1,N):
        # 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
    firstDecr = N - 1

    i = N - 2
    while(i >= 0):
        # if current value is greater than next,
        # update first decreasing
        if (arr[i] > arr[i + 1]):
            firstDecr = i
        right[i] = firstDecr
        i -= 1

# method returns true if arr[L..R] is in mountain form
def isSubarrayMountainForm(arr, left, right, L, R):

    # return true only if right at starting range is
    # greater than left at ending range
    return (right[L] >= left[R])

# Driver code
if __name__ == '__main__':
    arr = [2, 3, 2, 4, 4, 6, 3, 2]
    N = len(arr)

    left = [0 for i in range(N)]
    right = [0 for i in range(N)]
    preprocess(arr, N, left, right)

    L = 0
    R = 2
    if (isSubarrayMountainForm(arr, left, right, L, R)):
        print("Subarray is in mountain form")
    else:
        print("Subarray is not in mountain form")

    L = 1
    R = 3
    if (isSubarrayMountainForm(arr, left, right, L, R)):
        print("Subarray is in mountain form")
    else:
        print("Subarray is not in mountain form")

# This code is contributed by
# Surendra_Gangwar

C#

// C# program to check whether
// a subarray is in mountain
// form or not
using System;

class GFG
{

    // 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 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
    static public void Main ()
    {
        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))
            Console.WriteLine("Subarray is in " +
                                "mountain form");
        else
            Console.WriteLine("Subarray is not " +
                              "in mountain form");

        L = 1;
        R = 3;

        if (isSubarrayMountainForm(arr, left,
                                   right, L, R))
            Console.WriteLine("Subarray is in " +
                                "mountain form");
        else
            Console.WriteLine("Subarray is not " +
                              "in mountain form");
    }
}

// This code is contributed by aj_36

Javascript

<script>
    // Javascript program to check whether
    // a subarray is in mountain
    // form or not

    // Utility method to construct
    // left and right array
    function preprocess(arr, N, left, right)
    {
        // initialize first left
        // index as that index only
        left[0] = 0;
        let lastIncr = 0;

        for (let 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;
        let firstDecr = N - 1;

        for (let 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
    function isSubarrayMountainForm(arr, left, right, L, R)
    {
        // return true only if right at
        // starting range is greater
        // than left at ending range
        return (right[L] >= left[R]);
    }

    let arr = [2, 3, 2, 4, 4, 6, 3, 2];
    let N = arr.length;
    let left = new Array(N);
    let right = new Array(N);
    preprocess(arr, N, left, right);

    let L = 0;
    let R = 2;

    if (isSubarrayMountainForm(arr, left, right, L, R))
      document.write("Subarray is in " + "mountain form" + "</br>");
    else
      document.write("Subarray is not " + "in mountain form" + "</br>");

    L = 1;
    R = 3;

    if (isSubarrayMountainForm(arr, left, right, L, R))
      document.write("Subarray is in " + "mountain form");
    else
      document.write("Subarray is not " + "in mountain form");

</script>

• Output:

Subarray is in mountain form
Subarray is not in mountain form

• Complexity Analysis:
  • Time Complexity:O(n). 
    Only two traversals are needed so the time complexity is O(n).
  • Space Complexity:O(n). 
    Two extra space of length n is required so the space complexity is O(n).