From 8d4b7fdd6f9b3a88020fe1b7d0efe0f813fcf4da Mon Sep 17 00:00:00 2001 From: Michael Xu Date: Sun, 2 Jul 2023 19:34:23 +0800 Subject: [PATCH 1/4] feat: Add rust implementation for NO.2132 --- .../2132.Stamping the Grid/README.md | 66 +++++++++++++++++++ .../2132.Stamping the Grid/README_EN.md | 66 +++++++++++++++++++ .../2132.Stamping the Grid/Solution.rs | 61 +++++++++++++++++ 3 files changed, 193 insertions(+) create mode 100644 solution/2100-2199/2132.Stamping the Grid/Solution.rs diff --git a/solution/2100-2199/2132.Stamping the Grid/README.md b/solution/2100-2199/2132.Stamping the Grid/README.md index fb4299cd9f7c6..5b859dd03a8ba 100644 --- a/solution/2100-2199/2132.Stamping the Grid/README.md +++ b/solution/2100-2199/2132.Stamping the Grid/README.md @@ -183,6 +183,72 @@ public: }; ``` +### **Rust** + +```rust +impl Solution { + pub fn possible_to_stamp(grid: Vec>, stamp_height: i32, stamp_width: i32) -> bool { + let n: usize = grid.len(); + let m: usize = grid[0].len(); + + let mut prefix_vec: Vec> = vec![vec![0; m + 1]; n + 1]; + + // Initialize the prefix vector + for i in 0..n { + for j in 0..m { + prefix_vec[i + 1][j + 1] = prefix_vec[i][j + 1] + prefix_vec[i + 1][j] - prefix_vec[i][j] + grid[i][j]; + } + } + + let mut diff_vec: Vec> = vec![vec![0; m + 1]; n + 1]; + + // Construct the difference vector based on prefix sum vector + for i in 0..n { + for j in 0..m { + // If the value of the cell is one, just bypass this + if grid[i][j] == 1 { + continue; + } + // Otherwise, try stick the stamp + let x: usize = i + stamp_height as usize; + let y: usize = j + stamp_width as usize; + // Check the bound + if x <= n && y <= m { + // If the region can be sticked (All cells are empty, which means the sum will be zero) + if prefix_vec[x][y] - prefix_vec[x][j] - prefix_vec[i][y] + prefix_vec[i][j] == 0 { + // Update the difference vector + diff_vec[i][j] += 1; + diff_vec[x][y] += 1; + + diff_vec[x][j] -= 1; + diff_vec[i][y] -= 1; + } + } + } + } + + let mut check_vec: Vec> = vec![vec![0; m + 1]; n + 1]; + + // Check the prefix sum of difference vector, to determine if there is any empty cell left + for i in 0..n { + for j in 0..m { + // If the value of the cell is one, just bypass this + if grid[i][j] == 1 { + continue; + } + // Otherwise, check if the region is empty, by calculating the prefix sum of difference vector + check_vec[i + 1][j + 1] = check_vec[i][j + 1] + check_vec[i + 1][j] - check_vec[i][j] + diff_vec[i][j]; + if check_vec[i + 1][j + 1] == 0 { + return false; + } + } + } + + true + } +} +``` + ### **Go** ```go diff --git a/solution/2100-2199/2132.Stamping the Grid/README_EN.md b/solution/2100-2199/2132.Stamping the Grid/README_EN.md index 7127bd99b28d8..cfbb0a8cbdfad 100644 --- a/solution/2100-2199/2132.Stamping the Grid/README_EN.md +++ b/solution/2100-2199/2132.Stamping the Grid/README_EN.md @@ -163,6 +163,72 @@ public: }; ``` +### **Rust** + +```rust +impl Solution { + pub fn possible_to_stamp(grid: Vec>, stamp_height: i32, stamp_width: i32) -> bool { + let n: usize = grid.len(); + let m: usize = grid[0].len(); + + let mut prefix_vec: Vec> = vec![vec![0; m + 1]; n + 1]; + + // Initialize the prefix vector + for i in 0..n { + for j in 0..m { + prefix_vec[i + 1][j + 1] = prefix_vec[i][j + 1] + prefix_vec[i + 1][j] - prefix_vec[i][j] + grid[i][j]; + } + } + + let mut diff_vec: Vec> = vec![vec![0; m + 1]; n + 1]; + + // Construct the difference vector based on prefix sum vector + for i in 0..n { + for j in 0..m { + // If the value of the cell is one, just bypass this + if grid[i][j] == 1 { + continue; + } + // Otherwise, try