Liquid Accumulation — Problem Statement & Solution Guide
Problem Description
Given a sequence of vertical bars of different heights, calculate the total amount of liquid that can accumulate between them. The liquid can accumulate between two bars if the height of the liquid does not exceed the minimum height of the two bars. The liquid accumulation is calculated as the minimum height of each pair of bars minus the height difference between the two bars if the height differs.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Liquid Accumulation"
WHY DOES IT MATTER?
This pattern is essential because it teaches the concept of 'bounded accumulation' and how to optimize repeated range maximum queries. It is a foundational problem for understanding how to use pointers to reduce time complexity from quadratic to linear, a skill critical for handling large-scale data processing tasks.
OPTIMIZATION CHALLENGE
The key insight is that you do not need to know the exact maximum height on both sides for every index. You only need to know the minimum of the two maxima. By moving the pointer on the side with the smaller current height, you guarantee that the maximum on that side is the limiting factor, allowing you to calculate the water without looking at the other side's future values.
REAL-WORLD CONNECTION
In distributed systems, this is analogous to calculating buffer overflow risks in network packets. The 'bars' represent the capacity of different network segments, and the 'water' represents the backlog of packets. Understanding how much backlog can accumulate before overflow helps in designing better load balancing and buffer management strategies.
In an interview, start with the brute force O(n^2) approach to show you understand the problem definition. Then, pivot to the two-pointer approach, explaining the invariant: 'The water level at the current position is determined by the smaller of the two global maxima.' This demonstrates both logical reasoning and optimization skills.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The 'Liquid Accumulation' problem, commonly known as the Trapping Rain Water problem, is a classic example of a two-pointer or monotonic stack application. The core theoretical insight is that the amount of water trapped at any specific index i is determined by the minimum of the maximum heights to its left and right, minus the height of the bar at i. Mathematically, water[i] = min(max_left[i], max_right[i]) - height[i]. This formulation relies on the principle that water cannot rise higher than the shortest boundary on either side; thus, the limiting factor is always the smaller of the two global maxima surrounding the current position.
Naive approaches, such as iterating through each bar and scanning left and right to find the maximums, result in O(n^2) time complexity. This is inefficient for large datasets where n can reach 10^5 or higher. The optimal paradigm shifts from repeated scanning to maintaining state. By using two pointers moving from the ends towards the center, we can track the maximum height seen so far from both sides. Since the water level at any point is constrained by the smaller of the two global maxima, we only need to update the pointer on the side with the smaller current height. This ensures that we always know the limiting boundary for the current position, allowing us to calculate the trapped water in a single pass.
Alternatively, a monotonic stack approach offers an O(n) time and O(n) space solution by processing bars in order of height. When a bar higher than the top of the stack is encountered, it forms a 'basin' with the previous bars. The water trapped in this basin is calculated based on the difference between the current bar, the previous bar (now popped), and the new bottom of the stack. This method is particularly useful when the problem is extended to 2D grids or when the input is streamed, as it handles local minima and maxima dynamically without pre-computing global arrays.
Interview Questions on This Problem
Q1At a fintech platform like Stripe, how would you adapt the trapping rain water algorithm to handle a stream of incoming data points where you cannot store the entire array in memory?
For a streaming scenario, a two-pointer approach is not feasible because it requires random access to the ends of the array. Instead, a monotonic stack approach is preferred, but even that requires O(n) space. If memory is strictly constrained, one might use a sliding window approximation or a probabilistic data structure like a Count-Min Sketch to estimate the maximum heights to the left and right, accepting a small error margin. However, in most practical fintech scenarios, O(n) space is acceptable, so the monotonic stack is the standard answer, emphasizing that we process elements as they arrive and only keep track of the relevant 'basin' boundaries.
Q2At a high-growth startup like Airbnb, if the 'bars' represent server load over time, how would you modify the algorithm to find the maximum contiguous period where the load is below a certain threshold, rather than just the total accumulation?
