
Problem Statement
In the problem, we are provided with two distinct integer arrays, nums1
and nums2
, where nums1
is a subset of nums2
. The arrays are indexed starting from zero. For every element x
in the nums1
, we need to identify its position in nums2
and then find the first element in nums2
that is greater than this x
and appears to its right. If no such greater element exists, the result for this element should be -1
. The goal is to return an array where each element corresponds to the result of the above search for the respective element in nums1
.
Examples
Example 1
Input:
nums1 = [4,1,2], nums2 = [1,3,4,2]
Output:
[-1,3,-1]
Explanation:
The next greater element for each value of nums1 is as follows: - 4 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1. - 1 is underlined in nums2 = [1,3,4,2]. The next greater element is 3. - 2 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.
Example 2
Input:
nums1 = [2,4], nums2 = [1,2,3,4]
Output:
[3,-1]
Explanation:
The next greater element for each value of nums1 is as follows: - 2 is underlined in nums2 = [1,2,3,4]. The next greater element is 3. - 4 is underlined in nums2 = [1,2,3,4]. There is no next greater element, so the answer is -1.
Constraints
1 <= nums1.length <= nums2.length <= 1000
0 <= nums1[i], nums2[i] <= 104
- All integers in
nums1
andnums2
are unique. - All the integers of
nums1
also appear innums2
.
Approach and Intuition
Mapping Indexes: Begin by creating a map (or dictionary) to store the position of each element of
nums2
. This helps in quickly locating any element ofnums1
withinnums2
.Finding Next Greater Element:
- Traverse each element
x
innums1
. - Use the previously constructed map to find the index
j
ofx
innums2
. - Starting from the next index to
j
innums2
, look for the first element that is greater thanx
.
- Traverse each element
Handling None Case:
- If we reach the end of the list
nums2
without finding a greater element, record-1
as the result for that particular element ofnums1
.
- If we reach the end of the list
Example Explanation from the Provided Scenarios:
* In the first example, for 4
we look to its right in nums2
but find no greater values hence -1
is recorded.
By using a map to track the indices, we significantly reduce the need for repetitive searches within nums2
, thereby optimizing our approach. This is crucial given the constraints with lengths up to 1000. This mapping strategy followed by a simple scan for the greater element caters effectively to the problem requirements.
Solutions
- C++
class Solution {
public:
vector<int> findNextGreater(vector<int>& first, vector<int>& second) {
stack<int> elements;
unordered_map<int, int> result_map;
for (int value : second) {
while (!elements.empty() && value > elements.top()) {
result_map[elements.top()] = value;
elements.pop();
}
elements.push(value);
}
while (!elements.empty()) {
result_map[elements.top()] = -1;
elements.pop();
}
vector<int> result;
for (int value : first) {
result.push_back(result_map[value]);
}
return result;
}
};
The provided C++ solution finds the "Next Greater Element I" using a stack and hash mapping technique. Here’s a breakdown of the implementation:
- Initialize a stack to keep elements from the
second
list and an unordered map to store mappings of the next greater element. - Traverse each element in the
second
list:- If the stack is not empty and the current element is greater than the element on top of the stack, map the top element of the stack to the current element in the result map and then pop from the stack.
- Push the current element onto the stack.
- After processing all elements of the
second
list, pop remaining elements from the stack, assigning them a next greater value of-1
in the result map, as they do not have a greater element to their right. - Create a vector to store the results for elements in the
first
list, utilizing the previously built map to fill this vector with their corresponding next greater elements.
Finally, return this result vector, facilitating a straightforward access to the next greater elements for the queried indices of the first
list. This method ensures efficient computation of results, leveraging stack operations and hashmap for quick lookup.
- Java
public class Solution {
public int[] findNextGreaterElements(int[] firstArray, int[] secondArray) {
Stack<Integer> processingStack = new Stack<>();
HashMap<Integer, Integer> resultMap = new HashMap<>();
for (int index = 0; index < secondArray.length; index++) {
while (!processingStack.isEmpty() && secondArray[index] > processingStack.peek()) resultMap.put(
processingStack.pop(),
secondArray[index]
);
processingStack.push(secondArray[index]);
}
while (!processingStack.isEmpty()) resultMap.put(processingStack.pop(), -1);
int[] result = new int[firstArray.length];
for (int i = 0; i < firstArray.length; i++) {
result[i] = resultMap.get(firstArray[i]);
}
return result;
}
}
The Java program provided defines a class Solution
with a method findNextGreaterElements
intended to find the next greater element for each element in the first array (firstArray
) based on elements in the second array (secondArray
). The method employs a stack and hashmap to efficiently map each element in secondArray
to its next greater element, if it exists; otherwise, it maps to -1
.
Follow these steps to understand the resolution approach:
- Initialize a
Stack<Integer>
calledprocessingStack
and aHashMap<Integer, Integer>
calledresultMap
to keep track of each element and its next greater element. - Traverse each element in
secondArray
using an index-based loop. For each element:- While the stack is not empty and the current element is greater than the element at the top of the stack:
- Pop the stack and map this value in
resultMap
to the currentsecondArray
element.
- Pop the stack and map this value in
- Push the current element onto the stack.
- While the stack is not empty and the current element is greater than the element at the top of the stack:
- After iterating through
secondArray
, pop any remaining elements from the stack and map them to-1
inresultMap
, indicating no greater element exists. - Initialize an array
result
to store the results for thefirstArray
. - For each element in
firstArray
, retrieve its corresponding next greater element fromresultMap
, and store this in theresult
array. - Return the
result
array.
This solution efficiently determines and maps the next greater elements by leveraging a stack to keep track of the processed elements and a hashmap for quick look-up capabilities.
- Python
class Solution:
def nextLargerElement(self, first_list, second_list):
temp_stack = []
dict_mapping = {}
for element in second_list:
while temp_stack and element > temp_stack[-1]:
dict_mapping[temp_stack.pop()] = element
temp_stack.append(element)
while temp_stack:
dict_mapping[temp_stack.pop()] = -1
return [dict_mapping.get(item, -1) for item in first_list]
The Python solution for the "Next Greater Element I" problem uses a stack and dictionary to efficiently determine the next larger element for each number in the given first_list
using numbers from second_list
. Follow these steps to understand the implemented approach:
Initialize an empty list
temp_stack
to hold elements temporarily and a dictionarydict_mapping
to store the relationship between elements insecond_list
and their next greater element.Iterate through each element in
second_list
:- While
temp_stack
is not empty and the current element is greater than the last element intemp_stack
, map the element popped fromtemp_stack
to the current element indict_mapping
. - Push the current element onto
temp_stack
.
- While
After the loop, clear out any remaining elements in
temp_stack
:- Map each remaining element in
temp_stack
to-1
indict_mapping
, indicating no greater element exists for these items.
- Map each remaining element in
Construct the result for
first_list
by iterating through its elements:- For each item in
first_list
, usedict_mapping
to find the next greater element, defaulting to-1
if not found.
- For each item in
This solution ensures each element from both lists is processed efficiently, and the use of a stack helps in keeping track of elements for which a next larger element has yet to be determined.
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