Solutions

Author: trinhminhtriet

solutions/1200-1299/1277.count-square-submatrices-with-all-ones/Solution.cs

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+public class Solution {
+    public int CountSquares(int[][] matrix) {
+        int m = matrix.Length;
+        int n = matrix[0].Length;
+        int[,] f = new int[m, n];
+        int ans = 0;
+
+        for (int i = 0; i < m; i++) {
+            for (int j = 0; j < n; j++) {
+                if (matrix[i][j] == 0) {
+                    continue;
+                }
+                if (i == 0 || j == 0) {
+                    f[i, j] = 1;
+                } else {
+                    f[i, j] = Math.Min(f[i - 1, j - 1], Math.Min(f[i - 1, j], f[i, j - 1])) + 1;
+                }
+                ans += f[i, j];
+            }
+        }
+
+        return ans;
+    }
+}

solutions/1200-1299/1277.count-square-submatrices-with-all-ones/Solution.rs

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+impl Solution {
+    pub fn count_squares(matrix: Vec<Vec<i32>>) -> i32 {
+        let m = matrix.len();
+        let n = matrix[0].len();
+        let mut f = vec![vec![0; n]; m];
+        let mut ans = 0;
+
+        for i in 0..m {
+            for j in 0..n {
+                if matrix[i][j] == 0 {
+                    continue;
+                }
+                if i == 0 || j == 0 {
+                    f[i][j] = 1;
+                } else {
+                    f[i][j] =
+                        std::cmp::min(f[i - 1][j - 1], std::cmp::min(f[i - 1][j], f[i][j - 1])) + 1;
+                }
+                ans += f[i][j];
+            }
+        }
+
+        ans
+    }
+}

solutions/1500-1599/1504.count-submatrices-with-all-ones/Solution.js

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+/**
+ * @param {number[][]} mat
+ * @return {number}
+ */
+var numSubmat = function (mat) {
+    const m = mat.length;
+    const n = mat[0].length;
+    const g = Array.from({ length: m }, () => Array(n).fill(0));
+
+    for (let i = 0; i < m; i++) {
+        for (let j = 0; j < n; j++) {
+            if (mat[i][j]) {
+                g[i][j] = j === 0 ? 1 : 1 + g[i][j - 1];
+            }
+        }
+    }
+
+    let ans = 0;
+    for (let i = 0; i < m; i++) {
+        for (let j = 0; j < n; j++) {
+            let col = Infinity;
+            for (let k = i; k >= 0; k--) {
+                col = Math.min(col, g[k][j]);
+                ans += col;
+            }
+        }
+    }
+
+    return ans;
+};

solutions/1500-1599/1504.count-submatrices-with-all-ones/Solution.rs

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+impl Solution {
+    pub fn num_submat(mat: Vec<Vec<i32>>) -> i32 {
+        let m = mat.len();
+        let n = mat[0].len();
+        let mut g = vec![vec![0; n]; m];
+
+        for i in 0..m {
+            for j in 0..n {
+                if mat[i][j] == 1 {
+                    if j == 0 {
+                        g[i][j] = 1;
+                    } else {
+                        g[i][j] = 1 + g[i][j - 1];
+                    }
+                }
+            }
+        }
+
+        let mut ans = 0;
+        for i in 0..m {
+            for j in 0..n {
+                let mut col = i32::MAX;
+                let mut k = i as i32;
+                while k >= 0 && col > 0 {
+                    col = col.min(g[k as usize][j]);
+                    ans += col;
+                    k -= 1;
+                }
+            }
+        }
+        ans
+    }
+}

solutions/3600-3699/3656.determine-if-a-simple-graph-exists/README.md

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+---
+comments: true
+difficulty: Medium
+---
+
+<!-- problem:start -->
+
+# [3656. Determine if a Simple Graph Exists 🔒](https://leetcode.com/problems/determine-if-a-simple-graph-exists)
+
+[README](/solution/3600-3699/3656.Determine%20if%20a%20Simple%20Graph%20Exists/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>You are given an integer array <code>degrees</code>, where <code>degrees[i]</code> represents the desired degree of the <code>i<sup>th</sup></code> vertex.</p>
+
+<p>Your task is to determine if there exists an <strong>undirected simple</strong> graph with <strong>exactly</strong> these vertex degrees.</p>
+
+<p>A <strong>simple</strong> graph has no self-loops or parallel edges between the same pair of vertices.</p>
+
+<p>Return <code>true</code> if such a graph exists, otherwise return <code>false</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">degrees = [3,1,2,2]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">true</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p><img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3600-3699/3656.Determine%20if%20a%20Simple%20Graph%20Exists/images/screenshot-2025-08-13-at-24347-am.png" style="width: 200px; height: 132px;" />​​​​​​​</p>
+
+<p>One possible undirected simple graph is:</p>
+
+<ul>
+	<li>Edges: <code>(0, 1), (0, 2), (0, 3), (2, 3)</code></li>
+	<li>Degrees: <code>deg(0) = 3</code>, <code>deg(1) = 1</code>, <code>deg(2) = 2</code>, <code>deg(3) = 2</code>.</li>
+</ul>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">degrees = [1,3,3,1]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">false</span></p>
+
+<p><strong>Explanation:</strong>​​​​​​​</p>
+
+<ul>
+	<li><code>degrees[1] = 3</code> and <code>degrees[2] = 3</code> means they must be connected to all other vertices.</li>
+	<li>This requires <code>degrees[0]</code> and <code>degrees[3]</code> to be at least 2, but both are equal to 1, which contradicts the requirement.</li>
+	<li>Thus, the answer is <code>false</code>.</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n == degrees.length &lt;= 10<sup>​​​​​​​5</sup></code></li>
+	<li><code>0 &lt;= degrees[i] &lt;= n - 1</code></li>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+
+```
+
+#### Java
+
+```java
+
+```
+
+#### C++
+
+```cpp
+
+```
+
+#### Go
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->