Merge pull request #369 from Naman2251/main

Create day003sol02.cpp

Author: Dheeraj-06

Problems/Data-structures/Day-04/sol/Naman Pal/day004sol01.cpp

@@ -0,0 +1,45 @@
+// Submission Link: https://codeforces.com/contest/1997/submission/355835686
+
+
+/*
+So first I made the RBS with minimmum cost by putting closing bracket as soon as we get a opening bracket before it, i.e., closing bracket as early as 
+possible(greedy appraoch). After getting corect bracket sequence I put the indexes of '(' in vector o and indexes of ')' in vector c in order they occur. Both 
+vector must contain n/2 elements and now for final answer I simply added the difference of each element in o and c at same index to ans to calculate cost 
+which was initially zero and then printed the ans.
+*/
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    long long t;
+    cin>>t;
+    while(t--) {
+        long long n;
+        cin>>n;
+        string s;
+        cin>>s;
+        long long a=0;
+        for(long long i=0; i<n; i++) {
+            if(s[i]=='(') a++;
+            if(s[i]==')') a--;
+            if(s[i]=='_') {
+                if (a>0) {
+                    s[i]=')';
+                    a--;
+                }
+                else {
+                    s[i]='(';
+                    a++;
+                }
+            }
+        }
+        long long ans=0;
+        vector <long long> o, c;
+        for(long long i=0; i<n; i++) {
+            if(s[i]=='(') o.push_back(i);
+            else c.push_back(i);
+        }
+        for(long long i=0; i<n/2; i++) ans+=(c[i]-o[i]);
+        cout<<ans<<endl;
+    }
+}

Problems/Mathematics/Day-03/sol/Naman Pal/day003sol02.cpp

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+Submission Link: https://codeforces.com/contest/343/submission/355828800
+
+/*
+Problem: Find the mininmum number of 1 ohm resistance needed to make equivalent resistance a/b
+
+The Calkin-Wilf Tree
+This problem can be modeled using the Calkin-Wilf Tree, a binary tree that generates every positive rational number exactly once. 
+Starting from 1/1, the two possible operations are:
+    1. Series: a/b -> (a+b)/b (Moves to the Right Child)
+    2. Parallel: a/b -> a/(a+b) (Moves to the Left Child)
+
+Approach: If a resistance a/b can be obained with k resistors then it is clear that we can get resistance (a+b)/b(if 1 ohm in series) or 
+a/(a+b)(if 1 ohm in parellel) with k+1 resistance. So we just go in reverse direction if a is greater than b then the last step must have been adding resistors 
+in series so we add a/b to our answer and make a equal to a%b.. similarly if b is greater than a then The last step must have been adding resistors in parallel so 
+we add b/a in answer and make b=b%a. This process continues untile either a or b becomes zero as either of a and b can become zero only when one is 
+completely divisible by other and in this case we will get minimum number of resistor as the quotient.
+
+Time Complexity: O(log(min(a, b))) as it is equivalent to the Euclidean GCD algorithm
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+ 
+int main() {
+    long long a,b;
+    cin>>a>>b;
+    long long ans=0;
+    while(a>0 && b>0) {
+        if(a>b) {
+            ans+=a/b;
+            a=a%b;
+        }
+        else {
+            ans+=b/a;
+            b=b%a;
+        }
+    }
+    cout<<ans<<endl;
+}