Create 2818. Apply Operations to Maximize Score

2818. Apply Operations to Maximize Score

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+typedef long long ll;
+class Solution {
+public:
+    // left: greater or equal
+    // right: greater
+    ll mod = 1e9 + 7;
+    ll myPow(ll a, ll b) {
+        ll res = 1;
+        while (a) {
+            if (a & 1) res = (res * b) % mod;
+            a = a >> 1;
+            b = b * b % mod;
+        }
+        return res;
+    }
+    int maximumScore(vector<int>& nums, int k) {
+        int n = nums.size();
+        vector<int> prime(n, 0);
+        for (int i = 0; i < nums.size(); i++) {
+            int cnt = 0;
+            int num = nums[i];
+            for (int j = 2; j * j <= num; j++) {
+                cnt += (num % j) == 0;
+                while ((num % j) == 0) num /= j;
+            }
+            if (num != 1) cnt += 1;
+            prime[i] = cnt;
+        }
+        unordered_map<int, int> leftMap;
+        unordered_map<int, vector<int>> rightMap;
+        for (int i = n - 1; i >= 0; i--) {
+            rightMap[prime[i]].push_back(i);
+        }
+
+        vector<pair<ll, ll>> record;
+        for (ll i = 0; i < n; i++) {
+            rightMap[prime[i]].pop_back();
+            ll leftLen = i + 1;
+            for (int j = prime[i]; j <= 10; j++)
+                if (leftMap.count(j))
+                    leftLen = min(leftLen, i - leftMap[j]);
+            ll rightLen = n - i;
+            for (int j = prime[i] + 1; j <= 10; j++)
+                if (rightMap.count(j) && rightMap[j].size() > 0)
+                    rightLen = min(rightLen, rightMap[j].back() - i);
+            record.push_back({nums[i], leftLen * rightLen});
+            leftMap[prime[i]] = i;
+        }
+        sort(record.begin(), record.end(), greater<>());
+
+        ll ans = 1;
+        for (auto &x: record) {
+            ll tmp = myPow(min(x.second, 1ll * k), x.first);
+            ans = (ans * tmp) % mod;
+            k -= x.second;
+            if (k <= 0) break;
+        }
+        return ans;
+    }
+};