Trapping Rain Water

🎯

Trapping Rain Water

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Imagine a city skyline after a heavy rainstorm, where water collects between buildings of different heights. How much water is trapped between these buildings?

💡 This problem involves understanding how to calculate trapped water between bars of varying heights. Beginners often struggle because it requires thinking about constraints from both sides simultaneously, which is not straightforward with simple loops.

📋

Problem Statement

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining. Input: An array of integers height[] representing the elevation map. Output: An integer representing total units of trapped rain water.

1 ≤ n ≤ 10^50 ≤ height[i] ≤ 10^5

💡

Example

Input"[0,1,0,2,1,0,1,3,2,1,2,1]"

Output6

Water trapped at indices 2,4,5,6,9,10 sums to 6 units.

• All bars of same height → 0 trapped water
• Strictly increasing heights → 0 trapped water
• Strictly decreasing heights → 0 trapped water
• Single bar → 0 trapped water

⚠️

Common Mistakes

Not handling empty or single-element arrays

Runtime errors or incorrect trapped water calculation

✅ Add checks for array length before processing

Using min(left_max, right_max) without checking if it's greater than current height

Negative trapped water values added, leading to incorrect results

✅ Ensure trapped water is max(0, min(left_max, right_max) - height[i])

Incorrect pointer movement in two pointer approach

Infinite loops or wrong trapped water calculation

✅ Move pointer with smaller height inward each iteration

Not updating left_max or right_max correctly

Underestimation of trapped water

✅ Update max heights before calculating trapped water at each step

Using nested loops without optimization for large inputs

Time limit exceeded in interviews

✅ Explain brute force but implement optimized approach

🧠

Brute Force (Nested Loops)

💡 This approach is the most straightforward and helps build intuition by directly applying the problem definition, even though it is inefficient.

Intuition

For each bar, find the tallest bar on its left and right, then the water trapped is the minimum of these two heights minus the current bar's height.

Algorithm

1. For each bar, scan all bars to the left to find the maximum height.
2. Scan all bars to the right to find the maximum height.
3. Calculate trapped water at current bar as max(0, min(left_max, right_max) - current height).
4. Sum trapped water for all bars.

💡 The nested loops make it hard to see efficiency, but this directly follows the problem's logic.

</>

Code

def trap(height):
    n = len(height)
    water = 0
    for i in range(n):
        left_max = 0
        for j in range(i + 1):
            left_max = max(left_max, height[j])
        right_max = 0
        for j in range(i, n):
            right_max = max(right_max, height[j])
        water += max(0, min(left_max, right_max) - height[i])
    return water

# Driver code
if __name__ == '__main__':
    print(trap([0,1,0,2,1,0,1,3,2,1,2,1]))  # Output: 6

Line Notes

n = len(height)Determine the number of bars to iterate over.

for i in range(n)Iterate over each bar to calculate trapped water at that position.

for j in range(i + 1)Find the tallest bar to the left including current bar.

left_max = max(left_max, height[j])Update left_max to the highest bar found so far on the left.

for j in range(i, n)Find the tallest bar to the right including current bar.

right_max = max(right_max, height[j])Update right_max to the highest bar found so far on the right.

water += max(0, min(left_max, right_max) - height[i])Add trapped water at current bar if positive, avoid negative values.

return waterReturn total trapped water after processing all bars.

public class Solution {
    public static int trap(int[] height) {
        int n = height.length;
        int water = 0;
        for (int i = 0; i < n; i++) {
            int left_max = 0;
            for (int j = 0; j <= i; j++) {
                left_max = Math.max(left_max, height[j]);
            }
            int right_max = 0;
            for (int j = i; j < n; j++) {
                right_max = Math.max(right_max, height[j]);
            }
            water += Math.max(0, Math.min(left_max, right_max) - height[i]);
        }
        return water;
    }

    public static void main(String[] args) {
        int[] height = {0,1,0,2,1,0,1,3,2,1,2,1};
        System.out.println(trap(height)); // Output: 6
    }
}

