🚀 02-Aug-2021

Author: sureshmangs

SQL/175. Combine Two Tables.cpp

@@ -0,0 +1,41 @@
+Table: Person
+
++-------------+---------+
+| Column Name | Type    |
++-------------+---------+
+| PersonId    | int     |
+| FirstName   | varchar |
+| LastName    | varchar |
++-------------+---------+
+PersonId is the primary key column for this table.
+Table: Address
+
++-------------+---------+
+| Column Name | Type    |
++-------------+---------+
+| AddressId   | int     |
+| PersonId    | int     |
+| City        | varchar |
+| State       | varchar |
++-------------+---------+
+AddressId is the primary key column for this table.
+ 
+
+Write a SQL query for a report that provides the following information for each person in the Person table, 
+regardless if there is an address for each of those people:
+
+FirstName, LastName, City, State
+
+
+
+
+
+
+
+
+
+
+
+SELECT FirstName, LastName, City, State 
+FROM Person LEFT JOIN Address 
+ON Person.PersonId = Address.PersonId;

SQL/175. Combine Two Tables.txt

@@ -0,0 +1,51 @@
+Table: Person
+
++-------------+---------+
+| Column Name | Type    |
++-------------+---------+
+| PersonId    | int     |
+| FirstName   | varchar |
+| LastName    | varchar |
++-------------+---------+
+PersonId is the primary key column for this table.
+Table: Address
+
++-------------+---------+
+| Column Name | Type    |
++-------------+---------+
+| AddressId   | int     |
+| PersonId    | int     |
+| City        | varchar |
+| State       | varchar |
++-------------+---------+
+AddressId is the primary key column for this table.
+ 
+
+Write a SQL query for a report that provides the following information for each person in the Person table, 
+regardless if there is an address for each of those people:
+
+FirstName, LastName, City, State
+
+
+
+
+
+
+
+
+
+Solution:
+
+Since the PersonId in table Address is the foreign key of table Person, 
+we can join this two table to get the address information of a person.
+
+Considering there might not be an address information for every person, 
+we should use outer join instead of the default inner join.
+
+Note: Using where clause to filter the records will fail if there is no address 
+information for a person because it will not display the name information.
+
+
+SELECT FirstName, LastName, City, State 
+FROM Person LEFT JOIN Address 
+ON Person.PersonId = Address.PersonId;

algorithms/graph/topological_sorting.cpp renamed to algorithms/graph/topological_sorting_dfs.cpp

@@ -2,13 +2,19 @@
 Given a Directed Graph. Find any Topological Sorting of that Graph.
 
 Input:
-The first line of input takes the number of test cases then T test cases follow . Each test case contains two lines. The first  line of each test case  contains two integers E and V representing no of edges and the number of vertices. Then in the next line are E  pairs of integers u, v representing an edge from u to v in the graph.
+The first line of input takes the number of test cases then T test cases follow . 
+Each test case contains two lines. The first  line of each test case  contains two integers E and V 
+representing no of edges and the number of vertices. Then in the next line are E  pairs of integers u, v 
+representing an edge from u to v in the graph.
 
 Output:
 For each test case output will be 1 if the topological sort is done correctly else it will be 0.
 
 Your Task:
-You don't need to read input or print anything. Your task is to complete the function topoSort() which takes the adjacency list of the Graph and the number of vertices (N) as inputs are returns an array consisting of a the vertices in Topological order. As there are multiple Topological orders possible, you may return any of them.
+You don't need to read input or print anything. Your task is to complete the function topoSort() 
+which takes the adjacency list of the Graph and the number of vertices (N) as inputs are returns 
+an array consisting of a the vertices in Topological order. As there are multiple Topological 
+orders possible, you may return any of them.
 
