feat: update solutions to lc problems: No.2642~2646 (#2476)

Author: yanglbme

solution/2600-2699/2642.Design Graph With Shortest Path Calculator/README.md

@@ -312,48 +312,43 @@ class Graph {
 ```cs
 public class Graph {
     private int n;
-    private int[][] g;
+    private int[,] g;
     private readonly int inf = 1 << 29;
 
     public Graph(int n, int[][] edges) {
         this.n = n;
-        g = new int[n][];
-        for (int i = 0; i < n; i++)
-        {
-            g[i] = new int[n];
-            for (int j = 0; j < n; j++)
-            {
-                g[i][j] = inf;
+        g = new int[n, n];
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < n; j++) {
+                g[i, j] = inf;
             }
         }
-        foreach (int[] e in edges)
-        {
-            g[e[0]][e[1]] = e[2];
+        foreach (var e in edges) {
+            int f = e[0], t = e[1], c = e[2];
+            g[f, t] = c;
         }
     }
 
     public void AddEdge(int[] edge) {
-        g[edge[0]][edge[1]] = edge[2];
+        int f = edge[0], t = edge[1], c = edge[2];
+        g[f, t] = c;
     }
 
     public int ShortestPath(int node1, int node2) {
         int[] dist = new int[n];
         bool[] vis = new bool[n];
         Array.Fill(dist, inf);
         dist[node1] = 0;
-
-        for (int i = 0; i < n; i++)
-        {
+        for (int i = 0; i < n; ++i) {
             int t = -1;
-            for (int j = 0; j < n; j++)
-            {
-                if (!vis[j] && (t == -1 || dist[t] > dist[j]))
+            for (int j = 0; j < n; ++j) {
+                if (!vis[j] && (t == -1 || dist[t] > dist[j])) {
                     t = j;
+                }
             }
             vis[t] = true;
-            for (int j = 0; j < n; j++)
-            {
-                dist[j] = Math.Min(dist[j], dist[t] + g[t][j]);
+            for (int j = 0; j < n; ++j) {
+                dist[j] = Math.Min(dist[j], dist[t] + g[t, j]);
             }
         }
         return dist[node2] >= inf ? -1 : dist[node2];

solution/2600-2699/2642.Design Graph With Shortest Path Calculator/README_EN.md

@@ -51,7 +51,15 @@ g.shortestPath(0, 3); // return 6. The shortest path from 0 to 3 now is 0 ->
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Dijsktra's Algorithm
+
+In the initialization function, we first use the adjacency matrix $g$ to store the edge weights of the graph, where $g_{ij}$ represents the edge weight from node $i$ to node $j$. If there is no edge between $i$ and $j$, the value of $g_{ij}$ is $\infty$.
+
+In the `addEdge` function, we update the value of $g_{ij}$ to $edge[2]$.
+
+In the `shortestPath` function, we use Dijsktra's algorithm to find the shortest path from node $node1$ to node $node2$. Here, $dist[i]$ represents the shortest path from node $node1$ to node $i$, and $vis[i]$ indicates whether node $i$ has been visited. We initialize $dist[node1]$ to $0$, and the rest of $dist[i]$ are all $\infty$. Then we iterate $n$ times, each time finding the current unvisited node $t$ such that $dist[t]$ is the smallest. Then we mark node $t$ as visited, and then update the value of $dist[i]$ to $min(dist[i], dist[t] + g_{ti})$. Finally, we return $dist[node2]$. If $dist[node2]$ is $\infty$, it means that there is no path from node $node1$ to node $node2$, so we return $-1$.
+
+The time complexity is $O(n^2 \times q)$, and the space complexity is $O(n^2)$. Where $n$ is the number of nodes, and $q$ is the number of calls to the `shortestPath` function.
 
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@@ -300,48 +308,43 @@ class Graph {
 ```cs
 public class Graph {
     private int n;
-    private int[][] g;
+    private int[,] g;
     private readonly int inf = 1 << 29;
 
     public Graph(int n, int[][] edges) {
         this.n = n;
-        g = new int[n][];
-        for (int i = 0; i < n; i++)
-        {
-            g[i] = new int[n];
-            for (int j = 0; j < n; j++)
-            {
-                g[i][j] = inf;
+        g = new int[n, n];
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < n; j++) {
+                g[i, j] = inf;
             }
         }
-        foreach (int[] e in edges)
-        {
-            g[e[0]][e[1]] = e[2];
+        foreach (var e in edges) {
+            int f = e[0], t = e[1], c = e[2];
+            g[f, t] = c;
         }
     }
 
     public void AddEdge(int[] edge) {
-        g[edge[0]][edge[1]] = edge[2];
+        int f = edge[0], t = edge[1], c = edge[2];
+        g[f, t] = c;
     }
 
     public int ShortestPath(int node1, int node2) {
         int[] dist = new int[n];
         bool[] vis = new bool[n];
         Array.Fill(dist, inf);
         dist[node1] = 0;
-
-        for (int i = 0; i < n; i++)
-        {
+        for (int i = 0; i < n; ++i) {
             int t = -1;
-            for (int j = 0; j < n; j++)
-            {
-                if (!vis[j] && (t == -1 || dist[t] > dist[j]))
+            for (int j = 0; j < n; ++j) {
+                if (!vis[j] && (t == -1 || dist[t] > dist[j])) {
                     t = j;
+                }
             }
             vis[t] = true;
-            for (int j = 0; j < n; j++)
-            {
-                dist[j] = Math.Min(dist[j], dist[t] + g[t][j]);
+            for (int j = 0; j < n; ++j) {
+                dist[j] = Math.Min(dist[j], dist[t] + g[t, j]);
             }
         }
         return dist[node2] >= inf ? -1 : dist[node2];