Line Notes

int n = height.length;Get the total number of bars.

for (int i = 0; i < n; i++)Process each bar to compute trapped water.

for (int j = 0; j <= i; j++)Find max height to the left including current bar.

left_max = Math.max(left_max, height[j]);Update left_max with highest bar found so far.

for (int j = i; j < n; j++)Find max height to the right including current bar.

right_max = Math.max(right_max, height[j]);Update right_max with highest bar found so far.

water += Math.max(0, Math.min(left_max, right_max) - height[i]);Add trapped water at current bar if positive, avoid negative values.

return water;Return total trapped water after all bars processed.

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int trap(vector<int>& height) {
    int n = height.size();
    int water = 0;
    for (int i = 0; i < n; i++) {
        int left_max = 0;
        for (int j = 0; j <= i; j++) {
            left_max = max(left_max, height[j]);
        }
        int right_max = 0;
        for (int j = i; j < n; j++) {
            right_max = max(right_max, height[j]);
        }
        water += max(0, min(left_max, right_max) - height[i]);
    }
    return water;
}

int main() {
    vector<int> height = {0,1,0,2,1,0,1,3,2,1,2,1};
    cout << trap(height) << endl; // Output: 6
    return 0;
}

Line Notes

int n = height.size();Get number of bars in elevation map.

for (int i = 0; i < n; i++)Iterate over each bar to calculate trapped water.

for (int j = 0; j <= i; j++)Find max height to the left including current bar.

left_max = max(left_max, height[j]);Update left_max with highest bar found so far.

for (int j = i; j < n; j++)Find max height to the right including current bar.

right_max = max(right_max, height[j]);Update right_max with highest bar found so far.

water += max(0, min(left_max, right_max) - height[i]);Add trapped water at current bar if positive, avoid negative values.

return water;Return total trapped water after processing all bars.

function trap(height) {
    const n = height.length;
    let water = 0;
    for (let i = 0; i < n; i++) {
        let left_max = 0;
        for (let j = 0; j <= i; j++) {
            left_max = Math.max(left_max, height[j]);
        }
        let right_max = 0;
        for (let j = i; j < n; j++) {
            right_max = Math.max(right_max, height[j]);
        }
        water += Math.max(0, Math.min(left_max, right_max) - height[i]);
    }
    return water;
}

// Test
console.log(trap([0,1,0,2,1,0,1,3,2,1,2,1])); // Output: 6

Line Notes

const n = height.length;Determine number of bars in the input array.

for (let i = 0; i < n; i++)Loop through each bar to compute trapped water.

for (let j = 0; j <= i; j++)Find tallest bar to the left including current bar.

left_max = Math.max(left_max, height[j]);Update left_max with highest bar found so far.

for (let j = i; j < n; j++)Find tallest bar to the right including current bar.

right_max = Math.max(right_max, height[j]);Update right_max with highest bar found so far.

water += Math.max(0, Math.min(left_max, right_max) - height[i]);Add trapped water at current bar if positive, avoid negative values.

return water;Return total trapped water after processing all bars.

⏱

Complexity

TimeO(n^2)

SpaceO(1)

For each bar, we scan left and right bars to find max heights, resulting in nested loops and quadratic time.

💡 For n=20 bars, this means about 400 operations, which grows quickly and becomes impractical for large n.

Interview Verdict: TLE for large inputs; useful only to introduce problem and intuition.

This approach is too slow for large inputs but helps understand the problem deeply before optimizing.

🧠

Prefix and Suffix Max Arrays

💡 This approach precomputes maximum heights to the left and right for each bar to avoid repeated scanning, improving efficiency.

Intuition

If we know the tallest bar to the left and right of each bar, we can directly compute trapped water without nested loops.

