1. Index stack
The stack holds indices waiting for a warmer day.
const stack = []Typorithm · medium
Daily Temperatures
daily-temperaturesstacks-queuesmonotonic-stacknext-greaterindices
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Step 1 of 8 · intro card
Learn Daily Temperatures
Stacks & Queues
Keep a stack of indices with decreasing temperatures; a warmer day resolves all cooler ones below it.
Each day needs the distance to the next warmer day, a next-greater-element problem.
A monotonic stack of indices holds days still waiting for something warmer.
When today is warmer, it pops and resolves every cooler day, recording the index gap.
01
Stack holds waiting indices.
02
Warmer day resolves cooler ones.
03
Answer is the index gap.
function dailyTemperatures(temps) {
const result = new Array(temps.length).fill(0);
const stack = [];
for (let i = 0; i < temps.length; i += 1) {
while (stack.length > 0 && temps[i] > temps[stack[stack.length - 1]]) {
const prev = stack.pop();
result[prev] = i - prev;
}
stack.push(i);
}
return result;
}Read the implementation
Connect each part of the code to the concept before you type.
2. Warmer found
Today resolves every cooler day still on the stack.
while (stack.length > 0 && temps[i] > temps[stack[stack.length - 1]])3. Days waited
The gap in indices is the days until a warmer temperature.
result[prev] = i - prev4. Wait for warmer
Today's index waits until a warmer day appears.
stack.push(i)A complete learning loop
01
Understand
Read the concept map and follow the code breakdown.
02
Reconstruct
Type the implementation, repair mistakes and work through every recall exercise.
03
Verify
Use the final check without hints; revisit weak concepts in short practice sessions.