Next Greater Element I

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.

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:

1. Initialize a Stack<Integer> called processingStack and a HashMap<Integer, Integer> called resultMap to keep track of each element and its next greater element.
2. 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 current secondArray element.
   • Push the current element onto the stack.
3. After iterating through secondArray, pop any remaining elements from the stack and map them to -1 in resultMap, indicating no greater element exists.
4. Initialize an array result to store the results for the firstArray.
5. For each element in firstArray, retrieve its corresponding next greater element from resultMap, and store this in the result array.
6. 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.

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:

1. Initialize an empty list temp_stack to hold elements temporarily and a dictionary dict_mapping to store the relationship between elements in second_list and their next greater element.

2. Iterate through each element in second_list:

   • While temp_stack is not empty and the current element is greater than the last element in temp_stack, map the element popped from temp_stack to the current element in dict_mapping.
   • Push the current element onto temp_stack.

3. After the loop, clear out any remaining elements in temp_stack:

   • Map each remaining element in temp_stack to -1 in dict_mapping, indicating no greater element exists for these items.

4. Construct the result for first_list by iterating through its elements:

   • For each item in first_list, use dict_mapping to find the next greater element, defaulting to -1 if not found.

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.