Algorithm

1. Create an array left_max where left_max[i] is the tallest bar from start to i.
2. Create an array right_max where right_max[i] is the tallest bar from end to i.
3. For each bar, trapped water = max(0, min(left_max[i], right_max[i]) - height[i]).
4. Sum trapped water for all bars.

💡 Precomputing max arrays removes the need for nested loops, making the solution more efficient and easier to understand.

</>

Code

def trap(height):
    n = len(height)
    if n == 0:
        return 0
    left_max = [0] * n
    right_max = [0] * n
    left_max[0] = height[0]
    for i in range(1, n):
        left_max[i] = max(left_max[i-1], height[i])
    right_max[n-1] = height[n-1]
    for i in range(n-2, -1, -1):
        right_max[i] = max(right_max[i+1], height[i])
    water = 0
    for i in range(n):
        water += max(0, min(left_max[i], right_max[i]) - height[i])
    return water

# Driver code
if __name__ == '__main__':
    print(trap([0,1,0,2,1,0,1,3,2,1,2,1]))  # Output: 6

Line Notes

if n == 0:Handle empty input edge case to avoid errors.

left_max = [0] * nInitialize array to store max heights from left.

left_max[0] = height[0]Set first element as starting max from left.

for i in range(1, n)Fill left_max by comparing current height with previous max.

right_max = [0] * nInitialize array to store max heights from right.

right_max[n-1] = height[n-1]Set last element as starting max from right.

for i in range(n-2, -1, -1)Fill right_max by comparing current height with next max.

water += max(0, min(left_max[i], right_max[i]) - height[i])Calculate trapped water at each bar using precomputed maxes, avoid negative values.

public class Solution {
    public static int trap(int[] height) {
        int n = height.length;
        if (n == 0) return 0;
        int[] left_max = new int[n];
        int[] right_max = new int[n];
        left_max[0] = height[0];
        for (int i = 1; i < n; i++) {
            left_max[i] = Math.max(left_max[i-1], height[i]);
        }
        right_max[n-1] = height[n-1];
        for (int i = n-2; i >= 0; i--) {
            right_max[i] = Math.max(right_max[i+1], height[i]);
        }
        int water = 0;
        for (int i = 0; i < n; i++) {
            water += Math.max(0, Math.min(left_max[i], right_max[i]) - height[i]);
        }
        return water;
    }

    public static void main(String[] args) {
        int[] height = {0,1,0,2,1,0,1,3,2,1,2,1};
        System.out.println(trap(height)); // Output: 6
    }
}

Line Notes

if (n == 0) return 0;Handle empty input to avoid errors.

int[] left_max = new int[n];Create array to store max heights from left.

left_max[0] = height[0];Initialize first element for left max array.

for (int i = 1; i < n; i++)Fill left_max by comparing current height with previous max.

int[] right_max = new int[n];Create array to store max heights from right.

right_max[n-1] = height[n-1];Initialize last element for right max array.

for (int i = n-2; i >= 0; i--)Fill right_max by comparing current height with next max.

water += Math.max(0, Math.min(left_max[i], right_max[i]) - height[i]);Calculate trapped water at each bar using precomputed maxes, avoid negative values.

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int trap(vector<int>& height) {
    int n = height.size();
    if (n == 0) return 0;
    vector<int> left_max(n), right_max(n);
    left_max[0] = height[0];
    for (int i = 1; i < n; i++) {
        left_max[i] = max(left_max[i-1], height[i]);
    }
    right_max[n-1] = height[n-1];
    for (int i = n-2; i >= 0; i--) {
        right_max[i] = max(right_max[i+1], height[i]);
    }
    int water = 0;
    for (int i = 0; i < n; i++) {
        water += max(0, min(left_max[i], right_max[i]) - height[i]);
    }
    return water;
}

int main() {
    vector<int> height = {0,1,0,2,1,0,1,3,2,1,2,1};
    cout << trap(height) << endl; // Output: 6
    return 0;
}

Line Notes

if (n == 0) return 0;Handle empty input to avoid runtime errors.