solution/2600-2699/2642.Design Graph With Shortest Path Calculator/Solution.cs

@@ -1,47 +1,42 @@
 public class Graph {
     private int n;
-    private int[][] g;
+    private int[,] g;
     private readonly int inf = 1 << 29;
 
     public Graph(int n, int[][] edges) {
         this.n = n;
-        g = new int[n][];
-        for (int i = 0; i < n; i++)
-        {
-            g[i] = new int[n];
-            for (int j = 0; j < n; j++)
-            {
-                g[i][j] = inf;
+        g = new int[n, n];
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < n; j++) {
+                g[i, j] = inf;
             }
         }
-        foreach (int[] e in edges)
-        {
-            g[e[0]][e[1]] = e[2];
+        foreach (var e in edges) {
+            int f = e[0], t = e[1], c = e[2];
+            g[f, t] = c;
         }
     }
 
     public void AddEdge(int[] edge) {
-        g[edge[0]][edge[1]] = edge[2];
+        int f = edge[0], t = edge[1], c = edge[2];
+        g[f, t] = c;
     }
 
     public int ShortestPath(int node1, int node2) {
         int[] dist = new int[n];
         bool[] vis = new bool[n];
         Array.Fill(dist, inf);
         dist[node1] = 0;
-
-        for (int i = 0; i < n; i++)
-        {
+        for (int i = 0; i < n; ++i) {
             int t = -1;
-            for (int j = 0; j < n; j++)
-            {
-                if (!vis[j] && (t == -1 || dist[t] > dist[j]))
+            for (int j = 0; j < n; ++j) {
+                if (!vis[j] && (t == -1 || dist[t] > dist[j])) {
                     t = j;
+                }
             }
             vis[t] = true;
-            for (int j = 0; j < n; j++)
-            {
-                dist[j] = Math.Min(dist[j], dist[t] + g[t][j]);
+            for (int j = 0; j < n; ++j) {
+                dist[j] = Math.Min(dist[j], dist[t] + g[t, j]);
             }
         }
         return dist[node2] >= inf ? -1 : dist[node2];
@@ -53,4 +48,4 @@ public int ShortestPath(int node1, int node2) {
  * Graph obj = new Graph(n, edges);
  * obj.AddEdge(edge);
  * int param_2 = obj.ShortestPath(node1,node2);
- */
+ */

solution/2600-2699/2642.Design Graph With Shortest Path Calculator/Solution.ts

@@ -1,9 +1,8 @@
 class Graph {
     private g: number[][] = [];
-    private inf: number = 1 << 29;
 
     constructor(n: number, edges: number[][]) {
-        this.g = Array.from({ length: n }, () => Array(n).fill(this.inf));
+        this.g = Array.from({ length: n }, () => Array(n).fill(Infinity));
         for (const [f, t, c] of edges) {
             this.g[f][t] = c;
         }
@@ -16,9 +15,9 @@ class Graph {
 
     shortestPath(node1: number, node2: number): number {
         const n = this.g.length;
-        const dist: number[] = new Array(n).fill(this.inf);
+        const dist: number[] = Array(n).fill(Infinity);
         dist[node1] = 0;
-        const vis: boolean[] = new Array(n).fill(false);
+        const vis: boolean[] = Array(n).fill(false);
         for (let i = 0; i < n; ++i) {
             let t = -1;
             for (let j = 0; j < n; ++j) {
@@ -31,7 +30,7 @@ class Graph {
                 dist[j] = Math.min(dist[j], dist[t] + this.g[t][j]);
             }
         }
-        return dist[node2] >= this.inf ? -1 : dist[node2];
+        return dist[node2] >= Infinity ? -1 : dist[node2];
     }
 }
 

solution/2600-2699/2643.Row With Maximum Ones/README_EN.md

@@ -49,7 +49,11 @@
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Simulation
+
+We directly traverse the matrix, count the number of $1$s in each row, and update the maximum value and the corresponding row index. Note that if the number of $1$s in the current row is equal to the maximum value, we need to choose the row with the smaller index.
+
+The time complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns of the matrix, respectively. The space complexity is $O(1)$.
 
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solution/2600-2699/2644.Find the Maximum Divisibility Score/README_EN.md

@@ -58,7 +58,16 @@ Since divisors[0] and divisors[1] both have the maximum divisibility score, we r
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Enumeration
+
+We can enumerate each element $div$ in $divisors$, and calculate how many elements in $nums$ can be divided by $div$, denoted as $cnt$.
+
+-   If $cnt$ is greater than the current maximum divisibility score $mx$, then update $mx = cnt$, and update $ans = div$.
+-   If $cnt$ equals $mx$ and $div$ is less than $ans$, then update $ans = div$.
+
+Finally, return $ans$.
+
+The time complexity is $O(m \times n)$, where $m$ and $n$ are the lengths of $nums$ and $divisors$ respectively. The space complexity is $O(1)$.
 
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solution/2600-2699/2646.Minimize the Total Price of the Trips/README_EN.md

@@ -59,7 +59,16 @@ The total price sum of all trips is 1. It can be proven, that 1 is the minimum a
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Enumeration
+
+We can enumerate each element $div$ in $divisors$, and calculate how many elements in $nums$ can be divided by $div$, denoted as $cnt$.
+
+-   If $cnt$ is greater than the current maximum divisibility score $mx$, then update $mx = cnt$, and update $ans = div$.
+-   If $cnt$ equals $mx$ and $div$ is less than $ans$, then update $ans = div$.
+
+Finally, return $ans$.
+
+The time complexity is $O(m \times n)$, where $m$ and $n$ are the lengths of $nums$ and $divisors$ respectively. The space complexity is $O(1)$.
 
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