On an infinite number line (x-axis), we drop given squares in the order they are given.
The i-th square dropped (positions[i] = (left, side_length)) is a square with the left-most point being positions[i][0] and sidelength positions[i][1].
The square is dropped with the bottom edge parallel to the number line, and from a higher height than all currently landed squares. We wait for each square to stick before dropping the next.
The squares are infinitely sticky on their bottom edge, and will remain fixed to any positive length surface they touch (either the number line or another square). Squares dropped adjacent to each other will not stick together prematurely.
Return a list ans of heights. Each height ans[i] represents the current highest height of any square we have dropped, after dropping squares represented by positions[0], positions[1], ..., positions[i].
Example 1:
Input: [[1, 2], [2, 3], [6, 1]] Output: [2, 5, 5] Explanation:After the first drop of
positions[0] = [1, 2]: _aa _aa -------The maximum height of any square is 2.After the second drop of
positions[1] = [2, 3]: __aaa __aaa __aaa _aa__ _aa__ --------------The maximum height of any square is 5. The larger square stays on top of the smaller square despite where its center of gravity is, because squares are infinitely sticky on their bottom edge.After the third drop of
positions[1] = [6, 1]: __aaa __aaa __aaa _aa _aa___a --------------The maximum height of any square is still 5. Thus, we return an answer of[2, 5, 5].
Example 2:
Input: [[100, 100], [200, 100]] Output: [100, 100] Explanation: Adjacent squares don't get stuck prematurely - only their bottom edge can stick to surfaces.
Note:
1 <= positions.length <= 1000.1 <= positions[i][0] <= 10^8.1 <= positions[i][1] <= 10^6.Intuition
\nIntuitively, there are two operations: update, which updates our notion of the board (number line) after dropping a square; and query, which finds the largest height in the current board on some interval. We will work on implementing these operations.
Coordinate Compression
\nIn the below approaches, since there are only up to 2 * len(positions) critical points, namely the left and right edges of each square, we can use a technique called coordinate compression to map these critical points to adjacent integers, as shown in the code snippets below.
For brevity, these snippets are omitted from the remaining solutions.
\nPython
\ncoords = set()\nfor left, size in positions:\n coords.add(left)\n coords.add(left + size - 1)\nindex = {x: i for i, x in enumerate(sorted(coords))}\n
Java
\nSet<Integer> coords = new HashSet();\nfor (int[] pos: positions) {\n coords.add(pos[0]);\n coords.add(pos[0] + pos[1] - 1);\n}\nList<Integer> sortedCoords = new ArrayList(coords);\nCollections.sort(sortedCoords);\n\nMap<Integer, Integer> index = new HashMap();\nint t = 0;\nfor (int coord: sortedCoords) index.put(coord, t++);\n
Intuition
\nInstead of asking the question "what squares affect this query?", lets ask the question "what queries are affected by this square?"
\nAlgorithm
\nLet qans[i] be the maximum height of the interval specified by positions[i]. At the end, we\'ll return a running max of qans.
For each square positions[i], the maximum height will get higher by the size of the square we drop. Then, for any future squares that intersect the interval [left, right) (where left = positions[i][0], right = positions[i][0] + positions[i][1]), we\'ll update the maximum height of that interval.
Python
\nclass Solution(object):\n def fallingSquares(self, positions):\n qans = [0] * len(positions)\n for i, (left, size) in enumerate(positions):\n right = left + size\n qans[i] += size\n for j in xrange(i+1, len(positions)):\n left2, size2 = positions[j]\n right2 = left2 + size2\n if left2 < right and left < right2: #intersect\n qans[j] = max(qans[j], qans[i])\n\n ans = []\n for x in qans:\n ans.append(max(ans[-1], x) if ans else x)\n return ans\n
Java
\nclass Solution {\n public List<Integer> fallingSquares(int[][] positions) {\n int[] qans = new int[positions.length];\n for (int i = 0; i < positions.length; i++) {\n int left = positions[i][0];\n int size = positions[i][1];\n int right = left + size;\n qans[i] += size;\n\n for (int j = i+1; j < positions.length; j++) {\n int left2 = positions[j][0];\n int size2 = positions[j][1];\n int right2 = left2 + size2;\n if (left2 < right && left < right2) { //intersect\n qans[j] = Math.max(qans[j], qans[i]);\n }\n }\n }\n\n List<Integer> ans = new ArrayList();\n int cur = -1;\n for (int x: qans) {\n cur = Math.max(cur, x);\n ans.add(cur);\n }\n return ans;\n }\n}\n
Complexity Analysis
\nTime Complexity: , where is the length of positions. We use two for-loops, each of complexity .
