diff --git a/solution/1200-1299/1277.Count Square Submatrices with All Ones/README.md b/solution/1200-1299/1277.Count Square Submatrices with All Ones/README.md index 03ef99ee958af..602ef655eca41 100644 --- a/solution/1200-1299/1277.Count Square Submatrices with All Ones/README.md +++ b/solution/1200-1299/1277.Count Square Submatrices with All Ones/README.md @@ -33,7 +33,7 @@ tags:   [0,1,1,1] ] 输出：15 -解释： +解释： 边长为 1 的正方形有 10 个。 边长为 2 的正方形有 4 个。 边长为 3 的正方形有 1 个。 @@ -42,7 +42,7 @@ tags:

示例 2：

-
输入：matrix = 
+
输入：matrix =
 [
   [1,0,1],
   [1,1,0],
@@ -50,7 +50,7 @@ tags:
 ]
 输出：7
 解释：
-边长为 1 的正方形有 6 个。 
+边长为 1 的正方形有 6 个。
 边长为 2 的正方形有 1 个。
 正方形的总数 = 6 + 1 = 7.
 

@@ -71,7 +71,20 @@ tags:
 
 
 
-### 方法一
+### 方法一：动态规划
+
+我们定义 $f[i][j]$ 为以 $(i,j)$ 为右下角的正方形子矩阵的边长，初始时 $f[i][j] = 0$，答案为 $\sum_{i,j} f[i][j]$。
+
+考虑 $f[i][j]$ 如何进行状态转移。
+
+-   当 $\text{matrix}[i][j] = 0$ 时，有 $f[i][j] = 0$。
+-   当 $\text{matrix}[i][j] = 1$ 时，状态 $f[i][j]$ 的值取决于其上、左、左上三个位置的值：
+    -   如果 $i = 0$ 或 $j = 0$，则 $f[i][j] = 1$。
+    -   否则 $f[i][j] = \min(f[i-1][j-1], f[i-1][j], f[i][j-1]) + 1$。
+
+答案为 $\sum_{i,j} f[i][j]$。
+
+时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别为矩阵的行数和列数。
 
 
 
@@ -133,11 +146,14 @@ public:
         vector> f(m, vector(n));
         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)
+                if (matrix[i][j] == 0) {
+                    continue;
+                }
+                if (i == 0 || j == 0) {
                     f[i][j] = 1;
-                else
+                } else {
                     f[i][j] = min(f[i - 1][j - 1], min(f[i - 1][j], f[i][j - 1])) + 1;
+                }
                 ans += f[i][j];
             }
         }
@@ -176,18 +192,22 @@ func countSquares(matrix [][]int) int {
 
 ```ts
 function countSquares(matrix: number[][]): number {
-    const [m, n] = [matrix.length, matrix[0].length];
-    const f = Array.from({ length: m }, () => Array(n));
-    const dfs = (i: number, j: number): number => {
-        if (i === m || j === n || !matrix[i][j]) return 0;
-        f[i][j] ??= 1 + Math.min(dfs(i + 1, j), dfs(i, j + 1), dfs(i + 1, j + 1));
-        return f[i][j];
-    };
+    const m = matrix.length;
+    const n = matrix[0].length;
+    const f: number[][] = Array.from({ length: m }, () => Array(n).fill(0));
     let ans = 0;
 
     for (let i = 0; i < m; i++) {
         for (let j = 0; j < n; j++) {
-            ans += dfs(i, 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];
         }
     }
 
@@ -195,26 +215,92 @@ function countSquares(matrix: number[][]): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn count_squares(matrix: Vec>) -> 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
+    }
+}
+```
+
 #### JavaScript
 
 ```js
-function countSquares(matrix) {
-    const [m, n] = [matrix.length, matrix[0].length];
-    const f = Array.from({ length: m }, () => Array(n));
-    const dfs = (i, j) => {
-        if (i === m || j === n || !matrix[i][j]) return 0;
-        f[i][j] ??= 1 + Math.min(dfs(i + 1, j), dfs(i, j + 1), dfs(i + 1, j + 1));
-        return f[i][j];
-    };
+/**
+ * @param {number[][]} matrix
+ * @return {number}
+ */
+var countSquares = function (matrix) {
+    const m = matrix.length;
+    const n = matrix[0].length;
+    const f = Array.from({ length: m }, () => Array(n).fill(0));
     let ans = 0;
 
