What is the time complexity of Daily Temperatures?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Daily Temperatures?

This problem uses Array, Stack, Monotonic Stack. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Daily Temperatures?

Insight 1: Monotonic Stack is the ideal data structure. The stack will store indices of days in decreasing temperature order. Insight 2: Iterating through the temperatures array, we can compare the current temperature with the temperature at the index at the top of the stack. If the current temperature is warmer, it means we have found the next warmer day for the temperature at the top of the stack.

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Daily Temperatures - LeetCode 739 Solution

Q: Which companies ask LeetCode 739. Daily Temperatures in interviews?

This problem is frequently asked at Accenture, Agoda, Airwallex, Anduril, Flipkart and 15 more companies.

Daily Temperatures - Complete Solution Guide

Daily Temperatures is LeetCode problem 739, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 i th 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] C

Detailed Explanation

The problem asks us to find, for each day in a given list of daily temperatures, how many days we need to wait until we encounter a warmer temperature. If there is no warmer temperature in the future, we should record 0 for that day. The input is an array of integers representing daily temperatures, and the output is an array of the same size, where each element represents the number of days to wait for a warmer temperature.

Solution Approach

The solution uses a monotonic stack to efficiently find the next warmer temperature for each day. We iterate through the temperatures array, maintaining a stack of indices in decreasing temperature order. If we encounter a warmer temperature, we pop the indices from the stack until the stack is empty or the temperature at the top of the stack is greater than or equal to the current temperature. For each popped index, we calculate the number of days to wait and store it in the answer array. Finally, any remaining indices in the stack represent days for which no warmer temperature was found, and their corresponding values in the answer array remain 0.

Step-by-Step Algorithm

Step 1: Initialize an array 'answer' of the same size as 'temperatures' with all elements set to 0. This will store the waiting days for each day.
Step 2: Initialize an empty stack 'stack' to store indices of the temperatures.
Step 3: Iterate through the 'temperatures' array using index 'i' and temperature 'temp'.
Step 4: While the 'stack' is not empty and the current temperature 'temp' is greater than the temperature at the index on top of the 'stack':
Step 5: Pop the index 'prev_index' from the 'stack'.
Step 6: Calculate the waiting days as 'i - prev_index' and store it in 'answer[prev_index]'.
Step 7: Push the current index 'i' onto the 'stack'.
Step 8: After iterating through all the temperatures, return the 'answer' array.

Key Insights

Insight 1: Monotonic Stack is the ideal data structure. The stack will store indices of days in decreasing temperature order.
Insight 2: Iterating through the temperatures array, we can compare the current temperature with the temperature at the index at the top of the stack. If the current temperature is warmer, it means we have found the next warmer day for the temperature at the top of the stack.
Insight 3: The answer array can be pre-initialized with zeros, simplifying the case where no warmer temperature is found.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Stack, Monotonic Stack.

Companies

Asked at: Accenture, Agoda, Airwallex, Anduril, Flipkart, Grab, Huawei, Intuit, J.P. Morgan, Netflix, Okta, Palantir Technologies, PhonePe, SAP, Salesforce, ServiceNow, Swiggy, Visa, Walmart Labs, josh technology.

Daily Temperatures | LeetCode 739

Medium

Array Stack Monotonic Stack

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.Stack;

class Solution {
    public int[] dailyTemperatures(int[] temperatures) {
        int n = temperatures.length;
        int[] answer = new int[n];
        Stack<Integer> stack = new Stack<>();

        for (int i = 0; i < n; i++) {
            int temp = temperatures[i];
            while (!stack.isEmpty() && temp > temperatures[stack.peek()]) {
                int prev_index = stack.pop();
                answer[prev_index] = i - prev_index;
            }
            stack.push(i);
        }

        return answer;
    }
}

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Problem Info

Description

Problem 739: Daily Temperatures - Detailed explanation and solution approaches.

Topics

Array Stack Monotonic Stack

Companies

AccentureAgodaAirwallexAndurilFlipkartGrabHuaweiIntuit

Stats

Difficulty:Medium

Languages:4

Companies:20

Problem #:739

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize an array 'answer' of the same size as 'temperatures' with all elements set to 0. This will store the waiting days for each day.
2
Step 2: Initialize an empty stack 'stack' to store indices of the temperatures.
3
Step 3: Iterate through the 'temperatures' array using index 'i' and temperature 'temp'.
4
Step 4: While the 'stack' is not empty and the current temperature 'temp' is greater than the temperature at the index on top of the 'stack':
5
Step 5: Pop the index 'prev_index' from the 'stack'.
6
Step 6: Calculate the waiting days as 'i - prev_index' and store it in 'answer[prev_index]'.
7
Step 7: Push the current index 'i' onto the 'stack'.
8
Step 8: After iterating through all the temperatures, return the 'answer' array.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because each element is visited and processed a constant number of times. The stack operations (push and pop) take constant time each. The space complexity is O(n) because, in the worst-case scenario (decreasing temperatures), the stack can hold all the indices of the input array.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

For each day, iterate through the remaining days to find the next warmer temperature. This approach involves nested loops.

Time: O(n^2), where n is the length of the 'temperatures' array.Space: O(1). Constant extra space is used.

Trade-offs: The brute force approach is simple to understand but inefficient for larger arrays. The monotonic stack approach is more efficient but requires a better understanding of data structures.

✨

Key Insights

1Insight 1: Monotonic Stack is the ideal data structure. The stack will store indices of days in decreasing temperature order.
2Insight 2: Iterating through the temperatures array, we can compare the current temperature with the temperature at the index at the top of the stack. If the current temperature is warmer, it means we have found the next warmer day for the temperature at the top of the stack.
3Insight 3: The answer array can be pre-initialized with zeros, simplifying the case where no warmer temperature is found.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

Edge Cases

Edge case 1: An array with decreasing temperatures, like [75, 74, 73, 72, 71]. In this case, the stack will contain all indices until the end, and the answer will be [0, 0, 0, 0, 0]. The algorithm correctly handles this because the answer array is initialized to 0.
Edge case 2: An array with increasing temperatures, like [71, 72, 73, 74, 75]. The algorithm handles this correctly. Each temperature will immediately pop previous ones and be added to the stack. The output will be [1, 1, 1, 1, 0].

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using a stack, which leads to less efficient (e.g., brute-force) solutions.

✗Common mistake 2: Popping elements from the stack in the wrong order or without checking if the stack is empty, leading to incorrect results or runtime errors.

✗Common mistake 3: Incorrectly calculating the waiting days (e.g., `prev_index - i` instead of `i - prev_index`).

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Stock price analysis. Finding the number of days to wait until a stock price exceeds a certain threshold is analogous to this problem.

🏦Real-world scenario 2: Weather forecasting. Predicting when the temperature will rise above a certain level based on historical data is a similar application.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 20, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What is a monotonic stack and when should I use it?

A monotonic stack maintains elements in sorted order (increasing or decreasing). Use it for "next greater element," "largest rectangle," or "stock span" problems. It achieves O(n) time by avoiding redundant comparisons through maintaining order invariants.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Daily Temperatures"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether stack techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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