503. Next Greater Element II- Detailed Explanation
Problem Statement
Description:
You are given a circular array of integers. For each element in the array, you need to find its next greater element. The next greater element of a number x is the first greater number to its right when traversing the array in a circular fashion (i.e. after the last element, the search continues from the first element). If no such greater number exists, return -1 for that element.
Constraints:
- The length of the array is at least 1.
- The array is circular, meaning the search for the next greater element wraps around to the beginning.
Example 1:
- Input:
nums = [1, 2, 1] - Output:
[2, -1, 2] - Explanation:
- For index 0 (value 1), the next greater element is 2.
- For index 1 (value 2), there is no greater element.
- For index 2 (value 1), we wrap around to find that the next greater element is 2.
Example 2:
- Input:
nums = [3, 8, 4, 1, 2] - Output:
[8, -1, -1, 3, 3] - Explanation:
- For index 0 (value 3), the next greater element is 8.
- For index 1 (value 8), there is no greater element.
- For index 2 (value 4), there is no greater element to its right or when wrapping around.
- For index 3 (value 1), the next greater element is 3 (after wrapping around).
- For index 4 (value 2), the next greater element is 3 (after wrapping around).
Hints
- Hint 1: A brute-force approach would involve checking each element against every other element in circular order, but this leads to an O(n²) time complexity.
- Hint 2: Think about using a stack to keep track of indices of the elements for which the next greater element hasn’t been found yet.
- Hint 3: To handle the circular nature of the array, consider iterating through the array twice (or use modulo arithmetic) so that every element’s next greater element is properly checked.
Approaches
Approach 1: Brute Force (Less Optimal)
Explanation:
- Idea:
For each element, iterate over the next elements (wrapping around at the end) until a greater element is found. - Steps:
- For each index
i, traverse fromi+1up toi+n(using modulo operation to wrap). - If an element greater than
nums[i]is found, record it and break out of the loop. - If no greater element is found, set the result for that index to
-1.
- For each index
- Complexity:
- Time: O(n²) in the worst-case.
- Space: O(1) extra space (excluding the output array).
Approach 2: Optimal Stack-Based Method
Explanation:
- Idea:
Use a stack to maintain indices of elements for which the next greater element is not yet determined. Process the array twice (simulate the circular behavior) and update the result as soon as a greater element is found. - Steps:
-
Initialize a result array with -1 for all elements.
-
Iterate through the array twice (total of 2*n iterations). Use the modulo operator (
i % n) to simulate the circular array. -
For each index (computed as
i % n):- While the stack is not empty and the current element is greater than the element at the index on the top of the stack, pop from the stack and set the current element as the next greater element for that popped index.
- If the current iteration index is within the original array range (i.e. less than n), push the index onto the stack.
-
The stack will naturally be empty for indices where a next greater element is found; remaining indices will retain -1.
-
- Complexity:
- Time: O(n) — each element is pushed and popped at most once.
- Space: O(n) for the stack and the result array.
Python Code
Below is the Python implementation using the optimal stack-based approach:
Java Code
Below is the Java implementation using the stack-based method.
Complexity Analysis
-
Time Complexity:
In the optimal stack-based approach, each element is processed at most twice (once during the push phase and possibly popped later). Thus, the overall time complexity is O(n). -
Space Complexity:
The space complexity is O(n) due to the result array and the stack that can hold up to n indices.
Step-by-Step Walkthrough & Visual Example
Consider the input nums = [1, 2, 1]:
-
Initialization:
result = [-1, -1, -1]stack = []n = 3
-
First Pass (i from 0 to 2):
- i = 0:
current_index = 0, value = 1.- Stack is empty → push index 0.
stack = [0]
- i = 1:
current_index = 1, value = 2.- Compare
nums[1] (2)withnums[stack.top()]which isnums[0] (1). - Since 2 > 1, pop index 0 and set
result[0] = 2. resultbecomes[2, -1, -1],stack = [].- Push index 1 →
stack = [1]
- i = 2:
current_index = 2, value = 1.- Compare
nums[2] (1)withnums[stack.top()]which isnums[1] (2). - Since 1 is not greater than 2, push index 2 →
stack = [1, 2].
- i = 0:
-
Second Pass (i from 3 to 5):
- i = 3:
current_index = 0(3 % 3 = 0), value = 1.- Stack top is index 2 (value 1). Since 1 is not greater than 1, nothing changes.
- i = 4:
current_index = 1, value = 2.- Stack top is index 2 (value 1). Since 2 > 1, pop index 2 → set
result[2] = 2.
Now,stack = [1]. - Compare again: now stack top is index 1 (value 2). Since 2 is not greater than 2, stop.
- i = 5:
current_index = 2, value = 1.- Stack top is index 1 (value 2). Since 1 is not greater than 2, nothing changes.
- End of iterations; indices still in the stack (like index 1) do not have a next greater element, so remain -1.
- i = 3:
-
Final Output:
result = [2, -1, 2]
Common Mistakes, Edge Cases & Alternative Variations
Common Mistakes:
- Not Simulating Circularity:
Failing to iterate over the array twice (or using modulo arithmetic) results in missing the next greater element that occurs before the current index. - Incorrect Stack Handling:
Pushing and popping indices incorrectly may lead to wrong results. Always ensure you compare the current element with the element corresponding to the index at the top of the stack. - Off-by-One Errors:
Care must be taken with array indices when using modulo operations.
Edge Cases:
-
Single Element Array:
If the array contains only one element, the output should be[-1]because no greater element exists. -
All Elements Equal:
If all numbers are the same, every element’s next greater element should be-1. -
Strictly Decreasing Order:
Every element will have-1as the next greater element because no element to the right (or in the wrapped-around part) is greater.
Alternative Variations:
-
Next Greater Element I:
A similar problem where the array is not circular. -
Circular Arrays in Other Contexts:
Many problems require handling circular data; understanding the modulo trick is widely applicable.
Related Problems for Further Practice
-
Next Greater Element I:
Find the next greater element in a non-circular array. -
Daily Temperatures:
For each day, find how many days you have to wait for a warmer temperature. -
Stock Span Problem:
Calculate how many consecutive days before each day the price was less or equal. -
Monotonic Stack Problems:
Explore other problems that can be solved with similar stack techniques.
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