Falling Squares

3 min readUpdated 27 May 2025

Problem Statement

The challenge presents a scenario where squares are sequentially dropped onto a 2D plane along the X-axis, each with specified starting positions and side lengths. The 2D array positions provides details for each square dropping operation, where positions[i] = [lefti, sideLengthi] defines the starting left-edge X-coordinate and the side length of the ith square. The simulation of the drop involves the square descending vertically until it lands either on another square (not by merely touching sides, but fully landing on top) or directly on the X-axis.

The task requires calculating the maximum height of the stacks formed by these squares after each square is dropped. The result should be an array ans, where each element ans[i] reflects the new height of the tallest stack after the ith square has been added. This problem combines spatial reasoning with incremental construction and querying of data structures to simulate and solve the sequence of drops effectively.

Examples

Example 1

Input:

EXAMPLE_ENTITY

positions = [[1,2],[2,3],[6,1]]

Output:

EXAMPLE_ENTITY

[2,5,5]

Explanation:

EXAMPLE_ENTITY

After the first drop, the tallest stack is square 1 with a height of 2.
After the second drop, the tallest stack is squares 1 and 2 with a height of 5.
After the third drop, the tallest stack is still squares 1 and 2 with a height of 5.
Thus, we return an answer of [2, 5, 5].

Example 2

Input:

EXAMPLE_ENTITY

positions = [[100,100],[200,100]]

Output:

EXAMPLE_ENTITY

[100,100]

Explanation:

EXAMPLE_ENTITY

After the first drop, the tallest stack is square 1 with a height of 100.
After the second drop, the tallest stack is either square 1 or square 2, both with heights of 100.
Thus, we return an answer of [100, 100].
Note that square 2 only brushes the right side of square 1, which does not count as landing on it.

Constraints

1 <= positions.length <= 1000
1 <= lefti <= 108
1 <= sideLengthi <= 106

Approach and Intuition

The simulation of this problem requires careful placement and tracking of squares. Here's a step-by-step breakdown of the intuitive approach to solve this problem:

Initialize Heights Tracking: Use a data structure, typically a map or dictionary, to maintain the heights at each relevant X-coordinate on the axis. This allows for efficient updates and queries.
Simulate Each Drop: For each square given in the positions array:
- Determine its landing position based on the current stack heights. Calculate the maximum height it will rest upon by checking the range of X-coordinates it would cover based on its left position and side length.
- Update this maximum height by adding the square's own side length to account for its contribution to the stack height.
Update Data Structures: After determining where a square will land and how high it will pile up, update the height tracking data structure for each relevant X-coordinate that the square covers.
Record Maximum Height: Post each drop, scan through the height tracking structure to find the maximum stack height and record this in the result array.
Result Compilation: After processing each drop, compile the results from each drop into the output array and return it.

This approach ensures an accurate simulation of each drop and proper tracking of resultant heights, adapting dynamically to each new square's landing and the evolving structure of the stack on the plane. This methodology supports the underlying need for correct, efficient height calculations and maximally utilizes the structural properties given in the constraints.

Solutions

JAVA
class BoxFalls {
    // Process squares after they fall in positions provided
    public List<Integer> calculateFalls(int[][] placements) {
        TreeMap<Integer, Integer> compressed = new TreeMap<>();
        SegmentTree segTree = new SegmentTree(compressed.size());
        int maxHeight = 0;
        List<Integer> results = new ArrayList();
    
        for (int[] placement : placements) {
            int start = compressed.get(placement[0]);
            int end = compressed.get(placement[0] + placement[1] - 1);
            int height = segTree.rangeQuery(start, end) + placement[1];
            segTree.rangeUpdate(start, end, height);
            maxHeight = Math.max(maxHeight, height);
            results.add(maxHeight);
        }
        return results;
    }
}
    
class SegmentTree {
    int size, height;
    int[] tree, refresh;
    
    public SegmentTree(int size) {
        this.size = size;
        height = 1;
        while ((1 << height) < size) {
            height++;
        }
        tree = new int[2 * size];
        refresh = new int[size];
    }
    
    private void applyUpdate(int x, int value) {
        tree[x] = Math.max(tree[x], value);
        if (x < size) {
            refresh[x] = Math.max(refresh[x], value);
        }
    }
    