     for (let i = 0; i < m; i++) {
         for (let j = 0; j < n; j++) {
-            ans += dfs(i, 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;
+};
+```
+
+#### C#
+
+```cs
+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;
+    }
 }
 ```
 
diff --git a/solution/1200-1299/1277.Count Square Submatrices with All Ones/README_EN.md b/solution/1200-1299/1277.Count Square Submatrices with All Ones/README_EN.md
index 2df7a529fba59..d21234d30f959 100644
--- a/solution/1200-1299/1277.Count Square Submatrices with All Ones/README_EN.md	
+++ b/solution/1200-1299/1277.Count Square Submatrices with All Ones/README_EN.md	
@@ -33,7 +33,7 @@ tags:
   [0,1,1,1]
 ]
 Output: 15
-Explanation: 
+Explanation:
 There are 10 squares of side 1.
 There are 4 squares of side 2.
 There is  1 square of side 3.
@@ -43,16 +43,16 @@ Total number of squares = 10 + 4 + 1 = 15.
 
Example 2:

 
 
-Input: matrix = 
+Input: matrix =
 [
   [1,0,1],
   [1,1,0],
   [1,1,0]
 ]
 Output: 7
-Explanation: 
-There are 6 squares of side 1.  
-There is 1 square of side 2. 
+Explanation:
+There are 6 squares of side 1.
+There is 1 square of side 2.
 Total number of squares = 6 + 1 = 7.
 

 
@@ -71,7 +71,20 @@ Total number of squares = 6 + 1 = 7.
 
 
 
-### Solution 1
+### Solution 1: Dynamic Programming
+
+We define $f[i][j]$ as the side length of the square submatrix with $(i,j)$ as the bottom-right corner. Initially $f[i][j] = 0$, and the answer is $\sum_{i,j} f[i][j]$.
+
+Consider how to perform state transition for $f[i][j]$.
+
+-   When $\text{matrix}[i][j] = 0$, we have $f[i][j] = 0$.
+-   When $\text{matrix}[i][j] = 1$, the value of state $f[i][j]$ depends on the values of the three positions above, left, and top-left:
+    -   If $i = 0$ or $j = 0$, then $f[i][j] = 1$.
+    -   Otherwise $f[i][j] = \min(f[i-1][j-1], f[i-1][j], f[i][j-1]) + 1$.
+
+The answer is $\sum_{i,j} f[i][j]$.
+
+Time complexity $O(m \times n)$, space complexity $O(m \times n)$. Where $m$ and $n$ are the number of rows and columns of the matrix respectively.
 
 
 
@@ -133,11 +146,14 @@ public:
         vector> f(m, vector(n));
         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)
+                if (matrix[i][j] == 0) {
+                    continue;
+                }
+                if (i == 0 || j == 0) {
                     f[i][j] = 1;
-                else
+                } else {
                     f[i][j] = min(f[i - 1][j - 1], min(f[i - 1][j], f[i][j - 1])) + 1;
+                }
                 ans += f[i][j];
             }
         }
@@ -176,18 +192,22 @@ func countSquares(matrix [][]int) int {
 
 ```ts
 function countSquares(matrix: number[][]): number {
-    const [m, n] = [matrix.length, matrix[0].length];
-    const f = Array.from({ length: m }, () => Array(n));
-    const dfs = (i: number, j: number): number => {
-        if (i === m || j === n || !matrix[i][j]) return 0;
-        f[i][j] ??= 1 + Math.min(dfs(i + 1, j), dfs(i, j + 1), dfs(i + 1, j + 1));
-        return f[i][j];
-    };
+    const m = matrix.length;
+    const n = matrix[0].length;
+    const f: number[][] = Array.from({ length: m }, () => Array(n).fill(0));
     let ans = 0;
 
     for (let i = 0; i < m; i++) {
         for (let j = 0; j < n; j++) {
-            ans += dfs(i, 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];
         }
     }
 
@@ -195,26 +215,92 @@ function countSquares(matrix: number[][]): number {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn count_squares(matrix: Vec>) -> 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
+    }
+}
+```
+
 #### JavaScript
 
 ```js
-function countSquares(matrix) {
-    const [m, n] = [matrix.length, matrix[0].length];
-    const f = Array.from({ length: m }, () => Array(n));
-    const dfs = (i, j) => {
-        if (i === m || j === n || !matrix[i][j]) return 0;
-        f[i][j] ??= 1 + Math.min(dfs(i + 1, j), dfs(i, j + 1), dfs(i + 1, j + 1));
-        return f[i][j];
-    };
+/**
+ * @param {number[][]} matrix
+ * @return {number}
+ */
+var countSquares = function (matrix) {
+    const m = matrix.length;
+    const n = matrix[0].length;
+    const f = Array.from({ length: m }, () => Array(n).fill(0));
     let ans = 0;
 
     for (let i = 0; i < m; i++) {
         for (let j = 0; j < n; j++) {
-            ans += dfs(i, 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;
+};
+```
+
+#### C#
+
+```cs
+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;
+    }
 }
 ```
 