Space Complexity: , the space used by qans and ans.
Intuition and Algorithm
\nLet N = len(positions). After mapping the board to a board of length at most , we can brute force the answer by simulating each square\'s drop directly.
Our answer is either the current answer or the height of the square that was just dropped, and we\'ll update it appropriately.
\nPython
\nclass Solution(object):\n def fallingSquares(self, positions):\n #Coordinate Compression\n #index = ...\n\n heights = [0] * len(index)\n def query(L, R):\n return max(heights[i] for i in xrange(L, R+1))\n\n def update(L, R, h):\n for i in xrange(L, R+1):\n heights[i] = max(heights[i], h)\n\n best = 0\n ans = []\n for left, size in positions:\n L = index[left]\n R = index[left + size - 1]\n h = query(L, R) + size\n update(L, R, h)\n best = max(best, h)\n ans.append(best)\n\n return ans\n
Java
\nclass Solution {\n int[] heights;\n\n public int query(int L, int R) {\n int ans = 0;\n for (int i = L; i <= R; i++) {\n ans = Math.max(ans, heights[i]);\n }\n return ans;\n }\n\n public void update(int L, int R, int h) {\n for (int i = L; i <= R; i++) {\n heights[i] = Math.max(heights[i], h);\n }\n }\n\n public List<Integer> fallingSquares(int[][] positions) {\n //Coordinate Compression\n //HashMap<Integer, Integer> index = ...;\n //int t = ...;\n\n heights = new int[t];\n int best = 0;\n List<Integer> ans = new ArrayList();\n\n for (int[] pos: positions) {\n int L = index.get(pos[0]);\n int R = index.get(pos[0] + pos[1] - 1);\n int h = query(L, R) + pos[1];\n update(L, R, h);\n best = Math.max(best, h);\n ans.add(best);\n }\n return ans;\n }\n}\n
Complexity Analysis
\nTime Complexity: , where is the length of positions. We use two for-loops, each of complexity (because of coordinate compression.)
Space Complexity: , the space used by heights.
Intuition
\nWhenever we perform operations (like update and query) on some interval in a domain, we could segment that domain with size into blocks of size .
Then, instead of a typical brute force where we update our array heights representing the board, we will also hold another array blocks, where blocks[i] represents the elements heights[B*i], heights[B*i + 1], ..., heights[B*i + B-1]. This allows us to write to the array in operations.
Algorithm
\nLet\'s get into the details. We actually need another array, blocks_read. When we update some element i in block b = i / B, we\'ll also update blocks_read[b]. If later we want to read the entire block, we can read from here (and stuff written to the whole block in blocks[b].)
When we write to a block, we\'ll write in blocks[b]. Later, when we want to read from an element i in block b = i / B, we\'ll read from heights[i] and blocks[b].
Our process for managing query and update will be similar. While left isn\'t a multiple of B, we\'ll proceed with a brute-force-like approach, and similarly for right. At the end, [left, right+1) will represent a series of contiguous blocks: the interval will have length which is a multiple of B, and left will also be a multiple of B.
