941. Valid Mountain Array

Problem Description

The problem requires us to determine if a given array arr of integers represents a mountain array or not. By definition, an array is considered a mountain array if it meets the following criteria:

Its length is at least 3.
There exists a peak element arr[i], such that:
- Elements before arr[i] are in strictly increasing order. This means arr[0] < arr[1] < ... < arr[i - 1] < arr[i].
- Elements after arr[i] are in strictly decreasing order. This means arr[i] > arr[i + 1] > ... > arr[arr.length - 1].

In essence, the array should rise to a single peak and then consistently decrease from that point, simulating the shape of a mountain.

Intuition

The idea for solving this problem is to find the peak of the potential mountain and to verify that there is a clear increasing sequence before the peak and a clear decreasing sequence after the peak.

To ascertain whether the array arr is a mountain array, we can walk from both ends of the array towards the center:

We start from the beginning of the array and keep moving right as long as each element is greater than the previous one (the array is ascending). We do this to find the increasing slope of the mountain.
Simultaneously, we start from the end of the array and keep moving left as long as each element is less than the previous one (the array is descending). We do this to find the decreasing slope of the mountain.

If after traversing, we find that both pointers meet at the same index (the peak), and it's neither the first nor the last index (since the peak cannot be at the ends), we can conclude that it is a valid mountain array. Otherwise, if the pointers do not meet or the peak is at the first or last index, the array is not a mountain array.

This approach works because for an array to be a mountain array, there must be one highest peak, and all other elements to the left and to the right of the peak must strictly increase and decrease respectively, which leaves no room for plateaus or multiple peaks.

Solution Approach

The solution approach utilizes a two-pointer technique, which is an algorithmic pattern where we use two indices to iterate over the data structure—in this case, the array—to solve the problem with better time or space complexity. The solution does not require any additional data structures and operates directly on the input array, which makes it an in-place algorithm with O(1) extra space complexity.

Here’s the step-by-step process of the solution approach:

Initial Check: First, we check if the array arr has at least three elements. If not, it cannot form a mountain array, so we return False.
Finding the Increasing Part: We set a pointer l at the start of the array. We move l to the right (l += 1) as long as arr[l] < arr[l + 1]. This loop will stop when l points to the peak of the mountain or when arr is no longer strictly increasing.
Finding the Decreasing Part: We set a pointer r at the end of the array. Similarly, we move r to the left (r -= 1) as long as arr[r] < arr[r - 1]. This loop will stop when r points to the peak of the mountain or when arr is no longer strictly decreasing.
Checking the Peak: After exiting both loops, if l and r are pointing to the same index, it means that we have potentially found the peak of the mountain. But for a valid mountain array, the peak cannot be the first or last element. Therefore, if l and r meet at the same index and it's not at the ends of the array, we can conclude that the array is a valid mountain array and return True.
Result: If the above conditions are not met, we return False since the array does not satisfy the criteria for being a mountain array.

The solution has a time complexity of O(n) where n is the length of the array since in the worst case we could traverse the entire array once with each pointer.

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Example Walkthrough

Let's take a small example with the array arr = [1, 3, 5, 4, 2] to illustrate the solution approach.

Initial Check: First, we check if arr has at least three elements. In this case, the array has five elements, so we can proceed.
Finding the Increasing Part: We set a pointer l at the start of the array (l = 0). We move l to the right while arr[l] < arr[l + 1].
- l = 0: we check if arr[0] < arr[1] which means 1 < 3. It's true, so we increment l to 1.
- l = 1: we check if arr[1] < arr[2] which means 3 < 5. It's true, so we increment l to 2.
- l = 2: we check if arr[2] < arr[3] which means 5 < 4. It's false, so we stop. Now l is pointing to the peak at index 2.
Finding the Decreasing Part: Similarly, we set a pointer r at the end of the array (r = 4). We move r to the left while arr[r] < arr[r - 1].
- r = 4: we check if arr[4] < arr[3] which means 2 < 4. It's true, so we decrement r to 3.
- r = 3: we check if arr[3] < arr[2] which means 4 < 5. It's true, so we decrement r to 2.
- Now r is pointing to the same peak index as l, which is index 2.
Checking the Peak: Both l and r are pointing to the same index (2). We check if this is not the first or last index of the array. Since 2 is neither 0 nor 4, the conditions are satisfied.
Result: Since all conditions are met (l and r met at the same index, not at the ends, and ascending/descending sequences are correct), we conclude that the given array arr = [1, 3, 5, 4, 2] is a valid mountain array and return True.

This example demonstrated a successful case where the array meets the criteria for a mountain array. If, for instance, arr had been [1, 2, 2, 1], the solution would have returned False because there is no strictly increasing or strictly decreasing sequence due to the repeated 2s.

