Merge pull request #387 from Aaryan-Degama/d4q2

D4q2

Author: Dheeraj-06

Problems/Data-structures/Day-04/sol/Aaryan/Solution2.cpp

@@ -0,0 +1,63 @@
+// link : https://codeforces.com/problemset/submission/1912/355879512
+
+#include <bits/stdc++.h>
+using namespace std;
+
+using ll = long long;
+using vi = vector<pair<ll, ll>>;
+
+int main() {
+    ll init_pow, k;
+    cin >> init_pow >> k;
+
+    vi seg;
+
+    for (ll grp = 0; grp < k; grp++) {
+        ll n;
+        cin >> n;
+
+        ll cur = 0, min_pref = 0;
+
+        for (ll i = 0; i < n; i++) {
+            ll x;
+            cin >> x;
+
+            cur += x;
+            min_pref = max(min_pref, -cur);
+
+            if (cur > 0) {
+                seg.push_back({min_pref, cur});
+                cur = 0;
+            }
+        }
+    }
+
+    sort(seg.begin(), seg.end());
+
+    for (auto p : seg) {
+        if (p.first > init_pow) break;
+        init_pow += p.second;
+    }
+
+    cout << init_pow << '\n';
+    return 0;
+}
+
+
+/*
+Explanation:
+
+Each group is split into profitable segments. While reading a group, `cur`
+stores the running sum and `min_pref` stores the maximum deficit
+(-prefix sum), i.e., the minimum power required to start that segment.
+Whenever `cur > 0`, the segment is stored as (min_pref, cur) and `cur` is reset.
+
+All segments are sorted by required power and processed greedily.
+If the required power of a segment exceeds current power, we stop;
+otherwise, we add its gain.
+
+Time Complexity: O(N log N)
+Space Complexity: O(N).
+
+
+*/