vector<int> left_max(n), right_max(n);Create arrays to store max heights from left and right.

left_max[0] = height[0];Initialize first element of left max array.

for (int i = 1; i < n; i++)Fill left_max by comparing current height with previous max.

right_max[n-1] = height[n-1];Initialize last element of right max array.

for (int i = n-2; i >= 0; i--)Fill right_max by comparing current height with next max.

water += max(0, min(left_max[i], right_max[i]) - height[i]);Calculate trapped water at each bar using precomputed maxes, avoid negative values.

function trap(height) {
    const n = height.length;
    if (n === 0) return 0;
    const left_max = new Array(n).fill(0);
    const right_max = new Array(n).fill(0);
    left_max[0] = height[0];
    for (let i = 1; i < n; i++) {
        left_max[i] = Math.max(left_max[i-1], height[i]);
    }
    right_max[n-1] = height[n-1];
    for (let i = n-2; i >= 0; i--) {
        right_max[i] = Math.max(right_max[i+1], height[i]);
    }
    let water = 0;
    for (let i = 0; i < n; i++) {
        water += Math.max(0, Math.min(left_max[i], right_max[i]) - height[i]);
    }
    return water;
}

// Test
console.log(trap([0,1,0,2,1,0,1,3,2,1,2,1])); // Output: 6

Line Notes

if (n === 0) return 0;Handle empty input to prevent errors.

const left_max = new Array(n).fill(0);Initialize array to store max heights from left.

left_max[0] = height[0];Set first element as starting max from left.

for (let i = 1; i < n; i++)Fill left_max by comparing current height with previous max.

const right_max = new Array(n).fill(0);Initialize array to store max heights from right.

right_max[n-1] = height[n-1];Set last element as starting max from right.

for (let i = n-2; i >= 0; i--)Fill right_max by comparing current height with next max.

water += Math.max(0, Math.min(left_max[i], right_max[i]) - height[i]);Calculate trapped water at each bar using precomputed maxes, avoid negative values.

⏱

Complexity

TimeO(n)

SpaceO(n)

We traverse the array multiple times but each traversal is linear, and we use extra arrays of size n.

💡 For n=20 bars, this means about 60 operations plus extra memory for two arrays of size 20.

Interview Verdict: Accepted; efficient for large inputs but uses extra space.

This approach is a common interview solution that balances clarity and efficiency.

🧠

Two Pointer Approach (Optimal)

💡 This approach uses two pointers and constant extra space to achieve linear time, making it the most optimal and elegant solution.

Intuition

Water trapped depends on the smaller of the tallest bars on left and right. By moving pointers inward from both ends, we can calculate trapped water on the fly.

Algorithm

1. Initialize two pointers at start and end of array, and variables to track max heights on both sides.
2. While left pointer is less than right pointer, compare heights at pointers.
3. Move the pointer with smaller height inward, update max height on that side, and add trapped water if current bar is lower than max.
4. Sum trapped water during traversal and return total.

💡 This approach cleverly uses the insight that trapped water depends on the smaller boundary, allowing a single pass with constant space.

</>

Code

def trap(height):
    left, right = 0, len(height) - 1
    left_max, right_max = 0, 0
    water = 0
    while left < right:
        if height[left] < height[right]:
            if height[left] >= left_max:
                left_max = height[left]
            else:
                water += left_max - height[left]
            left += 1
        else:
            if height[right] >= right_max:
                right_max = height[right]
            else:
                water += right_max - height[right]
            right -= 1
    return water

# Driver code
if __name__ == '__main__':
    print(trap([0,1,0,2,1,0,1,3,2,1,2,1]))  # Output: 6

Line Notes

left, right = 0, len(height) - 1Initialize pointers at both ends of the array.

left_max, right_max = 0, 0Track max heights seen so far from left and right.

while left < right:Traverse until pointers meet.

if height[left] < height[right]:Decide which pointer to move based on smaller height.

