Merge pull request #355 from Aiyaan-Mahajan/DAY-4-Q2

Add files via upload

Author: Dheeraj-06

Problems/Data-structures/Day-04/sol/Aiyaan_Mahajan/Q2sol.cpp

@@ -0,0 +1,100 @@
+/*
+DAY 4
+Q2 Accumulator Apex
+.*/
+/*
+We can only take elements from the left of each list, so every valid move is some prefix of a list.
+
+For each prefix we know:
+
+pref = sum of the prefix
+
+mn = minimum prefix sum while walking through it
+
+A prefix is usable only if:
+
+x + mn >= 0
+
+
+If we take it, we end with:
+
+x += pref
+
+
+We consider all prefixes from all lists and always pick the one that safely gives the most benefit.
+For each list we only keep the best prefix we have already used.
+
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+struct Item {
+    long long pref;
+    long long minPref;
+    int idx;
+};
+
+struct Cmp {
+    bool operator()(const Item& a, const Item& b) const {
+        if (a.minPref != b.minPref) return a.minPref < b.minPref;
+        return a.pref < b.pref;
+    }
+};
+
+int main() {
+ 
+
+    long long x;
+    int k;
+    cin >> x >> k;
+
+    vector<vector<long long>> lists(k);
+    for (int i = 0; i < k; ++i) {
+        int len;
+        cin >> len;
+        lists[i].resize(len);
+        for (int j = 0; j < len; ++j) cin >> lists[i][j];
+    }
+
+    priority_queue<Item, vector<Item>, Cmp> pq;
+    vector<vector<long long>> pref(k);
+
+    for (int i = 0; i < k; ++i) {
+        long long s = 0, mn = 0;
+        pref[i].push_back(0);
+        for (auto v : lists[i]) {
+            s += v;
+            mn = min(mn, s);
+            pref[i].push_back(s);
+            pq.push({s, mn, i});
+        }
+    }
+
+    long long best = x;
+    vector<long long> taken(k, 0);
+
+    while (!pq.empty()) {
+        auto cur = pq.top();
+        pq.pop();
+
+        if (best + cur.minPref < 0) break;
+
+        if (cur.pref > taken[cur.idx]) {
+            best -= taken[cur.idx];
+            best += cur.pref;
+            taken[cur.idx] = cur.pref;
+        }
+    }
+
+    cout << best << "\n";
+    return 0;
+}
+
+
+
+
+//TC =O(Nlog(N)) and SC = O(N)
+/*
+My submission : https://codeforces.com/contest/1912/submission/355690272
+*/