    private void pullUp(int x) {
        while (x > 1) {
            x >>= 1;
            tree[x] = Math.max(tree[2 * x], tree[2 * x + 1]);
            tree[x] = Math.max(tree[x], refresh[x]);
        }
    }
    
    private void pushDown(int x) {
        for (int h = height; h > 0; h--) {
            int y = x >> h;
            if (refresh[y] > 0) {
                applyUpdate(2 * y, refresh[y]);
                applyUpdate(2 * y + 1, refresh[y]);
                refresh[y] = 0;
            }
        }
    }
    
    public void rangeUpdate(int left, int right, int value) {
        left += size;
        right += size;
        int start = left, end = right;
        while (left <= right) {
            if ((left & 1) == 1) {
                applyUpdate(left++, value);
            }
            if ((right & 1) == 0) {
                applyUpdate(right--, value);
            }
            left >>= 1;
            right >>= 1;
        }
        pullUp(start);
        pullUp(end);
    }
    
    public int rangeQuery(int left, int right) {
        left += size;
        right += size;
        int result = 0;
        pushDown(left);
        pushDown(right);
        while (left <= right) {
            if ((left & 1) == 1) {
                result = Math.max(result, tree[left++]);
            }
            if ((right & 1) == 0) {
                result = Math.max(result, tree[right--]);
            }
            left >>= 1;
            right >>= 1;
        }
        return result;
    }
}
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The provided Java code defines a solution for calculating the resulting peak heights of falling squares whenever each square is placed on a given position with certain dimensions on a plane. The main components of this solution involve using a SegmentTree class to handle range updates and queries efficiently, and a BoxFalls class managing the primary logic.

In the BoxFalls class:
- Method calculateFalls accepts a matrix of placements where each element contains coordinates that dictate where a square starts and its size.
- A TreeMap compressed is utilized to map the squares’ positions but stays uninstantiated in this sketch.
- As squares are placed, heights at each range impacted by a square are updated via a segment tree. This tree is initialized with a size corresponding to the uncompressed input scale.
- The method updates the segment tree to maintain and retrieve the maximum height for given segments efficiently and updates a list with the new peak height after each insertion.
In the SegmentTree class:
- The constructor initializes the segment tree and its accompanying data structures based on the size.
- applyUpdate applies a value update across a specific range.
- pushDown and pullUp methods help maintain segment tree invariants during updates, ensuring that internal nodes have correct values post-update.
- rangeQuery and rangeUpdate are public methods that handle querying and updating ranges of indices in the tree with efficient optimization techniques like lazy propagation.

This architecture allows managing and querying dynamic range maximums efficiently, essential for computing peak heights as new squares are added sequentially. This approach significantly reduces the complexity compared to naive methods, making it suitable for scenarios with multiple updates and queries.

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Falling Squares

3 min readUpdated 27 May 2025

Problem Statement

Examples

Example 1

Input:

EXAMPLE_ENTITY

positions = [[1,2],[2,3],[6,1]]

Output:

EXAMPLE_ENTITY

[2,5,5]

Explanation:

EXAMPLE_ENTITY

After the first drop, the tallest stack is square 1 with a height of 2.
After the second drop, the tallest stack is squares 1 and 2 with a height of 5.
After the third drop, the tallest stack is still squares 1 and 2 with a height of 5.
Thus, we return an answer of [2, 5, 5].

Example 2

Input:

EXAMPLE_ENTITY

positions = [[100,100],[200,100]]

Output:

EXAMPLE_ENTITY

[100,100]

Explanation:

EXAMPLE_ENTITY

After the first drop, the tallest stack is square 1 with a height of 100.
After the second drop, the tallest stack is either square 1 or square 2, both with heights of 100.
Thus, we return an answer of [100, 100].
Note that square 2 only brushes the right side of square 1, which does not count as landing on it.

Constraints

1 <= positions.length <= 1000
1 <= lefti <= 108
1 <= sideLengthi <= 106

Approach and Intuition

The simulation of this problem requires careful placement and tracking of squares. Here's a step-by-step breakdown of the intuitive approach to solve this problem:

Initialize Heights Tracking: Use a data structure, typically a map or dictionary, to maintain the heights at each relevant X-coordinate on the axis. This allows for efficient updates and queries.
Simulate Each Drop: For each square given in the positions array:
- Determine its landing position based on the current stack heights. Calculate the maximum height it will rest upon by checking the range of X-coordinates it would cover based on its left position and side length.
- Update this maximum height by adding the square's own side length to account for its contribution to the stack height.
Update Data Structures: After determining where a square will land and how high it will pile up, update the height tracking data structure for each relevant X-coordinate that the square covers.
Record Maximum Height: Post each drop, scan through the height tracking structure to find the maximum stack height and record this in the result array.
Result Compilation: After processing each drop, compile the results from each drop into the output array and return it.

