MakeMyTrip Coding Round

GE Healthcare/Managerial/.vscode/settings.json

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+{
+    "files.associations": {
+        "*.rmd": "markdown",
+        "xstring": "cpp"
+    }
+}

MakeMyTrip/.vscode/c_cpp_properties.json

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+{
+    "configurations": [
+        {
+            "name": "Win32",
+            "includePath": [
+                "${workspaceFolder}/**"
+            ],
+            "defines": [
+                "_DEBUG",
+                "UNICODE",
+                "_UNICODE"
+            ],
+            "windowsSdkVersion": "10.0.17763.0",
+            "compilerPath": "C:/MinGW/bin/gcc.exe",
+            "cStandard": "c17",
+            "cppStandard": "c++17",
+            "intelliSenseMode": "windows-msvc-x64"
+        }
+    ],
+    "version": 4
+}

MakeMyTrip/longestSubsequence.cpp

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+// Problem Description : Longest Special Subsequences
+// You are given a string S of the length n consisting of only lowercase alphabets. If the two consecutive characters in a subsequence have a difference that is no more than k, then it is called a Special Subsequence. Find the length of the longest special subsequence.
+
+// INPUT FORMAT :
+
+// Two space separated integers n and k
+// String S
+// OUTPUT FORMAT :
+// Print the length of the longest special subsequence in a new line.
+
+// CONSTRAINTS :
+// 1<= n <= (10^5)
+// 1<=k<=26
+
+// Sample Input :
+// 7 2
+// afcbedf
+
+// Sample Output :
+// 4
+
+// Explaination :
+// One of the longest special subsequence present in afcbedf is a,c,b,d. It is special because |a-c|<=2 , |c-b|<=2 and |b-d|<=2.
+
+#include<bits/stdc++.h>
+using namespace std;
+
+int longest_special_subseq(int n, int k, string s){
+    vector<int> dp(n, 0);
+    int max_length[26] = {0};
+
+    for (int i = 0; i < n; i++){
+        int curr = s[i] - 'a';
+        int lower = max(0, curr - k);
+        int upper = min(25, curr + k);
+        for (int j = lower; j < upper + 1; j++){
+            dp[i] = max(dp[i], max_length[j] + 1);
+        }
+        max_length[curr] = max(dp[i], max_length[curr]);
+    }
+
+    int ans = 0;
+    for(int i:dp) ans = max(i, ans);
+    return ans;
+}
+
+int main(){
+    int t, n, k;
+    cin >> t;
+    while(t--){
+        cin >> n >> k;
+        string s;
+        cin >> s;
+        cout << longest_special_subseq(n, k, s) << endl;
+    }
+}
+
+// Sample Input: 
+// 1
+// 7 2 afcbedg
+
+// Sample Output:
+// 4

MakeMyTrip/safePaths.cpp

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+// Problem Description - SAFE PATHS
+// There are n cities numbered from 1 to n and there are n-1 bidirectional roads such that all cities are connected . There are k trees, each one is in a different city. You are currently in the 1st city. You want to visit city X, such that neither X nor the the cities in the path from 1 to X has a tree. Find out how many such X can you visit (X != 1).
+
+// INPUT FORMAT :
+// First line contains two space separted integers n , k representing the number of cities and the number of trees respectively.
+// Next n-1 lines : Two integers ui and vi meaning there is a road between cities ui and vi.
+// Next line : k integers p1,p2,p3....pk where there is a tree at pi city.
+
+// OUTPUT FORMAT :
+// Print a single integer denoting the number of cities you can visit.
+
+// Input Constraints :
+
+// 1<=n<=10^5
+// 0<=k<n
+// 1<=ui,vi<=n; ui != vi
+// 2<=pi<=n
+
+// Sample Input :
+// 6 3
+// 1 2
+// 1 6
+// 2 3
+// 2 4
+// 2 5
+// 2 3 4
+
+// Sample Output :
+// 1
+
+// Explaination :
+// There are total 5 possiblities :
+
+// 1 -> 2
+// 1 -> 2 -> 3
+// 1 -> 2 -> 4
+// 1 -> 2 -> 5
+// 1 -> 6
+// In the first 4 cases , the second city has a tree, so he can't go to 2nd , 3rd , 4th and 5th city. But in 5th case, none of the city on the simple path has a theif so he can go to 6th city.
+
+// Working Solution :
+
+// The idea is to visit those cities from 1 that don't have a tree by doing a BFS or a DFS.
+// Here is a working CPP code that uses a BFS based traversal approach.
+
+
+#include<bits/stdc++.h>
+using namespace std;
+
+int main(){
+	int n,k;
+	cin>>n>>k;
+	vector<int> adj[n+1];
+	for(int i=0;i<n-1;i++){
+		int a,b;
+		cin>>a>>b;
+		adj[a].push_back(b);
+		adj[b].push_back(a);
+	}
+	vector<int> vis(n+1);
+	vector<int> A(k);
+	for(int i=0;i<k;i++)
+		cin>>A[k], vis[A[k]] = 1;
+	if(vis[1]==1){
+		cout << 0 << endl;
+		return 0;
+	}
+	queue<int> q;
+	q.push(1);
+	vis[1] = 1;
+	int ans = 0;
+	while(!q.empty()){
+		int f = q.front();
+		q.pop();
+		for(auto itr:adj[f]){
+			if(vis[itr]==0){
+				vis[itr] = 1;
+				ans++;
+				q.push(itr);
+			}
+		}
+	}
+	cout << ans << endl;
+}
+
+// int solution(vector<int> thieves,vector<vector<int>> edges){
+//     int n = thieves.size();
+//     vector<int> adj[n+1];
+//     for(int i=0;i<n-1;i++){
+//         int a = edges[i][0];
+//         int b = edges[i][1];
+//         adj[a].push_back(b);
+//         adj[b].push_back(a);
+//     }
+//     vector<int> vis(n+1);
+//     vector<int> A(thieves.begin(),thieves.end());
+//     for(int i=0;i<thieves.size();i++)
+//         vis[thieves[i]] = 1;
+//     if(vis[1]==1){
+//         return 0;
+//     }
+//     queue<int> q;
+//     q.push(1);
+//     vis[1] = 1;
+//     int ans = 0;
+//     while(!q.empty()){
+//         int f = q.front();
+//         q.pop();
+//         for(auto itr:adj[f]){
+//             if(vis[itr]==0){
+//                 vis[itr] = 1;
+//                 ans++;
+//                 q.push(itr);
+//             }
+//         }
+//     }
+//     return ans;
+// }
+
+// int main(){
+//     ios::sync_with_stdio(false);
+//     cin.tie(0);
+//     int n,k;
+//     cin>>n>>k;
+//     vector<vector<int> >edges(n-1,vector<int>(2));
+//     for(int i_edges=0;i_edges<n-1;i_edges++){
+//         for(int j_edges=0;j_edges<2;j_edges++){
+//             cin>>edges[i_edges][j_edges];
+//         }
+//     }
+//     vector<int> thieves(k);
+//     for(int i_thieves=0;i_thieves<k;i_thieves++){
+//         cin>>thieves[i_thieves];
+//     }
+//     int out_;
+//     out_ = solution(thieves,edges);
+//     cout<<out_<<endl;
+// }