stick the stamp + let x: usize = i + stamp_height as usize; + let y: usize = j + stamp_width as usize; + // Check the bound + if x <= n && y <= m { + // If the region can be sticked (All cells are empty, which means the sum will be zero) + if prefix_vec[x][y] - prefix_vec[x][j] - prefix_vec[i][y] + prefix_vec[i][j] == 0 { + // Update the difference vector + diff_vec[i][j] += 1; + diff_vec[x][y] += 1; + + diff_vec[x][j] -= 1; + diff_vec[i][y] -= 1; + } + } + } + } + + let mut check_vec: Vec> = vec![vec![0; m + 1]; n + 1]; + + // Check the prefix sum of difference vector, to determine if there is any empty cell left + for i in 0..n { + for j in 0..m { + // If the value of the cell is one, just bypass this + if grid[i][j] == 1 { + continue; + } + // Otherwise, check if the region is empty, by calculating the prefix sum of difference vector + check_vec[i + 1][j + 1] = check_vec[i][j + 1] + check_vec[i + 1][j] - check_vec[i][j] + diff_vec[i][j]; + if check_vec[i + 1][j + 1] == 0 { + return false; + } + } + } + + true + } +} +``` + ### **Go** ```go diff --git a/solution/2100-2199/2132.Stamping the Grid/Solution.rs b/solution/2100-2199/2132.Stamping the Grid/Solution.rs new file mode 100644 index 0000000000000..5677a3207ae15 --- /dev/null +++ b/solution/2100-2199/2132.Stamping the Grid/Solution.rs @@ -0,0 +1,61 @@ +impl Solution { + pub fn possible_to_stamp(grid: Vec>, stamp_height: i32, stamp_width: i32) -> bool { + let n: usize = grid.len(); + let m: usize = grid[0].len(); + + let mut prefix_vec: Vec> = vec![vec![0; m + 1]; n + 1]; + + // Initialize the prefix vector + for i in 0..n { + for j in 0..m { + prefix_vec[i + 1][j + 1] = prefix_vec[i][j + 1] + prefix_vec[i + 1][j] - prefix_vec[i][j] + grid[i][j]; + } + } + + let mut diff_vec: Vec> = vec![vec![0; m + 1]; n + 1]; + + // Construct the difference vector based on prefix sum vector + for i in 0..n { + for j in 0..m { + // If the value of the cell is one, just bypass this + if grid[i][j] == 1 { + continue; + } + // Otherwise, try stick the stamp + let x: usize = i + stamp_height as usize; + let y: usize = j + stamp_width as usize; + // Check the bound + if x <= n && y <= m { + // If the region can be sticked (All cells are empty, which means the sum will be zero) + if prefix_vec[x][y] - prefix_vec[x][j] - prefix_vec[i][y] + prefix_vec[i][j] == 0 { + // Update the difference vector + diff_vec[i][j] += 1; + diff_vec[x][y] += 1; + + diff_vec[x][j] -= 1; + diff_vec[i][y] -= 1; + } + } + } + } + + let mut check_vec: Vec> = vec![vec![0; m + 1]; n + 1]; + + // Check the prefix sum of difference vector, to determine if there is any empty cell left + for i in 0..n { + for j in 0..m { + // If the value of the cell is one, just bypass this + if grid[i][j] == 1 { + continue; + } + // Otherwise, check if the region is empty, by calculating the prefix sum of difference vector + check_vec[i + 1][j + 1] = check_vec[i][j + 1] + check_vec[i + 1][j] - check_vec[i][j] + diff_vec[i][j]; + if check_vec[i + 1][j + 1] == 0 { + return false; + } + } + } + + true + } +} \ No newline at end of file From 9e9dce1a67f23832abd092fdb35589bf96275779 Mon Sep 17 00:00:00 2001 From: Michael Xu Date: Sun, 2 Jul 2023 19:46:01 +0800 Subject: [PATCH 2/4] feat: Improve implementation logic for NO.2132 --- solution/2100-2199/2132.Stamping the Grid/README.md | 10 ++++++++++ 1 file changed, 10 insertions(+) diff --git a/solution/2100-2199/2132.Stamping the Grid/README.md b/solution/2100-2199/2132.Stamping the Grid/README.md index 5b859dd03a8ba..5165bd6055f5f 100644 --- a/solution/2100-2199/2132.Stamping the Grid/README.md +++ b/solution/2100-2199/2132.Stamping the Grid/README.md @@ -60,6 +60,16 @@ **方法一：二维前缀和 + 二维差分** +根据题给的约束，很容易推出，一个格子能贴邮票的条件为，在加上邮票的长和宽后，右下角不越界的情况下，当前子区域中所有格子的和为0。 + +那么显然我们可以维护一个二维的前缀和数组，在 `O(1)` 的时间复杂度下就可以判断每次遍历到的格子是否能贴邮票。 + +而因为贴邮票的操作，可以概括为将当前子区域的所有格子的值都置为 `1`，很自然的就能想到用一个二维的差分数组来维护贴邮票后的状态。 + +最后只要对该差分数组再求一次二维前缀和，只要当前格子的和为 `0`，就意味着存在没有覆盖完全的情况，直接返回 `false` 即可。 + +需要注意的是二维数组的下标关系，具体参考如下。 + `s[i + 1][j + 1]` 表示第 i 行第 j 列左上部分所有元素之和，其中 i, j 下标从 0 开始。 则 `s[i + 1][j + 1] = s[i + 1][j] + s[i][j + 1] - s[i][j] + nums[i][j]`。 From e96c163149928941a6dfbec43cbd0d21b73d0ac0 Mon Sep 17 00:00:00 2001 From: Michael Xu Date: Sun, 2 Jul 2023 19:49:06 +0800 Subject: [PATCH 3/4] feat: Improve implementation logic for NO.2132 (EN version) --- .../2132.Stamping the Grid/README_EN.md | 18 ++++++++++++++++++ 1 file changed, 18 insertions(+) diff --git a/solution/2100-2199/2132.Stamping the Grid/README_EN.md b/solution/2100-2199/2132.Stamping the Grid/README_EN.md index cfbb0a8cbdfad..bd13f4bf1551b 100644 --- a/solution/2100-2199/2132.Stamping the Grid/README_EN.md +++ b/solution/2100-2199/2132.Stamping the Grid/README_EN.md @@ -50,6 +50,24 @@ ## Solutions +Based on the constraints provided by the question, it's easy to infer that a cell can accommodate a stamp if, after adding the length and width of the stamp, the bottom-right corner does not exceed the boundary, and the sum of all cells in the current sub-area is zero. + +Apparently, we can maintain a two-dimensional prefix sum array, and in `O(1)` time complexity, we can judge whether each cell traversed can accommodate a stamp. + +Since the action of affixing a stamp can be generalized to setting the values of all cells in the current sub-area to `1`, it's natural to think of maintaining the state after stamp affixing using a two-dimensional difference array. + +Finally, just calculate the two-dimensional prefix sum for this difference array again. + +If the sum of the current cell is `0`, it means there are cases that have not been completely covered, and you can directly return `false`. + +It's worth noting the subscript relationship of the two-dimensional array, which is as follows. + +`s[i + 1][j + 1]` represents the sum of all elements in the upper left part of the i-th row and j-th column, where the subscript i, j starts from 0. + +So `s[i + 1][j + 1] = s[i + 1][j] + s[i][j + 1] - s[i][j] + nums[i][j]`. + +For a sub-matrix with (x1, y1) as the upper left corner and (x2, y2) as the bottom right corner, the sum `sub = s[x2 + 1][y2 + 1] - s[x2 + 1][y1] - s[x1][y2 + 1] + s[x1][y1]`. + ### **Python3** From 26d0992b52ef6db1e49d54fd917b301f313d1376 Mon Sep 17 00:00:00 2001 From: Yang Libin Date: Sun, 2 Jul 2023 20:03:06 +0800 Subject: [PATCH 4/4] Update README.md --- solution/2100-2199/2132.Stamping the Grid/README.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/solution/2100-2199/2132.Stamping the Grid/README.md b/solution/2100-2199/2132.Stamping the Grid/README.md index 5165bd6055f5f..60c4e4c7774ed 100644 --- a/solution/2100-2199/2132.Stamping the Grid/README.md +++ b/solution/2100-2199/2132.Stamping the Grid/README.md @@ -60,7 +60,7 @@ **方法一：二维前缀和 + 二维差分** -根据题给的约束，很容易推出，一个格子能贴邮票的条件为，在加上邮票的长和宽后，右下角不越界的情况下，当前子区域中所有格子的和为0。 +根据题给的约束，很容易推出，一个格子能贴邮票的条件为，在加上邮票的长和宽后，右下角不越界的情况下，当前子区域中所有格子的和为 0。 那么显然我们可以维护一个二维的前缀和数组，在 `O(1)` 的时间复杂度下就可以判断每次遍历到的格子是否能贴邮票。 @@ -208,7 +208,7 @@ impl Solution { for j in 0..m { prefix_vec[i + 1][j + 1] = prefix_vec[i][j + 1] + prefix_vec[i + 1][j] - prefix_vec[i][j] + grid[i][j]; } - } + } let mut diff_vec: Vec> = vec![vec![0; m + 1]; n + 1];