leetcode weekly 322

Author: Pankajcoder1

LeetCode/2490. Circular Sentence.cpp

@@ -0,0 +1,73 @@
+/*
+    written by Pankaj Kumar.
+    country:-INDIA
+*/
+typedef long long ll ;
+
+
+/* ascii value
+A=65,Z=90,a=97,z=122
+*/
+
+/*  --------------------MAIN PROGRAM----------------------------*/
+// to run ctrl+b
+const ll INF=1e18;
+const ll mod1=1e9+7;
+const ll mod2=998244353;
+// Techniques :
+// divide into cases, brute force, pattern finding
+// sort, greedy, binary search, two pointer
+// transform into graph
+
+// Experience :
+// Cp is nothing but only observation and mathematics.
+
+
+//Add main code here
+
+class Solution
+{
+public:
+    bool isCircularSentence(string sentence)
+    {
+        vector<string> v;
+        string temp="";
+        for (auto x : sentence)
+        {
+            if(x==' '){
+                if(temp!=""){
+                    v.push_back(temp);
+                    temp="";
+                }
+            }
+            else{
+                temp+=x;
+            }
+        }
+        if(temp!=""){
+            v.push_back(temp);
+        }
+        // for(auto x:v){
+        //     cout<<"x is "<<x<<endl;
+        // }
+        int n=v.size();
+        for(int i=0;i+1<n;i++){
+            if(v[i+1][0]!=v[i][v[i].size()-1]){
+                return false;
+            }
+        }
+        if(v[0][0]!=v[n-1][v[n-1].size()-1]){
+            return false;
+        }
+        return true;
+    }
+};
+
+/* -----------------END OF PROGRAM --------------------*/
+/*
+* stuff you should look before submission
+* constraint and time limit
+* int overflow
+* special test case (n=0||n=1||n=2)
+* don't get stuck on one approach if you get wrong answer
+*/

LeetCode/2491. Divide Players Into Teams of Equal Skill.cpp

@@ -0,0 +1,71 @@
+/*
+    written by Pankaj Kumar.
+    country:-INDIA
+*/
+typedef long long ll ;
+
+
+/* ascii value
+A=65,Z=90,a=97,z=122
+*/
+
+/*  --------------------MAIN PROGRAM----------------------------*/
+// to run ctrl+b
+const ll INF=1e18;
+const ll mod1=1e9+7;
+const ll mod2=998244353;
+// Techniques :
+// divide into cases, brute force, pattern finding
+// sort, greedy, binary search, two pointer
+// transform into graph
+
+// Experience :
+// Cp is nothing but only observation and mathematics.
+
+
+//Add main code here
+
+class Solution
+{
+public:
+    long long dividePlayers(vector<int> &skill)
+    {
+        map<int,int> m;
+        long long totalSum=0;
+        for(auto x:skill){
+            m[x]++;
+            totalSum+=x;
+        }
+        int n=skill.size();
+        n/=2;
+        if(totalSum%n){
+            return -1;
+        }
+        long long sum=totalSum/n;
+        
+        long long group=0,ans=0;
+        for(auto x:skill){
+            if(m[x]>0 && m[sum-x]>0){
+                group++;
+                ans+=(x*(sum-x));
+                m[x]--;
+                m[sum-x]--;
+            }
+        }
+        if(group==n){
+            return ans;
+        }
+        else{
+            return -1;
+        }
+    }
+};
+
+/* -----------------END OF PROGRAM --------------------*/
+/*
+* stuff you should look before submission
+* constraint and time limit
+* int overflow
+* special test case (n=0||n=1||n=2)
+* don't get stuck on one approach if you get wrong answer
+*/

LeetCode/2492. Minimum Score of a Path Between Two Cities.cpp

@@ -0,0 +1,76 @@
+/*
+    written by Pankaj Kumar.
+    country:-INDIA
+*/
+typedef long long ll ;
+
+
+/* ascii value
+A=65,Z=90,a=97,z=122
+*/
+
+/*  --------------------MAIN PROGRAM----------------------------*/
+// to run ctrl+b
+const ll INF=1e18;
+const ll mod1=1e9+7;
+const ll mod2=998244353;
+// Techniques :
+// divide into cases, brute force, pattern finding
+// sort, greedy, binary search, two pointer
+// transform into graph
+
+// Experience :
+// Cp is nothing but only observation and mathematics.
+
+
+//Add main code here
+
+class Solution
+{
+public:
+    int mini=INT_MAX;
+    set<int> v;
+    void dfs(int root,vector<vector<int>> &graph, vector<int> &visited){
+        if(visited[root]==1)
+            return;
+        visited[root]=1;
+        for(auto x:graph[root]){
+            if(visited[x]==0){
+                v.insert(x);
+                dfs(x,graph,visited);
+            }
+        }
+        return ;
+    }
+
+    int minScore(int n, vector<vector<int>> &roads)
+    {
+        map<pair<int,int>,int> m;
+        for(auto x:roads){
+            m[{min(x[0],x[1]),max(x[0],x[1])}]=x[2];
+        }
+        vector<vector<int>> graph(n+1);
+        for(auto x:roads){
+            graph[x[0]].push_back(x[1]);
+            graph[x[1]].push_back(x[0]);
+        }
+        vector<int> visited(n+1,0);
+        dfs(1,graph,visited);
+        for (auto y : m)
+        {
+            if(v.find(y.first.first)!=v.end() || v.find(y.first.second)!=v.end()){
+                mini=min(mini,y.second);
+            }
+        }
+        return mini;
+    }
+};
+
+/* -----------------END OF PROGRAM --------------------*/
+/*
+* stuff you should look before submission
+* constraint and time limit
+* int overflow
+* special test case (n=0||n=1||n=2)
+* don't get stuck on one approach if you get wrong answer
+*/