diff --git a/solution/1200-1299/1277.Count Square Submatrices with All Ones/Solution.cpp b/solution/1200-1299/1277.Count Square Submatrices with All Ones/Solution.cpp
index 75f9da14ffd31..49aca7700e297 100644
--- a/solution/1200-1299/1277.Count Square Submatrices with All Ones/Solution.cpp	
+++ b/solution/1200-1299/1277.Count Square Submatrices with All Ones/Solution.cpp	
@@ -6,14 +6,17 @@ class Solution {
         vector> f(m, vector(n));
         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)
+                if (matrix[i][j] == 0) {
+                    continue;
+                }
+                if (i == 0 || j == 0) {
                     f[i][j] = 1;
-                else
+                } else {
                     f[i][j] = min(f[i - 1][j - 1], min(f[i - 1][j], f[i][j - 1])) + 1;
+                }
                 ans += f[i][j];
             }
         }
         return ans;
     }
-};
\ No newline at end of file
+};
diff --git a/solution/1200-1299/1277.Count Square Submatrices with All Ones/Solution.cs b/solution/1200-1299/1277.Count Square Submatrices with All Ones/Solution.cs
new file mode 100644
index 0000000000000..608c1a65ada57
--- /dev/null
+++ b/solution/1200-1299/1277.Count Square Submatrices with All Ones/Solution.cs	
@@ -0,0 +1,24 @@
+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;
+    }
+}
diff --git a/solution/1200-1299/1277.Count Square Submatrices with All Ones/Solution.js b/solution/1200-1299/1277.Count Square Submatrices with All Ones/Solution.js
index 422c84c168eff..2dda7776c0e02 100644
--- a/solution/1200-1299/1277.Count Square Submatrices with All Ones/Solution.js	
+++ b/solution/1200-1299/1277.Count Square Submatrices with All Ones/Solution.js	
@@ -1,18 +1,26 @@
-function countSquares(matrix) {
-    const [m, n] = [matrix.length, matrix[0].length];
-    const f = Array.from({ length: m }, () => Array(n));
-    const dfs = (i, j) => {
-        if (i === m || j === n || !matrix[i][j]) return 0;
-        f[i][j] ??= 1 + Math.min(dfs(i + 1, j), dfs(i, j + 1), dfs(i + 1, j + 1));
-        return f[i][j];
-    };
+/**
+ * @param {number[][]} matrix
+ * @return {number}
+ */
+var countSquares = function (matrix) {
+    const m = matrix.length;
+    const n = matrix[0].length;
+    const f = Array.from({ length: m }, () => Array(n).fill(0));
     let ans = 0;
 
     for (let i = 0; i < m; i++) {
         for (let j = 0; j < n; j++) {
-            ans += dfs(i, 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;
-}
+};
diff --git a/solution/1200-1299/1277.Count Square Submatrices with All Ones/Solution.rs b/solution/1200-1299/1277.Count Square Submatrices with All Ones/Solution.rs
new file mode 100644
index 0000000000000..57a7faf6f20f7
--- /dev/null
+++ b/solution/1200-1299/1277.Count Square Submatrices with All Ones/Solution.rs	
@@ -0,0 +1,25 @@
+impl Solution {
+    pub fn count_squares(matrix: Vec>) -> 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
+    }
+}
diff --git a/solution/1200-1299/1277.Count Square Submatrices with All Ones/Solution.ts b/solution/1200-1299/1277.Count Square Submatrices with All Ones/Solution.ts
index 15f02bbe04c42..985938d954e55 100644
--- a/solution/1200-1299/1277.Count Square Submatrices with All Ones/Solution.ts	
+++ b/solution/1200-1299/1277.Count Square Submatrices with All Ones/Solution.ts	
@@ -1,16 +1,20 @@
 function countSquares(matrix: number[][]): number {
-    const [m, n] = [matrix.length, matrix[0].length];
-    const f = Array.from({ length: m }, () => Array(n));
-    const dfs = (i: number, j: number): number => {
-        if (i === m || j === n || !matrix[i][j]) return 0;
-        f[i][j] ??= 1 + Math.min(dfs(i + 1, j), dfs(i, j + 1), dfs(i + 1, j + 1));
-        return f[i][j];
-    };
+    const m = matrix.length;
+    const n = matrix[0].length;
+    const f: number[][] = Array.from({ length: m }, () => Array(n).fill(0));
     let ans = 0;
 
