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
function maximumScore(nums: number[], k: number): number {
    const mod = 10 ** 9 + 7;
    const n = nums.length;
    const arr: number[][] = Array(n)
        .fill(0)
        .map(() => Array(3).fill(0));
    const left: number[] = Array(n).fill(-1);
    const right: number[] = Array(n).fill(n);
    for (let i = 0; i < n; ++i) {
        arr[i] = [i, primeFactors(nums[i]), nums[i]];
    }
    const stk: number[] = [];
    for (const [i, f, _] of arr) {
        while (stk.length && arr[stk.at(-1)!][1] < f) {
            stk.pop();
        }
        if (stk.length) {
            left[i] = stk.at(-1)!;
        }
        stk.push(i);
    }
    stk.length = 0;
    for (let i = n - 1; i >= 0; --i) {
        const f = arr[i][1];
        while (stk.length && arr[stk.at(-1)!][1] <= f) {
            stk.pop();
        }
        if (stk.length) {
            right[i] = stk.at(-1)!;
        }
        stk.push(i);
    }
    arr.sort((a, b) => b[2] - a[2]);
    let ans = 1n;
    for (const [i, _, x] of arr) {
        const l = left[i];
        const r = right[i];
        const cnt = (i - l) * (r - i);
        if (cnt <= k) {
            ans = (ans * qpow(BigInt(x), cnt, mod)) % BigInt(mod);
            k -= cnt;
        } else {
            ans = (ans * qpow(BigInt(x), k, mod)) % BigInt(mod);
            break;
        }
    }
    return Number(ans);
}

function primeFactors(n: number): number {
    let i = 2;
    const s: Set<number> = new Set();
    while (i * i <= n) {
        while (n % i === 0) {
            s.add(i);
            n = Math.floor(n / i);
        }
        ++i;
    }
    if (n > 1) {
        s.add(n);
    }
    return s.size;
}

function qpow(a: bigint, n: number, mod: number): bigint {
    let ans = 1n;
    for (; n; n >>>= 1) {
        if (n & 1) {
            ans = (ans * a) % BigInt(mod);
        }
        a = (a * a) % BigInt(mod);
    }
    return ans;
}