This transforms the problem from a summation problem to a range query problem. While the core logic of finding the 'basin' remains similar, the objective changes. You would still use a two-pointer or stack approach to identify regions where the load is bounded. However, instead of summing the water, you would track the length of the contiguous segment where height[i] < threshold and the boundaries are defined by the max heights. This requires maintaining a running count of the current valid segment and updating the global maximum whenever a segment ends. The key is to recognize that the 'water' logic identifies the valid region, but the metric changes from area to length.
Q3At a product company like Google, how would you extend this 1D problem to a 2D grid of heights to calculate the total water trapped in a 2D landscape?
The 2D extension is significantly more complex and typically solved using a Priority Queue (Min-Heap) combined with BFS/DFS. You start from the boundary cells of the grid, which cannot trap water, and push them into the min-heap. You then repeatedly pop the cell with the smallest height, which acts as the current 'water level' boundary. For each neighbor, if its height is less than the current water level, it can trap water equal to the difference. You update the neighbor's effective height to the water level and push it into the heap. This ensures that water flows from the lowest boundary inward, correctly accounting for the 2D topology. The time complexity becomes O(N log N) where N is the number of cells.
Examples
Input
[1, 8, 6, 2, 5, 4, 8, 3, 7]
Output
47
Explanation: Step-by-step: with input [1, 8, 6, 2, 5, 4, 8, 3, 7], we calculate the liquid accumulation between each pair of bars. We start by finding the maximum height of the left and right bars, then calculate the minimum height of each pair of bars minus the height difference between the two bars if the height differs. The total liquid accumulation is the sum of these calculations.
Input
[1, 2, 3, 4, 5]
Output
6
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], we calculate the liquid accumulation between each pair of bars. Since the height of the liquid does not exceed the minimum height of the two bars, the liquid accumulation is calculated as the minimum height of each pair of bars minus the height difference between the two bars if the height differs. The total liquid accumulation is the sum of these calculations.
Constraints
- 1 <= number of bars <= 10^5
- 1 <= height of each bar <= 10^4
Optimal Approach & Strategy
Use two pointers starting from the ends of the array, moving inward. Maintain the maximum height seen so far from both sides. Always move the pointer on the side with the smaller current height, as that side determines the water level. Calculate the trapped water for the current position and update the maximum height if necessary.
Brute Force Approach
Iterate through each bar, and for each bar, scan all bars to its left to find the maximum height and all bars to its right to find the maximum height. Calculate the water trapped at that bar as the minimum of these two maxima minus the current bar's height, and sum the results.
Verified Code Solutions
function solution(height) { let result = 0; for (let i = 0; i < height.length; i++) { for (let j = i + 1; j < height.length; j++) { let minHeight = Math.min(height[i], height[j]); let diff = Math.abs(height[i] - height[j]); result += minHeight - diff; } } return result; }class Solution { public: int solution(vector<int>& height) { int result = 0; for (int i = 0; i < height.size(); i++) { for (int j = i + 1; j < height.size(); j++) { int minHeight = min(height[i], height[j]); int diff = abs(height[i] - height[j]); result += minHeight - diff; } } return result; } }class Solution { public int solution(int[] height) { int result = 0; for (int i = 0; i < height.length; i++) { for (int j = i + 1; j < height.length; j++) { int minHeight = Math.min(height[i], height[j]); int diff = Math.abs(height[i] - height[j]); result += minHeight - diff; } } return result; } }def solution(height): result = 0; for i in range(len(height)): for j in range(i + 1, len(height)): min_height = min(height[i], height[j]); diff = abs(height[i] - height[j]); result += min_height - diff; return resultfunction solution(height) { let result = 0; for (let i = 0; i < height.length; i++) { for (let j = i + 1; j < height.length; j++) { let minHeight = Math.min(height[i], height[j]); let diff = Math.abs(height[i] - height[j]); result += minHeight - diff; } } return result; }Solve in Interative Editor
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