Solution Implementation

1class Solution:
2    def validMountainArray(self, arr: List[int]) -> bool:
3        # Get the length of the array
4        num_elements = len(arr)
5      
6        # A valid mountain array requires at least 3 elements
7        if num_elements < 3:
8            return False
9      
10        # Initialize two pointers starting from the beginning and the end of the array
11        left_index, right_index = 0, num_elements - 1
12      
13        # Move the left_index towards the right, stop before the peak
14        while left_index + 1 < num_elements - 1 and arr[left_index] < arr[left_index + 1]:
15            left_index += 1
16      
17        # Move the right_index towards the left, stop before the peak
18        while right_index - 1 > 0 and arr[right_index] < arr[right_index - 1]:
19            right_index -= 1
20      
21        # The mountain peak is valid if left and right pointers meet at the same index,
22        # implying exactly one peak is present.
23        return left_index == right_index
24

1class Solution {
2
3    // Method to check if the given array represents a valid mountain array.
4    public boolean validMountainArray(int[] arr) {
5        int length = arr.length; // Get the length of the array
6
7        // A mountain array must have at least 3 elements
8        if (length < 3) {
9            return false;
10        }
11
12        int left = 0;   // Starting index from the beginning of the array
13        int right = length - 1; // Starting index from the end of the array
14
15        // Move the left index towards the right as long as the next element is greater
16        // which is analogous to climbing up the mountain
17        while (left + 1 < length - 1 && arr[left] < arr[left + 1]) {
18            left++;
19        }
20
21        // Move the right index towards the left as long as the previous element is greater
22        // which is analogous to walking down the mountain
23        while (right - 1 > 0 && arr[right] < arr[right - 1]) {
24            right--;
25        }
26
27        // For the array to be a mountain array, we must have climbed up and then walked down the mountain,
28        // which means the indexes should meet at the peak, and the peak can't be the first or last element.
29        return left == right;
30    }
31}
32

1#include <vector>
2
3class Solution {
4public:
5    bool validMountainArray(std::vector<int>& arr) {
6        int arrSize = arr.size();
7      
8        // A mountain array must have at least 3 elements
9        if (arrSize < 3) return false;
10      
11        // Initialize pointers to start and end of the array
12        int leftIndex = 0, rightIndex = arrSize - 1;
13
14        // Move leftIndex rightward as long as the next element is greater
15        while (leftIndex + 1 < arrSize - 1 && arr[leftIndex] < arr[leftIndex + 1]) {
16            ++leftIndex;
17        }
18
19        // Move rightIndex leftward as long as the previous element is greater
20        while (rightIndex - 1 > 0 && arr[rightIndex] < arr[rightIndex - 1]) {
21            --rightIndex;
22        }
23
24        // A valid mountain array will have both pointers meet at the same peak index
25        return leftIndex == rightIndex;
26    }
27};
28

1function validMountainArray(arr: number[]): boolean {
2    let arrSize = arr.length;
3
4    // A mountain array must have at least 3 elements.
5    if (arrSize < 3) {
6        return false;
7    }
8
9    // Initialize pointers to start and end of the array.
10    let leftIndex = 0;
11    let rightIndex = arrSize - 1;
12
13    // Move leftIndex rightward as long as the next element is greater.
14    while (leftIndex + 1 < arrSize && arr[leftIndex] < arr[leftIndex + 1]) {
15        leftIndex++;
16    }
17
18    // Move rightIndex leftward as long as the previous element is greater.
19    while (rightIndex > 0 && arr[rightIndex - 1] > arr[rightIndex]) {
20        rightIndex--;
21    }
22
23    // A valid mountain array will have both pointers meet at the same peak index
24    // and that index cannot be the first or last index of the array.
25    return leftIndex == rightIndex && leftIndex !== 0 && rightIndex !== arrSize - 1;
26}
27

Time and Space Complexity

The time complexity of the given validMountainArray function can be considered to be O(n), where n is the length of the input array arr. This is because the function uses two while loops that iterate through the array from both ends, but each element is visited at most once. The first loop increments the variable l until the ascending part of the mountain is traversed, and the second loop decrements the variable r until the descending part of the mountain is traversed. In the worst case, these two loops together will scan all elements of the array once.

The space complexity of the function is O(1), as it only uses a fixed number of extra variables (n, l, r) that do not depend on the size of the input array.

Learn more about how to find time and space complexity quickly using problem constraints.

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Question 1 out of 10

What is an advantages of top-down dynamic programming vs bottom-up dynamic programming?

It's faster

It has better space complexity

Order of computer subproblems does not matter

It has better time complexity

I don’t know