Solution of problem 1 : CPP

.vscode/settings.json

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+{
+    "files.associations": {
+        "iostream": "cpp"
+    }
+}

Problem Statement 1/mohitProb1.cpp

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+#include <bits/stdc++.h>
+using namespace std;
+
+#define ll long long
+#define double long double
+#define pb push_back
+#define all(v) v.begin(), v.end()
+#define allr(v) v.rbegin(), v.rend()
+#define sz(v) (int)v.size()
+#define deb(x) cout << #x << "=" << x << endl;
+
+const int mod = 1e9 + 7;
+const int mod2 = 998244353;
+const double PI = 3.1415926535897932384626433832795;
+
+//EXPLAINATION :
+
+//the number n is between 0 < x < n, so it can't hold value equal to 0 and n.
+// so if we take example of n=4 (even), then only value x=1,2,3 exists, which means only Anuj won 1 and avni won 1, then  anuj won 2 , and avni has no more moves.
+// hence anuj won and true is returned
+
+// consider n=3(odd), then only values x=1,2 exists, which means anuj won 1, then avni won 1 and then moves are over. so anuj did'nt won last match
+// so he lost and false is returned. 
+
+
+int solve(int x)
+{
+    if(x%2==0){
+        // cout<<"yes"<<endl;
+        return true;
+    }
+    else{
+        // cout<<"no"<<endl;
+        return false;
+    }
+
+
+}
+
+
+int32_t main()
+{
+    cin.tie(0)->sync_with_stdio(0);
+    int n; cin>>n;
+    solve(n);
+
+}

Problem Statement 2/mohitProb2.cpp

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+#include <bits/stdc++.h>
+using namespace std;
+
+#define ll long long
+#define double long double
+#define pb push_back
+#define all(v) v.begin(), v.end()
+#define allr(v) v.rbegin(), v.rend()
+#define sz(v) (int)v.size()
+#define deb(x) cout << #x << "=" << x << endl;
+
+const int mod = 1e9 + 7;
+const int mod2 = 998244353;
+const double PI = 3.1415926535897932384626433832795;
+
+//EXPLAINATION :
+
+//here we have 5 characters, and we need to output pairs of string on increasing the value of n
+// for n=1, we get a,e,i,o,u, which means output : 5 (numbers)
+//for n=2, we get ["aa","ae","ai","ao","au","ee","ei","eo","eu","ii","io","iu","oo","ou","uu"]., we get 15 strings so output = 15
+
+// so simply, if we suppose every vowel equal to 1, and we sum it up, on every iteration till n==0, then we get the solution,
+// it is because it simply calculates the pairs, so for n=1, it got 5, so if we are finding n=2, the data will iterate from 2 to 0, so the answer will be
+//will be 15
+
+
+int solve(int x)
+{
+    int a=1,e=1,i=1,o=1,u=1;
+
+    while(x--){
+        a=a;
+        e= e +a;
+        i= i+e;
+        o= o+i;
+        u= u+o;
+
+
+    }
+    // cout<<a+e+i+o+u<<endl;
+
+    return a+e+i+o+u;    
+
+
+}
+
+
+int32_t main()
+{
+    cin.tie(0)->sync_with_stdio(0);
+    int n; cin>>n;
+    solve(n);
+
+}

Problem Statement 3/mohitProb3.cpp

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+#include <bits/stdc++.h>
+using namespace std;
+
+#define ll long long
+#define double long double
+#define pb push_back
+#define all(v) v.begin(), v.end()
+#define allr(v) v.rbegin(), v.rend()
+#define sz(v) (int)v.size()
+#define deb(x) cout << #x << "=" << x << endl;
+
+const int mod = 1e9 + 7;
+const int mod2 = 998244353;
+const double PI = 3.1415926535897932384626433832795;
+
+//EXPLAINATION :
+
+// in this problem, I used DP. basically I am creating an array that keeps the max solution stored.
+//in addition to it, I have added two variables, namely current_max and current_sol, so we are checking the value iteratively and
+//keeping the value stored
+// at the end we are returning the sum of elements of maximum array created
+
+
+
+int solve(vector<int>&v, int k)
+{
+    int n = v.size();
+    vector<int> letsUseDP(n+1, 0);
+
+		  for(int i=1; i<=n; i++){
+              int current_sol = 0;
+              int current_max = 0;
+
+
+			  for(int j=1; j<=k && i-j>=0; j++){
+				  current_max = max(current_max, v[i-j]);
+				  current_sol = max(current_sol, letsUseDP[i-j] + current_max * j);
+			  }
+			  letsUseDP[i] = current_sol;
+		  }
+
+        //   cout<<dp[n];
+		   return letsUseDP[n];
+
+
+
+}
+