     for (let i = 0; i < m; i++) {
         for (let j = 0; j < n; j++) {
-            ans += dfs(i, 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];
         }
     }
 
diff --git a/solution/2300-2399/2348.Number of Zero-Filled Subarrays/README_EN.md b/solution/2300-2399/2348.Number of Zero-Filled Subarrays/README_EN.md
index 65887ef4f9bf5..a1d563d27c336 100644
--- a/solution/2300-2399/2348.Number of Zero-Filled Subarrays/README_EN.md	
+++ b/solution/2300-2399/2348.Number of Zero-Filled Subarrays/README_EN.md	
@@ -29,7 +29,7 @@ tags:
 
 Input: nums = [1,3,0,0,2,0,0,4]
 Output: 6
-Explanation:
+Explanation: 
 There are 4 occurrences of [0] as a subarray.
 There are 2 occurrences of [0,0] as a subarray.
 There is no occurrence of a subarray with a size more than 2 filled with 0. Therefore, we return 6.

diff --git a/solution/3600-3699/3656.Determine if a Simple Graph Exists/README.md b/solution/3600-3699/3656.Determine if a Simple Graph Exists/README.md
new file mode 100644
index 0000000000000..fbd917ca343ee
--- /dev/null
+++ b/solution/3600-3699/3656.Determine if a Simple Graph Exists/README.md	
@@ -0,0 +1,109 @@
+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3656.Determine%20if%20a%20Simple%20Graph%20Exists/README.md
+---
+
+
+
+# [3656. 判断是否存在简单图 🔒](https://leetcode.cn/problems/determine-if-a-simple-graph-exists)
+
+[English Version](/solution/3600-3699/3656.Determine%20if%20a%20Simple%20Graph%20Exists/README_EN.md)
+
+## 题目描述
+
+
+
+
给定一个整数数组 degrees，其中 degrees[i] 表示第 i 个顶点的期望度数。

+
+
你的任务是确定是否存在一个 恰好 具有这些顶点度数的 无向简单 图。

+
+
一个 简单 图没有同一对顶点之间的自环或平行边。

+
+
如果存在这样的图，返回 true，否则返回 false。

+
+
 

+
+
示例 1：

+
+

+
输入：degrees = [3,1,2,2]

+
+
输出：true

+
+
解释：

+
+

+
+
一个可能的无向简单图是：

+
+
    
+	
  • 边：(0, 1), (0, 2), (0, 3), (2, 3)
    • 
+	
  • 度数：deg(0) = 3，deg(1) = 1，deg(2) = 2，deg(3) = 2。
    • 
+

+

+
+
示例 2：

+
+

+
输入：degrees = [1,3,3,1]

+
+
输出：false

+
+
解释：​​​​​​​

+
+
    
+	
  • degrees[1] = 3 和 degrees[2] = 3 意味着它们必须连接到所有其他顶点。
    • 
+	
  • 这需要 degrees[0] 和 degrees[3] 至少是 2，但它们都等于 1，这违反了需求。
    • 
+	
  • 因此，答案是 false。
    • 
+

+

+
+
 

+
+
提示：

+
+
    
+	
  • 1 <= n == degrees.length <= 10​​​​​​​5
    • 
+	
  • 0 <= degrees[i] <= n - 1
    • 
+

+
+
+
+## 解法
+
+
+
+### 方法一
+
+
+
+#### Python3
+
+```python
+
+```
+
+#### Java
+
+```java
+
+```
+
+#### C++
+
+```cpp
+
+```
+
+#### Go
+
+```go
+
+```
+
+
+
+
+
+
diff --git a/solution/3600-3699/3656.Determine if a Simple Graph Exists/README_EN.md b/solution/3600-3699/3656.Determine if a Simple Graph Exists/README_EN.md
new file mode 100644
index 0000000000000..8a8a24fef29ad
--- /dev/null
+++ b/solution/3600-3699/3656.Determine if a Simple Graph Exists/README_EN.md	
@@ -0,0 +1,107 @@
+---
+comments: true
+difficulty: Medium
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3656.Determine%20if%20a%20Simple%20Graph%20Exists/README_EN.md
+---
+
+
+
+# [3656. Determine if a Simple Graph Exists 🔒](https://leetcode.com/problems/determine-if-a-simple-graph-exists)
+
+[中文文档](/solution/3600-3699/3656.Determine%20if%20a%20Simple%20Graph%20Exists/README.md)
+
+## Description
+
+
+
+
You are given an integer array degrees, where degrees[i] represents the desired degree of the ith vertex.