Python
\nclass Solution(object):\n def fallingSquares(self, positions):\n #Coordinate compression\n #index = ...\n\n W = len(index)\n B = int(W**.5)\n heights = [0] * W\n blocks = [0] * (B+2)\n blocks_read = [0] * (B+2)\n\n def query(left, right):\n ans = 0\n while left % B and left <= right:\n ans = max(ans, heights[left], blocks[left / B])\n left += 1\n while right % B != B-1 and left <= right:\n ans = max(ans, heights[right], blocks[right / B])\n right -= 1\n while left <= right:\n ans = max(ans, blocks[left / B], blocks_read[left / B])\n left += B\n return ans\n\n def update(left, right, h):\n while left % B and left <= right:\n heights[left] = max(heights[left], h)\n blocks_read[left / B] = max(blocks_read[left / B], h)\n left += 1\n while right % B != B-1 and left <= right:\n heights[right] = max(heights[right], h)\n blocks_read[right / B] = max(blocks_read[right / B], h)\n right -= 1\n while left <= right:\n blocks[left / B] = max(blocks[left / B], h)\n left += B\n\n best = 0\n ans = []\n for left, size in positions:\n L = index[left]\n R = index[left + size - 1]\n h = query(L, R) + size\n update(L, R, h)\n best = max(best, h)\n ans.append(best)\n\n return ans\n
Java
\nclass Solution {\n int[] heights;\n int[] blocks;\n int[] blocks_read;\n int B;\n\n public int query(int left, int right) {\n int ans = 0;\n while (left % B > 0 && left <= right) {\n ans = Math.max(ans, heights[left]);\n ans = Math.max(ans, blocks[left / B]);\n left++;\n }\n while (right % B != B - 1 && left <= right) {\n ans = Math.max(ans, heights[right]);\n ans = Math.max(ans, blocks[right / B]);\n right--;\n }\n while (left <= right) {\n ans = Math.max(ans, blocks[left / B]);\n ans = Math.max(ans, blocks_read[left / B]);\n left += B;\n }\n return ans;\n }\n\n public void update(int left, int right, int h) {\n while (left % B > 0 && left <= right) {\n heights[left] = Math.max(heights[left], h);\n blocks_read[left / B] = Math.max(blocks_read[left / B], h);\n left++;\n }\n while (right % B != B - 1 && left <= right) {\n heights[right] = Math.max(heights[right], h);\n blocks_read[right / B] = Math.max(blocks_read[right / B], h);\n right--;\n }\n while (left <= right) {\n blocks[left / B] = Math.max(blocks[left / B], h);\n left += B;\n }\n }\n\n public List<Integer> fallingSquares(int[][] positions) {\n //Coordinate Compression\n //HashMap<Integer, Integer> index = ...;\n //int t = ...;\n\n heights = new int[t];\n B = (int) Math.sqrt(t);\n blocks = new int[B+2];\n blocks_read = new int[B+2];\n\n int best = 0;\n List<Integer> ans = new ArrayList();\n\n for (int[] pos: positions) {\n int L = index.get(pos[0]);\n int R = index.get(pos[0] + pos[1] - 1);\n int h = query(L, R) + pos[1];\n update(L, R, h);\n best = Math.max(best, h);\n ans.add(best);\n }\n return ans;\n }\n}\n
Complexity Analysis
\nTime Complexity: , where is the length of positions. Each query and update has complexity .
Space Complexity: , the space used by heights.
Intuition
\nIf we were familiar with the idea of a segment tree (which supports queries and updates on intervals), we can immediately crack the problem.
\nAlgorithm
\nSegment trees work by breaking intervals into a disjoint sum of component intervals, whose number is at most log(width). The motivation is that when we change an element, we only need to change log(width) many intervals that aggregate on an interval containing that element.
When we want to update an interval all at once, we need to use lazy propagation to ensure good run-time complexity. This topic is covered in more depth here.
\nWith such an implementation in hand, the problem falls out immediately.