if height[left] >= left_max:Update left_max if current bar is taller.

water += left_max - height[left]Add trapped water if current bar is lower than left_max.

left += 1Move left pointer inward.

else:Process right pointer similarly.

return waterReturn total trapped water after traversal.

public class Solution {
    public static int trap(int[] height) {
        int left = 0, right = height.length - 1;
        int left_max = 0, right_max = 0;
        int water = 0;
        while (left < right) {
            if (height[left] < height[right]) {
                if (height[left] >= left_max) {
                    left_max = height[left];
                } else {
                    water += left_max - height[left];
                }
                left++;
            } else {
                if (height[right] >= right_max) {
                    right_max = height[right];
                } else {
                    water += right_max - height[right];
                }
                right--;
            }
        }
        return water;
    }

    public static void main(String[] args) {
        int[] height = {0,1,0,2,1,0,1,3,2,1,2,1};
        System.out.println(trap(height)); // Output: 6
    }
}

Line Notes

int left = 0, right = height.length - 1;Initialize pointers at array ends.

int left_max = 0, right_max = 0;Track max heights from left and right.

while (left < right)Loop until pointers meet.

if (height[left] < height[right])Move pointer with smaller height inward.

if (height[left] >= left_max)Update left_max if current bar is taller.

water += left_max - height[left];Add trapped water if current bar is lower than left_max.

left++;Increment left pointer.

elseProcess right pointer similarly.

return water;Return total trapped water.

#include <iostream>
#include <vector>
using namespace std;

int trap(vector<int>& height) {
    int left = 0, right = height.size() - 1;
    int left_max = 0, right_max = 0;
    int water = 0;
    while (left < right) {
        if (height[left] < height[right]) {
            if (height[left] >= left_max) {
                left_max = height[left];
            } else {
                water += left_max - height[left];
            }
            left++;
        } else {
            if (height[right] >= right_max) {
                right_max = height[right];
            } else {
                water += right_max - height[right];
            }
            right--;
        }
    }
    return water;
}

int main() {
    vector<int> height = {0,1,0,2,1,0,1,3,2,1,2,1};
    cout << trap(height) << endl; // Output: 6
    return 0;
}

Line Notes

int left = 0, right = height.size() - 1;Initialize two pointers at array ends.

int left_max = 0, right_max = 0;Track max heights from left and right sides.

while (left < right)Loop until pointers meet.

if (height[left] < height[right])Move pointer with smaller height inward.

if (height[left] >= left_max)Update left_max if current bar is taller.

water += left_max - height[left];Add trapped water if current bar is lower than left_max.

left++;Increment left pointer.

elseProcess right pointer similarly.

return water;Return total trapped water after traversal.

function trap(height) {
    let left = 0, right = height.length - 1;
    let left_max = 0, right_max = 0;
    let water = 0;
    while (left < right) {
        if (height[left] < height[right]) {
            if (height[left] >= left_max) {
                left_max = height[left];
            } else {
                water += left_max - height[left];
            }
            left++;
        } else {
            if (height[right] >= right_max) {
                right_max = height[right];
            } else {
                water += right_max - height[right];
            }
            right--;
        }
    }
    return water;
}

// Test
console.log(trap([0,1,0,2,1,0,1,3,2,1,2,1])); // Output: 6

Line Notes

let left = 0, right = height.length - 1;Initialize two pointers at array ends.

let left_max = 0, right_max = 0;Track max heights from left and right.

while (left < right)Loop until pointers meet.

if (height[left] < height[right])Move pointer with smaller height inward.

if (height[left] >= left_max)Update left_max if current bar is taller.

water += left_max - height[left];Add trapped water if current bar is lower than left_max.

left++;Increment left pointer.

elseProcess right pointer similarly.

return water;Return total trapped water after traversal.

⏱

Complexity

TimeO(n)

SpaceO(1)

Single pass through the array with two pointers and constant extra space.

