LeetCode #739 — MEDIUM

Daily Temperatures

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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]

Constraints:

1 <= temperatures.length <= 10⁵
30 <= temperatures[i] <= 100

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Stack

Example 1

[73,74,75,71,69,72,76,73]

Example 2

[30,40,50,60]

Example 3

[30,60,90]

Core Insight

What unlocks the optimal approach

If the temperature is say, 70 today, then in the future a warmer temperature must be either 71, 72, 73, ..., 99, or 100. We could remember when all of them occur next.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #739: Daily Temperatures
class Solution {
    public int[] dailyTemperatures(int[] temperatures) {
        int n = temperatures.length;
        Deque<Integer> stk = new ArrayDeque<>();
        int[] ans = new int[n];
        for (int i = n - 1; i >= 0; --i) {
            while (!stk.isEmpty() && temperatures[stk.peek()] <= temperatures[i]) {
                stk.pop();
            }
            if (!stk.isEmpty()) {
                ans[i] = stk.peek() - i;
            }
            stk.push(i);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #739: Daily Temperatures
func dailyTemperatures(temperatures []int) []int {
	n := len(temperatures)
	ans := make([]int, n)
	stk := []int{}
	for i := n - 1; i >= 0; i-- {
		for len(stk) > 0 && temperatures[stk[len(stk)-1]] <= temperatures[i] {
			stk = stk[:len(stk)-1]
		}
		if len(stk) > 0 {
			ans[i] = stk[len(stk)-1] - i
		}
		stk = append(stk, i)
	}
	return ans
}

# Accepted solution for LeetCode #739: Daily Temperatures
class Solution:
    def dailyTemperatures(self, temperatures: List[int]) -> List[int]:
        stk = []
        n = len(temperatures)
        ans = [0] * n
        for i in range(n - 1, -1, -1):
            while stk and temperatures[stk[-1]] <= temperatures[i]:
                stk.pop()
            if stk:
                ans[i] = stk[-1] - i
            stk.append(i)
        return ans

// Accepted solution for LeetCode #739: Daily Temperatures
impl Solution {
    pub fn daily_temperatures(temperatures: Vec<i32>) -> Vec<i32> {
        let n = temperatures.len();
        let mut stk: Vec<usize> = Vec::new();
        let mut ans = vec![0; n];

        for i in (0..n).rev() {
            while let Some(&top) = stk.last() {
                if temperatures[top] <= temperatures[i] {
                    stk.pop();
                } else {
                    break;
                }
            }
            if let Some(&top) = stk.last() {
                ans[i] = (top - i) as i32;
            }
            stk.push(i);
        }

        ans
    }
}

// Accepted solution for LeetCode #739: Daily Temperatures
function dailyTemperatures(temperatures: number[]): number[] {
    const n = temperatures.length;
    const ans: number[] = Array(n).fill(0);
    const stk: number[] = [];
    for (let i = n - 1; ~i; --i) {
        while (stk.length && temperatures[stk.at(-1)!] <= temperatures[i]) {
            stk.pop();
        }
        if (stk.length) {
            ans[i] = stk.at(-1)! - i;
        }
        stk.push(i);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.