Discover how a simple stack can solve a tricky water-trapping puzzle in just one pass!
Why Trapping Rain Water Using Stack in DSA Javascript?
Imagine you have a row of buildings with different heights after a rainstorm. You want to find out how much water is trapped between these buildings. Doing this by looking at each building and guessing the water trapped is like trying to count drops of water in a puddle by hand--slow and confusing.
Manually checking each building and comparing it with others to find trapped water means repeating many checks and calculations. This is slow and easy to make mistakes, especially when buildings have varying heights and gaps.
Using a stack helps keep track of buildings that can trap water efficiently. The stack remembers the positions of buildings that might form boundaries for water. This way, you only pass through the buildings once, calculating trapped water as you go, making the process fast and less error-prone.
let totalWater = 0; for (let i = 0; i < heights.length; i++) { for (let j = i + 1; j < heights.length; j++) { // complex checks and calculations } }
const stack = []; let totalWater = 0; for (let i = 0; i < heights.length; i++) { while (stack.length && heights[i] > heights[stack[stack.length - 1]]) { // calculate trapped water } stack.push(i); }
This method lets you quickly and accurately find how much water is trapped between buildings in just one pass.
City planners can use this to estimate water accumulation between buildings after rain, helping design better drainage systems.
Manual checking is slow and error-prone.
Stack keeps track of building boundaries efficiently.
Calculates trapped water in one pass through the buildings.
Practice
Trapping Rain Water Using Stack algorithm?Solution
Step 1: Understand the role of stack in the algorithm
The stack stores indices of bars that are waiting to find a right boundary to trap water.Step 2: Clarify what the stack does not do
The stack does not store water amounts or sort bars; it helps find boundaries for trapped water.Final Answer:
To keep track of bars that are not yet processed for water trapping -> Option BQuick Check:
Stack tracks bars for trapping water [OK]
- Thinking stack stores water volume directly
- Assuming stack sorts the bars
- Believing stack reverses the array
Solution
Step 1: Identify the correct data structure for stack in JavaScript
JavaScript uses arrays as stacks by pushing and popping elements.Step 2: Check each option
const stack = []; initializes an empty array, which works as a stack. const stack = {}; is an object, not a stack. const stack = null; is null, no stack. const stack = new Stack(); assumes a Stack class which is not built-in.Final Answer:
const stack = []; -> Option AQuick Check:
Empty array for stack [OK]
- Using object {} instead of array
- Assuming Stack class exists by default
- Initializing stack as null
const height = [0,1,0,2,1,0,1,3,2,1,2,1];
function trap(height) {
let stack = [];
let water = 0;
for (let i = 0; i < height.length; i++) {
while (stack.length && height[i] > height[stack[stack.length - 1]]) {
let top = stack.pop();
if (!stack.length) break;
let distance = i - stack[stack.length - 1] - 1;
let boundedHeight = Math.min(height[i], height[stack[stack.length - 1]]) - height[top];
water += distance * boundedHeight;
}
stack.push(i);
}
return water;
}
console.log(trap(height));Solution
Step 1: Understand the input and algorithm flow
The input array represents bar heights. The stack-based algorithm calculates trapped water by finding boundaries and distances.Step 2: Trace the code logic mentally
The code correctly calculates trapped water for the given heights, which is a classic example resulting in 6 units of water trapped.Final Answer:
6 -> Option AQuick Check:
Classic trapping rain water example = 6 [OK]
- Miscounting trapped water units
- Ignoring stack empty check
- Wrong distance or bounded height calculation
function trap(height) {
let stack = [];
let water = 0;
for (let i = 0; i < height.length; i++) {
while (stack.length && height[i] >= height[stack[stack.length - 1]]) {
let top = stack.pop();
if (!stack.length) break;
let distance = i - stack[stack.length - 1] - 1;
let boundedHeight = Math.min(height[i], height[stack[stack.length - 1]]) - height[top];
water += distance * boundedHeight;
}
stack.push(i);
}
return water;
}Solution
Step 1: Analyze the while loop condition
The condition uses >= which causes incorrect popping when heights are equal, leading to wrong trapped water calculation.Step 2: Correct the condition
Changing >= to > ensures only strictly higher bars trigger popping, preserving correct boundaries.Final Answer:
The comparison should be >, not >= in the while condition -> Option CQuick Check:
Use > in while condition to avoid wrong pops [OK]
- Using >= causes incorrect water calculation
- Initializing stack as object instead of array
- Resetting water inside loop wrongly
[4, 2, 0, 3, 2, 5], what is the total trapped rain water calculated using the stack method?Solution
Step 1: Visualize the bars and trapped water
Bars heights: 4,2,0,3,2,5. Water trapped between bars is calculated by stack method considering boundaries and distances.Step 2: Calculate trapped water step-by-step
Water at index 1: 2, index 2: 4, index 3: 1, index 4: 2, total 9 units.Final Answer:
9 -> Option DQuick Check:
Sum trapped water units = 9 [OK]
- Miscounting trapped water units
- Ignoring zero height bars
- Wrong boundary identification
Ready to prove it?
Pass the test and this pattern moves unknown