💡 For n=20 bars, this means about 20 operations and no extra memory, making it very efficient.

Interview Verdict: Accepted; optimal solution for time and space.

This is the best approach to implement in interviews due to its efficiency and elegance.

📊

All Approaches - One-Glance Tradeoffs

💡 The two pointer approach is the best to code in 95% of interviews due to its optimal time and space complexity.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force	O(n^2)	O(1)	No	N/A	Mention only - never code
2. Prefix and Suffix Max Arrays	O(n)	O(n)	No	N/A	Good for explaining optimization
3. Two Pointer Approach	O(n)	O(1)	No	N/A	Code this approach

💼

Interview Strategy

💡 Use this guide to understand the problem deeply, practice coding all approaches, and prepare to explain tradeoffs clearly in interviews.

How to Present

Clarify the problem and constraints with the interviewer.Start with the brute force approach to show understanding.Explain inefficiencies and propose prefix/suffix max arrays.Finally, present the two pointer optimal solution.Write clean code and test with examples.

Time Allocation

Clarify: 2min → Approach: 5min → Code: 10min → Test & Discuss: 3min. Total ~20min

What the Interviewer Tests

Ability to understand problem constraints, optimize naive solutions, and implement efficient code with clear explanations.

Common Follow-ups

• What if the width of bars is not 1? → Multiply trapped water by width.
• Can you solve it using a stack? → Yes, stack-based approach exists.
• How to handle very large inputs efficiently? → Use two pointer approach.
• What if input is streaming? → Two pointer approach is still best with some modifications.

💡 These follow-ups test deeper understanding and ability to adapt solutions to variations.

🔍

Pattern Recognition

When to Use

1) Problem involves trapping or accumulating values between bars. 2) Input is an array representing heights or elevations. 3) Need to find max constraints from both left and right sides. 4) Goal is to calculate total trapped or accumulated quantity.

Signature Phrases

'compute how much water it can trap''elevation map''width of each bar is 1'

NOT This Pattern When

Problems involving only prefix sums or simple sliding windows without two-sided constraints.

Similar Problems

Container With Most Water - focuses on max area between bars, related to two pointer technique.Largest Rectangle in Histogram - uses stack to find max area, related to height constraints.Rain Water Trapped II - extension to 2D grid, similar concept of trapping water.

Practice

1. Consider the following Python code implementing the bidirectional BFS with state compression to find the shortest path visiting all nodes. Given the input graph = [[1,2,3],[0],[0],[0]], what is the final returned shortest path length?

easy

A. 3

B. 4

C. 2

D. 1

Solution

1. Step 1: Initialize queues and visited states

   For n=4 nodes, all_visited=15 (binary 1111). Each node starts in q1 with mask having only that node visited, and in q2 with mask having all nodes except that node visited.

2. Step 2: Trace BFS expansions

   From q1, starting at each node, BFS explores neighbors updating visited masks. The shortest path visiting all nodes in this star graph is 3 edges (e.g., 1->0->2->3).

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Star graph shortest path length visiting all nodes is 3 edges [OK]

Hint: Star graph shortest path visiting all nodes equals 3 edges [OK]

Common Mistakes:

• Confusing number of nodes with number of edges in path
• Off-by-one errors in BFS distance counting
• Misinterpreting bitmask updates

2. You are given a list of buildings represented as triplets (left, right, height). The goal is to output the skyline formed by these buildings as a list of key points where the height changes. Which algorithmic approach guarantees an optimal solution with time complexity O(n log n) for this problem?

easy

A. Divide and conquer approach that recursively merges skylines of building subsets

B. Greedy approach that iteratively picks the tallest building at each x-coordinate

C. Dynamic programming approach that stores maximum heights for subproblems

D. Brute force simulation by mapping heights on a large array and scanning

Solution

1. Step 1: Understand problem constraints

   The problem requires merging multiple building outlines efficiently, which suggests a divide and conquer strategy to combine partial skylines.

