Solution Explanation: Daily Temperatures

This problem asks us to find, for each day, the number of days we need to wait until a warmer temperature arrives. We can efficiently solve this using a monotonic stack.

Approach: Monotonic Stack

A monotonic stack (in this case, a decreasing stack) is ideal because it keeps track of indices of days with decreasing temperatures. When we encounter a warmer day, we can easily determine the waiting period for all previous cooler days stored in the stack.

Algorithm:

1. Initialization: Create an empty stack stk to store indices of days and an array ans of the same size as temperatures initialized with zeros. This ans array will hold our results.

2. Iteration: Iterate through the temperatures array from right to left (this is crucial for efficiently finding the next warmer day).

3. Stack Processing: For each day i:

   • While the stack is not empty and the temperature at the top of the stack (temperatures[stk.top()]) is less than or equal to the current day's temperature (temperatures[i]):
     • Pop the top index from the stack.
     • The difference between the popped index and the current index i represents the waiting period. Update ans accordingly.
   • Push the current day's index i onto the stack. This maintains the decreasing order of temperatures in the stack.

4. Return: After iterating through all days, ans contains the waiting periods for each day. Return ans.

Time and Space Complexity:

• Time Complexity: O(N), where N is the length of the temperatures array. Each element is pushed and popped at most once from the stack.
• Space Complexity: O(N) in the worst case, as the stack might store all indices if the temperatures are strictly decreasing.

Code Examples (with detailed comments):

Several code examples in various languages are presented below, showcasing the implementation of this algorithm. Each example follows the algorithm steps explained above.

Python:

def dailyTemperatures(temperatures):
    n = len(temperatures)
    ans = [0] * n  # Initialize result array with zeros
    stk = []       # Initialize an empty stack
 
    for i in range(n - 1, -1, -1):  # Iterate from right to left
        while stk and temperatures[stk[-1]] <= temperatures[i]: # While stack not empty and top element <= current element
            stk.pop()  # Pop the smaller temperature indices
        if stk:  # If stack is not empty after popping
            ans[i] = stk[-1] - i  # Waiting period = index difference
        stk.append(i)  # Push the current index onto stack
    return ans
 

Java:

import java.util.Deque;
import java.util.ArrayDeque;
 
class Solution {
    public int[] dailyTemperatures(int[] temperatures) {
        int n = temperatures.length;
        int[] ans = new int[n];
        Deque<Integer> stk = new ArrayDeque<>();
 
        for (int i = n - 1; i >= 0; i--) { // Iterate from right to left
            while (!stk.isEmpty() && temperatures[stk.peek()] <= temperatures[i]) { // Stack processing
                stk.pop();
            }
            if (!stk.isEmpty()) {
                ans[i] = stk.peek() - i;
            }
            stk.push(i); // Push index to stack
        }
        return ans;
    }
}

(Similar code structures would be used for C++, JavaScript, Go, TypeScript, and Rust, following the same algorithmic principles).

This detailed explanation and diverse code examples should provide a comprehensive understanding of solving the Daily Temperatures problem using the efficient monotonic stack approach. Remember that the key to this solution lies in the right-to-left traversal and the maintenance of a decreasing monotonic stack.