Introduction

In this blog post, we will explore how to solve the LeetCode problem "739" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an array of integers temperatures represents the daily temperatures, return an array answer such that answer[i] is the number of days you have to wait after the ith day to get a warmer temperature. If there is no future day for which this is possible, keep answer[i] == 0 instead. Example 1: Input: temperatures = [73,74,75,71,69,72,76,73] Output: [1,1,4,2,1,1,0,0] Example 2: Input: temperatures = [30,40,50,60] Output: [1,1,1,0] Example 3: Input: temperatures = [30,60,90] Output: [1,1,0] Constraints: 1 <= temperatures.length <= 105 30 <= temperatures[i] <= 100

Explanation

Here's an efficient solution to the daily temperatures problem:

• Monotonic Stack: Utilize a stack to keep track of indices of days with decreasing temperatures. This allows us to efficiently find the next warmer day for each day.
• Iterate and Compare: Iterate through the temperatures array. For each day, compare its temperature with the temperature of the day at the top of the stack. If the current day is warmer, pop the stack until the stack is empty or the current day isn't warmer than the top of the stack.

• Calculate Difference: When popping the stack, calculate the difference between the current day's index and the popped index, storing it in the answer array.

• Runtime Complexity: O(n), Storage Complexity: O(n)

Code

    def dailyTemperatures(temperatures):
    n = len(temperatures)
    answer = [0] * n
    stack = []  # Stack to store indices

    for i in range(n):
        while stack and temperatures[i] > temperatures[stack[-1]]:
            prev_index = stack.pop()
            answer[prev_index] = i - prev_index
        stack.append(i)

    return answer