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]
1 <= temperatures.length <= 10530 <= temperatures[i] <= 100
class Solution:
def dailyTemperatures(self, temperatures: List[int]) -> List[int]:
# Initialize the result array with zeros
result = [0] * len(temperatures)
# Stack to keep track of the indices of temperatures
stack = []
for i, temp in enumerate(temperatures):
# Pop indices from the stack as long as the current temperature is warmer
# than the temperatures at those indices
while stack and temperatures[stack[-1]] < temp:
prev_index = stack.pop()
result[prev_index] = i - prev_index
# Push the current index onto the stack
stack.append(i)
return resultInitially thought of O(n^2) where you check each element with all remaining elements to get the solution, but using stacks we can reduce it to O(N) solution. ( both in time and space complexity)