Next Greater Element I Leetcode Problem 496 [Python Solution]

If you're new to coding and looking to solve the Next Greater Element I problem on LeetCode, you've come to the right place.

In this blog post, we will walk you through the problem, explore different approaches to solve it, and provide a Python solution.

Let's dive in!

Problem Overview

The Next Greater Element I problem is a classic coding challenge.

It asks you to find the next greater element for each element in an array nums1 from a larger array nums2.

The catch is that nums1 is a subset of nums2.

The definition of the "next greater element" of some element x in an array is the first greater element to the right of x in the same array.

In simpler terms, you need to find the next element that is larger than the current element, and this element should appear to the right of the current element in the same array.

Here's a quick recap of the problem statement:

Input:
– Two distinct 0-indexed integer arrays nums1 and nums2, where nums1 is a subset of nums2.
– For each element x in nums1, you need to find the next greater element in nums2.

Output:
– An array ans of the same length as nums1, where ans[i] is the next greater element for nums1[i].

Now, let's look at an example to make this clearer.

Example:

Suppose we have the following input:

nums1 = [4, 1, 2]
nums2 = [1, 3, 4, 2]

The expected output should be:

[-1, 3, -1]

Here's how we get these results:

• For 4 in nums1, there is no next greater element in nums2, so we put -1.
• For 1 in nums1, the next greater element is 3 in nums2.
• For 2 in nums1, there is no next greater element in nums2, so we put -1.

That's the essence of the problem.

Now, let's explore different approaches to solving it.

Understanding the Constraints

Before we dive into the solutions, let's clarify the problem constraints:

• 1 <= nums1.length <= nums2.length <= 1000: This means the size of nums1 will be at least 1 and never exceed the size of nums2.
• 0 <= nums1[i], nums2[i] <= 10^4: All values in both arrays are integers between 0 and 10,000.
• All integers in nums1 also appear in nums2: Every element in nums1 is guaranteed to exist in nums2.

Now that we've got a solid understanding of the problem and its constraints, let's explore two different approaches to solve it.

Bruteforce Approach

One way to solve this problem is through a brute force approach.

In this approach, we iterate through nums1 and, for each element x, we iterate through nums2 to find the next greater element.

This approach involves nested loops, resulting in a time complexity of O(n * m), where n is the length of nums1, and m is the length of nums2.

Here's a Python solution for the brute force approach:

def nextGreaterElement(nums1, nums2):
    # Initialize a dictionary to map values to their indices in nums1
    nums1Idx = {n: i for i, n in enumerate(nums1)}
    # Initialize the result array with -1 values
    res = [-1] * len(nums1)
    
    stack = []
    for i in range(len(nums2)):
        cur = nums2[i]

        # While the stack exists and the current element is greater than the top of the stack
        while stack and cur > stack[-1]:
            val = stack.pop()  # Take the top value from the stack
            idx = nums1Idx[val]
            res[idx] = cur

        if cur in nums1Idx:
            stack.append(cur)
    
    return res

This code initializes a dictionary nums1Idx to map values from nums1 to their indices in the array.

We also initialize the result array with -1 values since that's the default when no next greater element is found.

We use a stack to keep track of elements from nums2 that might be the next greater element for elements in nums1.

We iterate through nums2, and for each element, we check if it's greater than the top element on the stack.

If it is, we pop elements from the stack, set the next greater element in the result, and continue this process until we find a value that's not greater.

This brute force approach will work, but it's not the most efficient solution due to the nested loops.

Efficient Approach with Monotonic Stack

To optimize our solution, we can use a monotonic stack, which is a key data structure for efficiently solving problems like this.

In this approach, we iterate through nums2 once and use a stack to keep track of elements that might be the next greater element for elements in nums1.

This approach has a time complexity of O(n + m), where n is the length of nums1, and m is the length of nums2.

Let's break down how the efficient approach works:

1. We start by creating an empty stack to store the indices of elements in nums2.

2. We iterate through nums2, and for each element cur, we check if the stack is not empty and if cur is greater than the element at the top of the stack.

If it is, we pop elements from the stack and set the next greater element in the result for the corresponding indices.

We repeat this process until cur is not greater than the top element on the stack.

1. If cur is in nums1 (i.e., it's one of the elements we're looking for), we add it to the stack.

2. After processing all elements in nums2, we return the result array.

Here's a Python solution for the efficient approach with a monotonic stack:

def nextGreaterElement(nums1, nums2):
    # Initialize a dictionary to map values to their indices in nums1
    nums1Idx = {n: i for i, n in enumerate(nums1)}
    # Initialize the result array with -1 values
    res = [-1] * len(nums1)

    stack = []
    for i in range(len(nums2)):
        cur = nums2[i]

        # While the stack exists and the current element is greater than the top of the stack
        while stack and cur > nums2[stack[-1]]:
            idx = stack.pop()
            if nums2[idx] in nums1Idx:
                res[nums1Idx[nums2[idx]]] = cur

        if cur in nums1Idx:
            stack.append(i)

    return res

This code follows

the steps mentioned earlier.

We use the same nums1Idx dictionary to map values from nums1 to their indices.

We initialize the result array with -1 values and use a stack to efficiently find the next greater elements.

Time and Space Complexity

Let's summarize the time and space complexity of the efficient approach:

• Time Complexity: O(n + m), where n is the length of nums1, and m is the length of nums2.

This is because we iterate through nums2 once, and the stack operations are done in constant time.

• Space Complexity: O(m), where m is the length of nums2.

The space complexity is determined by the size of the stack.

Reasoning Behind Our Approach

In the efficient approach, we use a monotonic stack to keep track of potential next greater elements efficiently.

This approach avoids the need for nested loops and achieves a linear time complexity.

By iterating through nums2 once and using a stack, we can quickly find the next greater elements for elements in nums1.

The key insight is that when we find the next greater element for an element, it applies to all the previous elements on the stack, as they are guaranteed to be less than the current element by definition.

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Conclusion

In this blog post, we tackled the Next Greater Element I problem on LeetCode.

We provided both a brute force solution and an efficient solution using a monotonic stack.

The efficient solution is recommended for larger inputs due to its superior time complexity of O(n + m).

We hope this guide has been helpful for beginners in coding and algorithmic problem-solving.

If you have any questions, comments, or suggestions, please feel free to leave them in the comments section.

Happy coding!

You can try this problem on LeetCode here.