How do you solve Daily Temperatures on LeetCode?

Daily Temperatures (LeetCode #739) can be solved using Brute Force. The optimal approach is Brute Force with O(n²) time complexity and O(1) space complexity. Key insight: The brute force approach involves checking each day's temperature against all future days to find the next warmer day. This is straightforward but inefficient for larger arrays.

What algorithm or data structure is used to solve Daily Temperatures?

Daily Temperatures uses the following concepts: Array, Stack, Monotonic Stack. The recommended patterns to study are: Stack, Array.

What are the interview tips for Daily Temperatures?

Explain your thought process clearly Consider edge cases early Practice stack-based problems to build intuition

What are common mistakes when solving Daily Temperatures?

Not using a stack when applicable Overlooking edge cases like the last element having no warmer day

Daily Temperatures — LeetCode #739 (Medium)

Tags: Array, Stack, Monotonic Stack

Related patterns: Stack, Array

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach involves checking each day's temperature against all future days to find the next warmer day. This is straightforward but inefficient for larger arrays.

The time complexity is O(n²) because for each day, we may potentially check all subsequent days. The space complexity is O(1) since we are using only a fixed amount of extra space.

Step 1: Initialize an answer array of the same length as temperatures with all zeros.
Step 2: For each day, iterate through the subsequent days to find the first day with a warmer temperature.
Step 3: If a warmer temperature is found, calculate the difference in days and store it in the answer array.

For temperatures = [73,74,75,71,69,72,76,73]: 1. i=0 (73): Check j=1 (74) -> answer[0]=1 2. i=1 (74): Check j=2 (75) -> answer[1]=1 3. i=2 (75): Check j=3 (71), j=4 (69), j=5 (72), j=6 (76) -> answer[2]=4 4. i=3 (71): Check j=4 (69), j=5 (72), j=6 (76) -> answer[3]=2 5. i=4 (69): Check j=5 (72) -> answer[4]=1 6. i=5 (72): Check j=6 (76) -> answer[5]=1 Final answer: [1,1,4,2,1,1,0,0]

Common Mistakes

Not using a stack when applicable
Overlooking edge cases like the last element having no warmer day

Interview Tips

Explain your thought process clearly
Consider edge cases early
Practice stack-based problems to build intuition

#739

Daily Temperatures

Medium

Array↗Stack↗Monotonic Stack↗Stack↗Array↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n²)Space O(1)

The brute force approach involves checking each day's temperature against all future days to find the next warmer day. This is straightforward but inefficient for larger arrays.

⚙️

Algorithm

3 steps

1Step 1: Initialize an answer array of the same length as temperatures with all zeros.
2Step 2: For each day, iterate through the subsequent days to find the first day with a warmer temperature.
3Step 3: If a warmer temperature is found, calculate the difference in days and store it in the answer array.

solution.py9 lines

1def dailyTemperatures(temperatures):
2    n = len(temperatures)
3    answer = [0] * n
4    for i in range(n):
5        for j in range(i + 1, n):
6            if temperatures[j] > temperatures[i]:
7                answer[i] = j - i
8                break
9    return answer

ℹ

Complexity note: The time complexity is O(n²) because for each day, we may potentially check all subsequent days. The space complexity is O(1) since we are using only a fixed amount of extra space.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.