Merge pull request #386 from Aaryan-Degama/main

d4q1 done

Author: Dheeraj-06

Problems/Data-structures/Day-04/sol/Aaryan/Solution1.cpp

@@ -0,0 +1,70 @@
+// link  :  https://codeforces.com/problemset/submission/1997/355846660
+
+#include <bits/stdc++.h>
+using namespace std;
+
+using ll = long long;
+#define fastio ios::sync_with_stdio(false); cin.tie(nullptr);
+
+void solve() {
+    ll n;
+    cin >> n;
+    string s;
+    cin >> s;
+
+    s[0] = '(';
+    for (ll i = 1; i < n; i++) {
+        if (i % 2 == 0) {
+            s[i] = (s[i - 1] == '(' ? ')' : '(');
+        }
+    }
+
+    stack<ll> st;
+    ll ans = 0;
+
+    for (ll i = 0; i < n; i++) {
+        if (s[i] == '(') {
+            st.push(i);
+        } else {
+            ans += i - st.top();
+            st.pop();
+        }
+    }
+
+    cout << ans << '\n';
+}
+
+int main() {
+    fastio;
+    int t;
+    cin >> t;
+    while (t--) solve();
+    return 0;
+}
+
+
+/*
+Explanation:
+
+We are given a string of length n consisting of '(' and ')'.
+First, we fix the string to follow a valid alternating pattern:
+- s[0] is always '('
+- For every even index i (> 0), we alternate the bracket based on s[i-1]
+
+After this, the string becomes a valid bracket sequence.
+
+To compute the answer:
+- Use a stack to store indices of '('
+- For every ')', match it with the latest '('
+- Add the distance (current index - matched index) to the answer
+
+This distance represents the contribution of that matched pair.
+
+The stack ensures correct pairing and minimal total cost.
+
+Time Complexity:
+O(n)
+
+Space Complexity:
+O(n)
+*/