Solutions

JAVA
class BoxFalls {
    // Process squares after they fall in positions provided
    public List<Integer> calculateFalls(int[][] placements) {
        TreeMap<Integer, Integer> compressed = new TreeMap<>();
        SegmentTree segTree = new SegmentTree(compressed.size());
        int maxHeight = 0;
        List<Integer> results = new ArrayList();
    
        for (int[] placement : placements) {
            int start = compressed.get(placement[0]);
            int end = compressed.get(placement[0] + placement[1] - 1);
            int height = segTree.rangeQuery(start, end) + placement[1];
            segTree.rangeUpdate(start, end, height);
            maxHeight = Math.max(maxHeight, height);
            results.add(maxHeight);
        }
        return results;
    }
}
    
class SegmentTree {
    int size, height;
    int[] tree, refresh;
    
    public SegmentTree(int size) {
        this.size = size;
        height = 1;
        while ((1 << height) < size) {
            height++;
        }
        tree = new int[2 * size];
        refresh = new int[size];
    }
    
    private void applyUpdate(int x, int value) {
        tree[x] = Math.max(tree[x], value);
        if (x < size) {
            refresh[x] = Math.max(refresh[x], value);
        }
    }
    
    private void pullUp(int x) {
        while (x > 1) {
            x >>= 1;
            tree[x] = Math.max(tree[2 * x], tree[2 * x + 1]);
            tree[x] = Math.max(tree[x], refresh[x]);
        }
    }
    
    private void pushDown(int x) {
        for (int h = height; h > 0; h--) {
            int y = x >> h;
            if (refresh[y] > 0) {
                applyUpdate(2 * y, refresh[y]);
                applyUpdate(2 * y + 1, refresh[y]);
                refresh[y] = 0;
            }
        }
    }
    
    public void rangeUpdate(int left, int right, int value) {
        left += size;
        right += size;
        int start = left, end = right;
        while (left <= right) {
            if ((left & 1) == 1) {
                applyUpdate(left++, value);
            }
            if ((right & 1) == 0) {
                applyUpdate(right--, value);
            }
            left >>= 1;
            right >>= 1;
        }
        pullUp(start);
        pullUp(end);
    }
    
    public int rangeQuery(int left, int right) {
        left += size;
        right += size;
        int result = 0;
        pushDown(left);
        pushDown(right);
        while (left <= right) {
            if ((left & 1) == 1) {
                result = Math.max(result, tree[left++]);
            }
            if ((right & 1) == 0) {
                result = Math.max(result, tree[right--]);
            }
            left >>= 1;
            right >>= 1;
        }
        return result;
    }
}
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In the BoxFalls class:
- Method calculateFalls accepts a matrix of placements where each element contains coordinates that dictate where a square starts and its size.
- A TreeMap compressed is utilized to map the squares’ positions but stays uninstantiated in this sketch.
- As squares are placed, heights at each range impacted by a square are updated via a segment tree. This tree is initialized with a size corresponding to the uncompressed input scale.
- The method updates the segment tree to maintain and retrieve the maximum height for given segments efficiently and updates a list with the new peak height after each insertion.
In the SegmentTree class:
- The constructor initializes the segment tree and its accompanying data structures based on the size.
- applyUpdate applies a value update across a specific range.
- pushDown and pullUp methods help maintain segment tree invariants during updates, ensuring that internal nodes have correct values post-update.
- rangeQuery and rangeUpdate are public methods that handle querying and updating ranges of indices in the tree with efficient optimization techniques like lazy propagation.

More to Explore

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Tech Talk: Implementing Fine-Grained Access Control with Vultr IAM

Tech Talk: Getting Started with Vultr IAM

Tech Talk: Simplifying Zero-Trust Networking with NetBird on Vultr Marketplace

0 Comments

Start the conversation

Be the first to comment and share your perspective with the community.