The next greater element of some element x in an array is the first greater element that is to the right of x in the same array.
Given an array of integers nums, return an array containing the indices of the next greater element for each number in nums.
If a number doesn't have a next greater element, denote that with -1.
Example 1:
Input: nums = [1, 3, 2, 4] Output: [1, 3, 3, -1] Explanation: The next greater element for each value of nums is as follows: - next greater element for 1 is 3 which is located at index 1 - next greater element for 3 is 4 which is located at index 3 - next greater element for 2 is 4 which is located at index 3 - 4 is the last and doesn't have a next greater element so we mark it with -1
Example 2:
Input: nums = [7, 3, 2, 6, 11, 9, 8, 10, 13] Output: [4, 3, 3, 4, 8, 7, 7, 8, -1]
The core challenge of this problem is to efficiently find the next greater element for each element in the array. This problem is significant in various applications such as stock price analysis, where you might want to know the next higher price after a given day.
Potential pitfalls include misunderstanding the requirement to find the first greater element to the right and not just any greater element.
To solve this problem, we can use a stack to keep track of the indices of the elements for which we are trying to find the next greater element. This approach ensures that we only traverse the array once, achieving an O(n) time complexity.
A naive solution would involve using two nested loops to compare each element with every other element to its right. This would result in an O(n^2) time complexity, which is not efficient for large arrays.
The optimized solution uses a stack to keep track of indices of elements for which we haven't found the next greater element yet. As we traverse the array, we compare the current element with the element at the index stored at the top of the stack. If the current element is greater, we pop the stack and record the current index as the next greater element for the popped index. We continue this until the stack is empty or the current element is not greater. Finally, we push the current index onto the stack.
#include <vector>
#include <stack>
#include <iostream>
std::vector<int> nextGreaterElements(const std::vector<int>& nums) {
int n = nums.size();
std::vector<int> result(n, -1); // Initialize result array with -1
std::stack<int> s; // Stack to store indices
for (int i = 0; i < n; ++i) {
// While stack is not empty and current element is greater than element at index stored at top of stack
while (!s.empty() && nums[i] > nums[s.top()]) {
result[s.top()] = i; // Set the result for the index at top of stack
s.pop(); // Pop the stack
}
s.push(i); // Push current index onto the stack
}
return result;
}
int main() {
std::vector<int> nums = {1, 3, 2, 4};
std::vector<int> result = nextGreaterElements(nums);
for (int index : result) {
std::cout << index << " ";
}
return 0;
}
The time complexity of this approach is O(n) because each element is pushed and popped from the stack at most once. The space complexity is also O(n) due to the stack and the result array.
Potential edge cases include:
These edge cases are handled naturally by the algorithm.
To test the solution comprehensively, consider the following test cases:
Using a testing framework like Google Test can help automate and manage these tests.
When approaching such problems, consider using data structures like stacks or queues that can help manage the elements efficiently. Practice solving similar problems to develop a deeper understanding of these techniques.
Understanding and solving the Next Greater Element problem is crucial for mastering array manipulation and stack usage. Practice and explore further to enhance your problem-solving skills.