\nPython
\nclass SegmentTree(object):\n def __init__(self, N, update_fn, query_fn):\n self.N = N\n self.H = 1\n while 1 << self.H < N:\n self.H += 1\n\n self.update_fn = update_fn\n self.query_fn = query_fn\n self.tree = [0] * (2 * N)\n self.lazy = [0] * N\n\n def _apply(self, x, val):\n self.tree[x] = self.update_fn(self.tree[x], val)\n if x < self.N:\n self.lazy[x] = self.update_fn(self.lazy[x], val)\n\n def _pull(self, x):\n while x > 1:\n x /= 2\n self.tree[x] = self.query_fn(self.tree[x*2], self.tree[x*2 + 1])\n self.tree[x] = self.update_fn(self.tree[x], self.lazy[x])\n\n def _push(self, x):\n for h in xrange(self.H, 0, -1):\n y = x >> h\n if self.lazy[y]:\n self._apply(y * 2, self.lazy[y])\n self._apply(y * 2+ 1, self.lazy[y])\n self.lazy[y] = 0\n\n def update(self, L, R, h):\n L += self.N\n R += self.N\n L0, R0 = L, R\n while L <= R:\n if L & 1:\n self._apply(L, h)\n L += 1\n if R & 1 == 0:\n self._apply(R, h)\n R -= 1\n L /= 2; R /= 2\n self._pull(L0)\n self._pull(R0)\n\n def query(self, L, R):\n L += self.N\n R += self.N\n self._push(L); self._push(R)\n ans = 0\n while L <= R:\n if L & 1:\n ans = self.query_fn(ans, self.tree[L])\n L += 1\n if R & 1 == 0:\n ans = self.query_fn(ans, self.tree[R])\n R -= 1\n L /= 2; R /= 2\n return ans\n\nclass Solution(object):\n def fallingSquares(self, positions):\n #Coordinate compression\n #index = ...\n\n tree = SegmentTree(len(index), max, max)\n best = 0\n ans = []\n for left, size in positions:\n L, R = index[left], index[left + size - 1]\n h = tree.query(L, R) + size\n tree.update(L, R, h)\n best = max(best, h)\n ans.append(best)\n\n return ans\n
Java
\nclass Solution {\n public List<Integer> fallingSquares(int[][] positions) {\n //Coordinate Compression\n //HashMap<Integer, Integer> index = ...;\n\n SegmentTree tree = new SegmentTree(sortedCoords.size());\n int best = 0;\n List<Integer> ans = new ArrayList();\n\n for (int[] pos: positions) {\n int L = index.get(pos[0]);\n int R = index.get(pos[0] + pos[1] - 1);\n int h = tree.query(L, R) + pos[1];\n tree.update(L, R, h);\n best = Math.max(best, h);\n ans.add(best);\n }\n return ans;\n }\n}\n\nclass SegmentTree {\n int N, H;\n int[] tree, lazy;\n\n SegmentTree(int N) {\n this.N = N;\n H = 1;\n while ((1 << H) < N) H++;\n tree = new int[2 * N];\n lazy = new int[N];\n }\n\n private void apply(int x, int val) {\n tree[x] = Math.max(tree[x], val);\n if (x < N) lazy[x] = Math.max(lazy[x], val);\n }\n\n private void pull(int x) {\n while (x > 1) {\n x >>= 1;\n tree[x] = Math.max(tree[x * 2], tree[x * 2 + 1]);\n tree[x] = Math.max(tree[x], lazy[x]);\n }\n }\n\n private void push(int x) {\n for (int h = H; h > 0; h--) {\n int y = x >> h;\n if (lazy[y] > 0) {\n apply(y * 2, lazy[y]);\n apply(y * 2 + 1, lazy[y]);\n lazy[y] = 0;\n }\n }\n }\n\n public void update(int L, int R, int h) {\n L += N; R += N;\n int L0 = L, R0 = R, ans = 0;\n while (L <= R) {\n if ((L & 1) == 1) apply(L++, h);\n if ((R & 1) == 0) apply(R--, h);\n L >>= 1; R >>= 1;\n }\n pull(L0); pull(R0);\n }\n\n public int query(int L, int R) {\n L += N; R += N;\n int ans = 0;\n push(L); push(R);\n while (L <= R) {\n if ((L & 1) == 1) ans = Math.max(ans, tree[L++]);\n if ((R & 1) == 0) ans = Math.max(ans, tree[R--]);\n L >>= 1; R >>= 1;\n }\n return ans;\n }\n}\n
Complexity Analysis
\nTime Complexity: , where is the length of positions. This is the run-time complexity of using a segment tree.
Space Complexity: , the space used by our tree.
\nAnalysis written by: @awice.
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