✨ (Solution): Problem 739. Daily Temperatures

solutions/739. Daily Temperatures/README.md

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+---
+comments: true
+difficulty: medium
+# Follow `Topics` tags
+tags:
+    - Array
+    - Stack
+    - Monotonic Stack
+---
+
+# [739. Daily Temperatures](https://leetcode.com/problems/daily-temperatures/description/)
+
+## Description
+
+You're given an array of integers `temperatures`, where each element represents the temperature of a specific day.
+Return a new array `answer` such that `answer[i]` indicates how many days you must wait after the i-th day to experience a warmer temperature.
+If there’s no future day with a warmer temperature, set `answer[i]` to `0`.
+
+**Example 1:**
+```
+Input: temperatures = [71,73,77,61,73,75,73]
+Output:  [1,1,3,1,1,0,0]
+```
+
+**Example 2:**
+```
+Input: temperatures = [10,20,30]
+Output: [1,1,0]
+```
+
+**Constraints:**
+
+* `1 <= n <= 8`
+
+## Solution
+
+We use a monotonic stack to track temperatures:
+1. Stack stores indices of temperatures in decreasing order
+2. When we find a warmer temperature, we pop indices from stack and calculate waiting days
+3. The waiting days are the difference between current index and popped index
+4. Any temperatures without a warmer day ahead remain in stack (result stays 0)
+
+```java
+class Solution {
+    public int[] dailyTemperatures(int[] temperatures) {
+        Stack<Integer> stack = new Stack<>();
+        int[] result = new int[temperatures.length];
+        for (int i = 0; i < temperatures.length; i++) {
+            while (!stack.isEmpty() && temperatures[i] > temperatures[stack.peek()]) {
+                int index = stack.pop();
+                result[index] = i - index;
+            }
+            stack.push(i);
+        }
+
+        return result;
+    }
+}
+```
+
+```python
+class Solution:
+    def dailyTemperatures(self, temperatures: list[int]) -> list[int]:
+        stack = []
+        res = [0] * len(temperatures)
+        for i, t in enumerate(temperatures):
+            while stack and t > temperatures[stack[-1]]:
+                idx = stack.pop()
+                res[idx] = i - idx
+            stack.append(i)
+        return res
+```
+
+## Complexity
+
+- Time complexity: $$O(n)$$
+<!-- Add time complexity here, e.g. $$O(n)$$ -->
+
+- Space complexity: $$O(n)$$
+<!-- Add space complexity here, e.g. $$O(n)$$ -->

solutions/739. Daily Temperatures/Solution.java

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+import java.util.*;
+
+class Solution {
+    public int[] dailyTemperatures(int[] temperatures) {
+        Stack<Integer> stack = new Stack<>();
+        int[] result = new int[temperatures.length];
+        for (int i = 0; i < temperatures.length; i++) {
+            while (!stack.isEmpty() && temperatures[i] > temperatures[stack.peek()]) {
+                int index = stack.pop();
+                result[index] = i - index;
+            }
+            stack.push(i);
+        }
+
+        return result;
+    }
+}

solutions/739. Daily Temperatures/Solution.py

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+class Solution:
+    def dailyTemperatures(self, temperatures: list[int]) -> list[int]:
+        stack = []
+        res = [0] * len(temperatures)
+        for i, t in enumerate(temperatures):
+            while stack and t > temperatures[stack[-1]]:
+                idx = stack.pop()
+                res[idx] = i - idx
+            stack.append(i)
+        return res