Rain Terraces / Trapping Rain Water
Trapping Rain Water asks how much water can sit above bars after rain. Water above index i is limited by the smaller of the tallest bar on its left and right.
Beginne mit diesem mentalen Modell
Trapping Rain Water asks how much water can sit above bars after rain. Water above index i is limited by the smaller of the tallest bar on its left and right.
So funktioniert es – ohne unnötigen Jargon
A two-pointer solution tracks leftMax and rightMax while moving the side with the smaller boundary.
This gives O(n) time and O(1) extra space.
Lies den Algorithmus als Folge von Entscheidungen
Restate constraints and edge cases.
Identify the pattern and invariant.
Walk through a small example before coding.
Read a tiny input by hand and write down the state after every important operation. For Rain Terraces / Trapping Rain Water, the goal is to make the invariant visible rather than to memorize code.
function trap(height) {
let l = 0, r = height.length - 1;
let leftMax = 0, rightMax = 0, water = 0;
while (l < r) {
if (height[l] <= height[r]) {
leftMax = Math.max(leftMax, height[l]);
water += leftMax - height[l++];
} else {
rightMax = Math.max(rightMax, height[r]);
water += rightMax - height[r--];
}
}
return water;
}Lerne die Syntax nicht auswendig. Folge dem Zustand, der sich in jeder Zeile ändert, und verbinde ihn mit der oben beschriebenen Invariante.
Varies by implementation / input constraints
Do not quote this complexity mechanically. Ask which operation or input dimension it describes, then derive the cost from the loops, recursion depth, data-structure operations and any preprocessing assumptions.
Gut geeignet
- Use the recognized pattern when its invariant matches the problem constraints.
Was typischerweise schiefgeht
- Jumping to code before defining the invariant.
- Optimizing before producing a correct baseline.
Struktur einer professionellen Antwort
Explain why moving the smaller boundary is safe: the smaller side already determines the maximum water level for that position.
What problem does Rain Terraces / Trapping Rain Water solve?
Trapping Rain Water asks how much water can sit above bars after rain. Water above index i is limited by the smaller of the tallest bar on its left and right.
Was sollte ich zur Komplexität erwähnen?
Its typical cost here is Varies by implementation / input constraints, but a complete answer should connect that cost to the representation, input constraints and the dominant operations.
Was ist der häufigste Fehler?
Jumping to code before defining the invariant.