grid47 Exploring patterns and algorithms
Aug 25, 2024
6 min read
Solution to LeetCode 739: Daily Temperatures Problem
Given an array of integers temperatures representing the daily temperatures, return an array where each element is the number of days you need to wait after that day to get a warmer temperature. If there is no future day for which this is possible, keep the answer as 0.
Problem
Approach
Steps
Complexity
Input: The input is an array of integers representing the daily temperatures.
Starts a while loop to check if the current temperature is greater than the previous stored temperature in the stack.
6 : Variable Assignment
pair<int, int> x = stk.top();
Pops the top element from the stack and assigns it to `x`.
7 : Stack Operation
stk.pop();
Pops the top element from the stack after processing it.
8 : Result Update
ans[x.second] = i-x.second;
Updates the result array to record the number of days until a warmer temperature.
9 : Stack Push
stk.push(make_pair(temp[i], i));
Pushes the current temperature and its index onto the stack for future comparisons.
10 : While Loop Setup
while(!stk.empty()) {
Starts a while loop to handle any remaining elements in the stack after processing all temperatures.
11 : Variable Assignment
auto x = stk.top();
Pops the top element of the stack into variable `x` to process any remaining days that didn't get a warmer temperature.
12 : Stack Operation
stk.pop();
Pops the top element from the stack.
13 : Result Update
ans[x.second] =0;
Sets the number of days to 0 for any temperatures that didn't have a warmer temperature.
14 : Return Statement
return ans;
Returns the result vector, which contains the number of days until a warmer temperature for each day.
Best Case: O(n), where n is the length of the temperatures array. This happens when the temperatures are strictly increasing.
Average Case: O(n), as we process each element in the temperatures array once.
Worst Case: O(n), where n is the length of the temperatures array. In the worst case, each element is processed once and added to the stack.
Description: The time complexity is O(n) because we only iterate through the array once, with each element being pushed and popped from the stack once.
Best Case: O(n), as the stack may store up to n elements in the worst case.
Worst Case: O(n), where n is the length of the temperatures array, due to the space required for the stack.
Description: The space complexity is O(n) because the stack can hold up to n elements in the worst case.