+
+
Your task is to determine if there exists an undirected simple graph with exactly these vertex degrees.

+
+
A simple graph has no self-loops or parallel edges between the same pair of vertices.

+
+
Return true if such a graph exists, otherwise return false.

+
+
 

+
Example 1:

+
+

+
Input: degrees = [3,1,2,2]

+
+
Output: true

+
+
Explanation:

+
+
​​​​​​​

+
+
One possible undirected simple graph is:

+
+
    
+	
  • Edges: (0, 1), (0, 2), (0, 3), (2, 3)
    • 
+	
  • Degrees: deg(0) = 3, deg(1) = 1, deg(2) = 2, deg(3) = 2.
    • 
+

+

+
+
Example 2:

+
+

+
Input: degrees = [1,3,3,1]

+
+
Output: false

+
+
Explanation:​​​​​​​

+
+
    
+	
  • degrees[1] = 3 and degrees[2] = 3 means they must be connected to all other vertices.
    • 
+	
  • This requires degrees[0] and degrees[3] to be at least 2, but both are equal to 1, which contradicts the requirement.
    • 
+	
  • Thus, the answer is false.
    • 
+

+

+
+
 

+
Constraints:

+
+
    
+	
  • 1 <= n == degrees.length <= 10​​​​​​​5
    • 
+	
  • 0 <= degrees[i] <= n - 1
    • 
+

+
+
+
+## Solutions
+
+
+
+### Solution 1
+
+
+
+#### Python3
+
+```python
+
+```
+
+#### Java
+
+```java
+
+```
+
+#### C++
+
+```cpp
+
+```
+
+#### Go
+
+```go
+
+```
+
+
+
+
+
+
diff --git a/solution/3600-3699/3656.Determine if a Simple Graph Exists/images/screenshot-2025-08-13-at-24347-am.png b/solution/3600-3699/3656.Determine if a Simple Graph Exists/images/screenshot-2025-08-13-at-24347-am.png
new file mode 100644
index 0000000000000..37c09ca697309
Binary files /dev/null and b/solution/3600-3699/3656.Determine if a Simple Graph Exists/images/screenshot-2025-08-13-at-24347-am.png differ
diff --git a/solution/README.md b/solution/README.md
index f3444add32b35..7f7bfb54d8c9e 100644
--- a/solution/README.md
+++ b/solution/README.md
@@ -3666,6 +3666,7 @@
 |  3653  |  [区间乘法查询后的异或 I](/solution/3600-3699/3653.XOR%20After%20Range%20Multiplication%20Queries%20I/README.md)  |    |  中等  |  第 463 场周赛  |
 |  3654  |  [删除可整除和后的最小数组和](/solution/3600-3699/3654.Minimum%20Sum%20After%20Divisible%20Sum%20Deletions/README.md)  |    |  中等  |  第 463 场周赛  |
 |  3655  |  [区间乘法查询后的异或 II](/solution/3600-3699/3655.XOR%20After%20Range%20Multiplication%20Queries%20II/README.md)  |    |  困难  |  第 463 场周赛  |
+|  3656  |  [判断是否存在简单图](/solution/3600-3699/3656.Determine%20if%20a%20Simple%20Graph%20Exists/README.md)  |    |  中等  |  🔒  |
 
 ## 版权
 
diff --git a/solution/README_EN.md b/solution/README_EN.md
index d4c7c90102062..73286191323dd 100644
--- a/solution/README_EN.md
+++ b/solution/README_EN.md
@@ -3664,6 +3664,7 @@ Press Control + F(or Command + F on
 |  3653  |  [XOR After Range Multiplication Queries I](/solution/3600-3699/3653.XOR%20After%20Range%20Multiplication%20Queries%20I/README_EN.md)  |    |  Medium  |  Weekly Contest 463  |
 |  3654  |  [Minimum Sum After Divisible Sum Deletions](/solution/3600-3699/3654.Minimum%20Sum%20After%20Divisible%20Sum%20Deletions/README_EN.md)  |    |  Medium  |  Weekly Contest 463  |
 |  3655  |  [XOR After Range Multiplication Queries II](/solution/3600-3699/3655.XOR%20After%20Range%20Multiplication%20Queries%20II/README_EN.md)  |    |  Hard  |  Weekly Contest 463  |
+|  3656  |  [Determine if a Simple Graph Exists](/solution/3600-3699/3656.Determine%20if%20a%20Simple%20Graph%20Exists/README_EN.md)  |    |  Medium  |  🔒  |
 
 ## Copyright