 Expected Time Complexity: O(V + E).
 Expected Auxiliary Space: O(V).
@@ -91,34 +97,51 @@ int main() {
 }// } Driver Code Ends
 
 
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
 // The Graph structure is as folows
 
 /*  Function which sorts the graph vertices in topological form
 *   N: number of vertices
 *   adj[]: input graph
 */
 
-stack<int> s;
+stack <int> s;
 
 void dfs(vector<int> adj[],int node, vector<bool> &vis){
-    vis[node]=true;
-    for(auto x: adj[node]){
+    vis[node] = true;
+    for (auto x: adj[node]){
         if(!vis[x])
             dfs(adj, x, vis);
     }
     s.push(node);
 }
 
-vector<int> topoSort(int V, vector<int> adj[]) {
-    vector<bool> vis(V, false);
-    for(int i=0;i<V;i++){
-        if(!vis[i])
+vector <int> topoSort(int V, vector<int> adj[]) {
+    vector <bool> vis(V, false);
+    for (int i = 0; i < V; i++){
+        if (!vis[i])
          dfs(adj, i, vis);
     }
-    vector<int>res;
-    while(!s.empty()){
+    vector <int> res;
+    
+    while (!s.empty()){
         res.push_back(s.top());
         s.pop();
     }
+    
     return res;
 }

backtracking/palindromic_partioning.cpp

@@ -0,0 +1,64 @@
+Given a string s, partition s such that every substring of the partition 
+is a palindrome. Return all possible palindrome partitioning of s.
+
+A palindrome string is a string that reads the same backward as forward.
+
+ 
+
+Example 1:
+
+Input: s = "aab"
+Output: [["a","a","b"],["aa","b"]]
+Example 2:
+
+Input: s = "a"
+Output: [["a"]]
+ 
+
+Constraints:
+
+1 <= s.length <= 16
+s contains only lowercase English letters.
+
+
+
+
+
+
+
+
+
+class Solution {
+public:
+    
+    bool isPalindrome(int start, int end, string &s) {
+        while (start < end) {
+            if (s[start++] != s[end--]) return false;
+        }
+        return true;
+    }
+    
+    void dfs(int start, string &s, vector<string> &curr, vector<vector<string>> &res) {
+        if (start == s.length()) {
+            res.push_back(curr);
+            return;
+        }
+        
+        for (int i = start; i < s.length(); i++) {
+            if (isPalindrome(start, i, s)) {
+                curr.push_back(s.substr(start, i - start + 1));
+                dfs(i + 1, s, curr, res);
+                curr.pop_back(); // backtrack
+            }
+        }
+    }
+    
+    vector<vector<string>> partition(string s) {
+        vector<vector<string>> res;
+        vector<string> curr;
+        
+        dfs(0, s, curr, res);
+        
+        return res;
+    }
+};

competitive programming/codeforces/1530A - Binary Decimal.cpp

@@ -0,0 +1,64 @@
+/*
+Let's call a number a binary decimal if it's a positive integer and all digits in its decimal notation are either 0 or 1. For example, 1010111 is a binary decimal, while 10201 and 787788 are not.
+
+Given a number n, you are asked to represent n as a sum of some (not necessarily distinct) binary decimals. Compute the smallest number of binary decimals required for that.
+
+Input
+The first line contains a single integer t (1=t=1000), denoting the number of test cases.
+
+The only line of each test case contains a single integer n (1=n=109), denoting the number to be represented.
+
+Output
+For each test case, output the smallest number of binary decimals required to represent n as a sum.
+
+Example
+inputCopy
+3
+121
+5
+1000000000
+outputCopy
+2
+5
+1
+Note
+In the first test case, 121 can be represented as 121=110+11 or 121=111+10.
+
+In the second test case, 5 can be represented as 5=1+1+1+1+1.
+
+In the third test case, 1000000000 is a binary decimal itself, thus the answer is 1.
+*/
+
+
+
+
+
+
+#include<bits/stdc++.h>
+using namespace std;
+ 
+void solve() {
+	int n;
+	cin >> n;
+	
+	int res = 0;
+	
+	while (n) {
+		res = max(res, n % 10);
+		n /= 10;
+	}
+	
+	cout << res << "\n";
+}
+ 
+int main(){
+    ios_base::sync_with_stdio(false);
+	cin.tie(NULL);
+	
+	int t = 1;
+	cin >> t;
+	
+	while (t--)  solve();
+	
+    return 0;
+}