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added leetcode problem 200 number of island solution in c++. solved issue keshavsingh4522#4483
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#include<stdio.h> | ||
#include<conio.h> | ||
void main() | ||
#include <stdio.h> | ||
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int main() | ||
{ | ||
clrscr(); | ||
int i,n,a[100],search; | ||
printf("Enter number of elements : "); | ||
scanf("%d",&n); | ||
printf("\nEnter %d elements : \n",n); | ||
for(i=0;i<n;i++) | ||
{ | ||
scanf("%d",&a[i]); | ||
} | ||
printf("\nEnter element to be searched : "); | ||
scanf("%d",&search); | ||
for(i=0;i<n;i++) | ||
{ | ||
if(a[i]==search) | ||
{ | ||
printf("\nFound at %d .",m+1); | ||
break; | ||
} | ||
} | ||
if(i==n) | ||
{ | ||
printf("\nElement not found."); | ||
} | ||
getch(); | ||
} | ||
int arr[50]; | ||
int n, number, ind = -1; | ||
printf("Enter the number of elements : "); | ||
scanf("%d", &n); | ||
printf("Enter Array Elements : "); | ||
for (int i = 0; i < n; i++) | ||
{ | ||
scanf("%d", &arr[i]); | ||
} | ||
printf("\nEnter the Number to be searched : "); | ||
scanf("%d", &number); | ||
for (int i = 0; i < n; i++) | ||
{ | ||
if (number == arr[i]) | ||
{ | ||
ind = i; | ||
break; | ||
} | ||
} | ||
if (ind != -1) | ||
{ | ||
printf("\nElement %d found at %d position", number, ind + 1); | ||
} | ||
else | ||
{ | ||
printf("Element not found :)\n"); | ||
} | ||
return 0; | ||
} |
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#include<bits/stdc++.h> | ||
#include<ext/pb_ds/assoc_container.hpp> | ||
#include<ext/pb_ds/tree_policy.hpp> | ||
#pragma GCC optimize("Ofast") | ||
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma") | ||
#pragma GCC optimize("unroll-loops") | ||
const unsigned int M = 1000000007; | ||
using namespace std; | ||
// Check | ||
using namespace __gnu_pbds; | ||
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> T_set; // PBDS_set | ||
typedef tree<int,null_type,less_equal<int>,rb_tree_tag,tree_order_statistics_node_update> T_multiset; // PBDS_multiset | ||
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void solve() | ||
{ | ||
int n ,m,u,v; | ||
cin>>n>>m; | ||
vector<list<int>> adj(n+1); | ||
vector<bool> vis(n+1,false); | ||
for(int i = 0; i < n ; i++ ){ | ||
cin>>u>>v; | ||
adj[u].push_back(v); | ||
adj[v].push_back(u); | ||
} | ||
queue<int> temp; | ||
temp.push(1); | ||
vis[1] = true; | ||
while(!temp.empty()){ | ||
int curr = temp.front(); | ||
cout<<curr<<" "; | ||
temp.pop(); | ||
for(int elem : adj[curr]){ | ||
if(!vis[elem]){ | ||
temp.push(elem); | ||
vis[elem] = true; | ||
} | ||
} | ||
} | ||
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} | ||
int main() | ||
{ | ||
ios_base::sync_with_stdio(false); | ||
cout.tie(NULL); | ||
cin.tie(NULL); | ||
solve(); | ||
return 0; | ||
} | ||
// Program to print BFS traversal from a given | ||
// source vertex. BFS(int s) traverses vertices | ||
// reachable from s. | ||
#include<iostream> | ||
#include <list> | ||
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using namespace std; | ||
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// This class represents a directed graph using | ||
// adjacency list representation | ||
class Graph | ||
{ | ||
int V; // No. of vertices | ||
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// Pointer to an array containing adjacency | ||
// lists | ||
list<int> *adj; | ||
public: | ||
Graph(int V); // Constructor | ||
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// function to add an edge to graph | ||
void addEdge(int v, int w); | ||
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// prints BFS traversal from a given source s | ||
void BFS(int s); | ||
}; | ||
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Graph::Graph(int V) | ||
{ | ||
this->V = V; | ||
adj = new list<int>[V]; | ||
} | ||
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void Graph::addEdge(int v, int w) | ||
{ | ||
adj[v].push_back(w); // Add w to v’s list. | ||
} | ||
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void Graph::BFS(int s) | ||
{ | ||
// Mark all the vertices as not visited | ||
bool *visited = new bool[V]; | ||
for(int i = 0; i < V; i++) | ||
visited[i] = false; | ||
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// Create a queue for BFS | ||
list<int> queue; | ||
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// Mark the current node as visited and enqueue it | ||
visited[s] = true; | ||
queue.push_back(s); | ||
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// 'i' will be used to get all adjacent | ||
// vertices of a vertex | ||
list<int>::iterator i; | ||
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while(!queue.empty()) | ||
{ | ||
// Dequeue a vertex from queue and print it | ||
s = queue.front(); | ||
cout << s << " "; | ||
queue.pop_front(); | ||
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// Get all adjacent vertices of the dequeued | ||
// vertex s. If a adjacent has not been visited, | ||
// then mark it visited and enqueue it | ||
for (i = adj[s].begin(); i != adj[s].end(); ++i) | ||
{ | ||
if (!visited[*i]) | ||
{ | ||
visited[*i] = true; | ||
queue.push_back(*i); | ||
} | ||
} | ||
} | ||
} | ||
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// Driver program to test methods of graph class | ||
int main() | ||
{ | ||
// Create a graph given in the above diagram | ||
Graph g(4); | ||
g.addEdge(0, 1); | ||
g.addEdge(0, 2); | ||
g.addEdge(1, 2); | ||
g.addEdge(2, 0); | ||
g.addEdge(2, 3); | ||
g.addEdge(3, 3); | ||
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cout << "Following is Breadth First Traversal " | ||
<< "(starting from vertex 2) \n"; | ||
g.BFS(2); | ||
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return 0; | ||
} |
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class Solution { | ||
public: | ||
int numIslands(vector<vector<char>>& grid) { | ||
if (grid.empty() || grid[0].empty()) { | ||
return 0; | ||
} | ||
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int numIslands = 0; | ||
int numRows = grid.size(); | ||
int numCols = grid[0].size(); | ||
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for (int i = 0; i < numRows; i++) { | ||
for (int j = 0; j < numCols; j++) { | ||
if (grid[i][j] == '1') { | ||
numIslands++; | ||
dfs(grid, i, j); | ||
} | ||
} | ||
} | ||
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return numIslands; | ||
} | ||
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private: | ||
void dfs(vector<vector<char>>& grid, int row, int col) { | ||
int numRows = grid.size(); | ||
int numCols = grid[0].size(); | ||
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if (row < 0 || col < 0 || row >= numRows || col >= numCols || grid[row][col] == '0') { | ||
return; | ||
} | ||
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grid[row][col] = '0'; // Mark the current cell as visited | ||
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// Explore adjacent cells in all four directions | ||
dfs(grid, row + 1, col); | ||
dfs(grid, row - 1, col); | ||
dfs(grid, row, col + 1); | ||
dfs(grid, row, col - 1); | ||
} | ||
}; |