Apply Operations to Maximize Score
<p>You are given an array <code>nums</code> of <code>n</code> positive integers and an integer <code>k</code>.</p>

<p>Initially, you start with a score of <code>1</code>. You have to maximize your score by applying the following operation at most <code>k</code> times:</p>

<ul>
	<li>Choose any <strong>non-empty</strong> subarray <code>nums[l, ..., r]</code> that you haven&#39;t chosen previously.</li>
	<li>Choose an element <code>x</code> of <code>nums[l, ..., r]</code> with the highest <strong>prime score</strong>. If multiple such elements exist, choose the one with the smallest index.</li>
	<li>Multiply your score by <code>x</code>.</li>
</ul>

<p>Here, <code>nums[l, ..., r]</code> denotes the subarray of <code>nums</code> starting at index <code>l</code> and ending at the index <code>r</code>, both ends being inclusive.</p>

<p>The <strong>prime score</strong> of an integer <code>x</code> is equal to the number of distinct prime factors of <code>x</code>. For example, the prime score of <code>300</code> is <code>3</code> since <code>300 = 2 * 2 * 3 * 5 * 5</code>.</p>

<p>Return <em>the <strong>maximum possible score</strong> after applying at most </em><code>k</code><em> operations</em>.</p>

<p>Since the answer may be large, return it modulo <code>10<sup>9 </sup>+ 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [8,3,9,3,8], k = 2
<strong>Output:</strong> 81
<strong>Explanation:</strong> To get a score of 81, we can apply the following operations:
- Choose subarray nums[2, ..., 2]. nums[2] is the only element in this subarray. Hence, we multiply the score by nums[2]. The score becomes 1 * 9 = 9.
- Choose subarray nums[2, ..., 3]. Both nums[2] and nums[3] have a prime score of 1, but nums[2] has the smaller index. Hence, we multiply the score by nums[2]. The score becomes 9 * 9 = 81.
It can be proven that 81 is the highest score one can obtain.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [19,12,14,6,10,18], k = 3
<strong>Output:</strong> 4788
<strong>Explanation:</strong> To get a score of 4788, we can apply the following operations: 
- Choose subarray nums[0, ..., 0]. nums[0] is the only element in this subarray. Hence, we multiply the score by nums[0]. The score becomes 1 * 19 = 19.
- Choose subarray nums[5, ..., 5]. nums[5] is the only element in this subarray. Hence, we multiply the score by nums[5]. The score becomes 19 * 18 = 342.
- Choose subarray nums[2, ..., 3]. Both nums[2] and nums[3] have a prime score of 2, but nums[2] has the smaller index. Hence, we multipy the score by nums[2]. The score becomes 342 * 14 = 4788.
It can be proven that 4788 is the highest score one can obtain.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length == n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= k &lt;= min(n * (n + 1) / 2, 10<sup>9</sup>)</code></li>
</ul>
Stack; Greedy; Array; Math; Number Theory; Sorting; Monotonic Stack
class Solution {
public:
    int maximumScore(vector<int>& nums, int k) {
        const int mod = 1e9 + 7;
        int n = nums.size();
        vector<tuple<int, int, int>> arr(n);
        for (int i = 0; i < n; ++i) {
            arr[i] = {i, primeFactors(nums[i]), nums[i]};
        }
        vector<int> left(n, -1);
        vector<int> right(n, n);
        stack<int> stk;
        for (auto [i, f, _] : arr) {
            while (!stk.empty() && get<1>(arr[stk.top()]) < f) {
                stk.pop();
            }
            if (!stk.empty()) {
                left[i] = stk.top();
            }
            stk.push(i);
        }
        stk = stack<int>();
        for (int i = n - 1; ~i; --i) {
            int f = get<1>(arr[i]);
            while (!stk.empty() && get<1>(arr[stk.top()]) <= f) {
                stk.pop();
            }
            if (!stk.empty()) {
                right[i] = stk.top();
            }
            stk.push(i);
        }
        sort(arr.begin(), arr.end(), [](const auto& lhs, const auto& rhs) {
            return get<2>(rhs) < get<2>(lhs);
        });
        long long ans = 1;
        auto qpow = [&](long long a, int n) {
            long long ans = 1;
            for (; n; n >>= 1) {
                if (n & 1) {
                    ans = ans * a % mod;
                }
                a = a * a % mod;
            }
            return ans;
        };
        for (auto [i, _, x] : arr) {
            int l = left[i], r = right[i];
            long long cnt = 1LL * (i - l) * (r - i);
            if (cnt <= k) {
                ans = ans * qpow(x, cnt) % mod;
                k -= cnt;
            } else {
                ans = ans * qpow(x, k) % mod;
                break;
            }
        }
        return ans;
    }

    int primeFactors(int n) {
        int i = 2;
        unordered_set<int> ans;
        while (i <= n / i) {
            while (n % i == 0) {
                ans.insert(i);
                n /= i;
            }
            ++i;
        }
        if (n > 1) {
            ans.insert(n);
        }
        return ans.size();
    }
};