2. Step 2: Compare approaches

   Greedy and brute force approaches are either incorrect or inefficient (O(n * maxRight)). Dynamic programming does not naturally fit this problem. Divide and conquer achieves O(n log n) by recursively merging skylines.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Divide and conquer merges partial results efficiently [OK]

Hint: Divide and conquer merges partial skylines efficiently [OK]

Common Mistakes:

• Believing greedy approach works optimally
• Thinking DP applies naturally here
• Assuming brute force is efficient enough

3. What is the worst-case time complexity of the optimal backtracking solution for the Expression Add Operators problem, given a numeric string of length n?

medium

A. O(n^3) due to nested loops and substring parsing

B. O(2^n) because each digit can either be concatenated or separated by an operator

C. O(4^n) because each digit can be followed by 4 choices: +, -, *, or no operator (concatenate)

D. O(3^n) since only three operators are inserted between digits

Solution

1. Step 1: Identify branching factor per digit

   At each digit (except the first), there are 4 choices: add '+', '-', '*', or concatenate (no operator).

2. Step 2: Calculate total combinations

   With n digits, the number of expressions is roughly 4^(n-1), which is O(4^n) in worst case.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   4 choices per digit lead to exponential 4^n growth [OK]

Hint: Count operator choices plus concatenation per digit [OK]

Common Mistakes:

• Confusing 3 operators only, ignoring concatenation
• Assuming polynomial complexity

4. Suppose you want to compute the minimum edit distance between two strings but now you are allowed to reuse characters from the second string any number of times (i.e., unlimited insertions of any character from word2). Which modification to the DP approach correctly handles this variant?

hard

A. Modify DP to allow insertions without advancing j, but keep other transitions unchanged

B. Use a greedy approach to match characters from word1 to any in word2 without DP

C. Change DP to allow insertions that advance j, but allow deletions to skip characters in word2

D. Use the same DP but allow insert operation to not advance j index

Solution

1. Step 1: Understand reuse means insertions can be repeated without advancing j

   Normally, insertion advances j by 1; here, we can insert the same character multiple times, so insertion does not advance j.

2. Step 2: Modify DP transitions accordingly

   Allow insert operation to keep j fixed while increasing i, other transitions remain the same.

3. Step 3: Confirm other options fail

   Use the same DP but allow insert operation to not advance j index changes DP but does not specify keeping other transitions unchanged; Change DP to allow insertions that advance j, but allow deletions to skip characters in word2 incorrectly allows deletions to skip word2 chars; Use a greedy approach to match characters from word1 to any in word2 without DP is greedy and incorrect.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Allowing insertion without advancing j models reuse correctly [OK]

Hint: Reuse means insertion does not advance j [OK]

Common Mistakes:

• Advancing j on insertion breaks reuse
• Using greedy approach fails correctness

5. Suppose the problem is modified so that each element in nums can be used multiple times for XOR queries (i.e., unlimited reuse), and queries remain the same. Which of the following changes to the optimized approach is correct to handle this variant efficiently?

hard

A. No change needed; the existing trie approach already supports unlimited reuse

B. Use a dynamic programming approach over XOR states instead of a trie

C. Modify the trie to store counts of inserted elements and adjust maxXor to consider reuse

D. Sort queries and nums but do not insert elements multiple times; reuse is implicit

Solution

1. Step 1: Understand reuse impact

   Unlimited reuse means elements can be XORed multiple times, changing the problem semantics.

2. Step 2: Adjust trie to track counts

   Trie nodes must store counts of inserted elements to allow multiple traversals representing reuse.

3. Step 3: Modify maxXor accordingly

   maxXor must consider paths with repeated elements, enabled by counts, to find max XOR with reuse.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Trie with counts supports reuse; naive trie does not [OK]

Hint: Reuse requires tracking counts in trie nodes [OK]

Common Mistakes:

• Assuming no change needed
• Switching to DP unnecessarily
• Ignoring reuse in insertion logic