Final Array State After K Multiplication Operations II
<p>You are given an integer array <code>nums</code>, an integer <code>k</code>, and an integer <code>multiplier</code>.</p>

<p>You need to perform <code>k</code> operations on <code>nums</code>. In each operation:</p>

<ul>
	<li>Find the <strong>minimum</strong> value <code>x</code> in <code>nums</code>. If there are multiple occurrences of the minimum value, select the one that appears <strong>first</strong>.</li>
	<li>Replace the selected minimum value <code>x</code> with <code>x * multiplier</code>.</li>
</ul>

<p>After the <code>k</code> operations, apply <strong>modulo</strong> <code>10<sup>9</sup> + 7</code> to every value in <code>nums</code>.</p>

<p>Return an integer array denoting the <em>final state</em> of <code>nums</code> after performing all <code>k</code> operations and then applying the modulo.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [2,1,3,5,6], k = 5, multiplier = 2</span></p>

<p><strong>Output:</strong> <span class="example-io">[8,4,6,5,6]</span></p>

<p><strong>Explanation:</strong></p>

<table>
	<tbody>
		<tr>
			<th>Operation</th>
			<th>Result</th>
		</tr>
		<tr>
			<td>After operation 1</td>
			<td>[2, 2, 3, 5, 6]</td>
		</tr>
		<tr>
			<td>After operation 2</td>
			<td>[4, 2, 3, 5, 6]</td>
		</tr>
		<tr>
			<td>After operation 3</td>
			<td>[4, 4, 3, 5, 6]</td>
		</tr>
		<tr>
			<td>After operation 4</td>
			<td>[4, 4, 6, 5, 6]</td>
		</tr>
		<tr>
			<td>After operation 5</td>
			<td>[8, 4, 6, 5, 6]</td>
		</tr>
		<tr>
			<td>After applying modulo</td>
			<td>[8, 4, 6, 5, 6]</td>
		</tr>
	</tbody>
</table>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [100000,2000], k = 2, multiplier = 1000000</span></p>

<p><strong>Output:</strong> <span class="example-io">[999999307,999999993]</span></p>

<p><strong>Explanation:</strong></p>

<table>
	<tbody>
		<tr>
			<th>Operation</th>
			<th>Result</th>
		</tr>
		<tr>
			<td>After operation 1</td>
			<td>[100000, 2000000000]</td>
		</tr>
		<tr>
			<td>After operation 2</td>
			<td>[100000000000, 2000000000]</td>
		</tr>
		<tr>
			<td>After applying modulo</td>
			<td>[999999307, 999999993]</td>
		</tr>
	</tbody>
</table>
</div>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= multiplier &lt;= 10<sup>6</sup></code></li>
</ul>
Array; Simulation; Heap (Priority Queue)
func getFinalState(nums []int, k int, multiplier int) []int {
	if multiplier == 1 {
		return nums
	}
	n := len(nums)
	pq := make(hp, n)
	for i, x := range nums {
		pq[i] = pair{x, i}
	}
	heap.Init(&pq)
	m := slices.Max(nums)
	for ; k > 0 && pq[0].x < m; k-- {
		x := pq[0]
		heap.Pop(&pq)
		x.x *= multiplier
		heap.Push(&pq, x)
	}
	const mod int = 1e9 + 7

	for i := range nums {
		p := heap.Pop(&pq).(pair)
		x, j := p.x, p.i
		power := k / n
		if i < k%n {
			power++
		}
		nums[j] = (x % mod) * qpow(multiplier, power, mod) % mod
	}
	return nums
}

func qpow(a, n, mod int) int {
	ans := 1 % mod
	a = a % mod
	for n > 0 {
		if n&1 == 1 {
			ans = (ans * a) % mod
		}
		a = (a * a) % mod
		n >>= 1
	}
	return int(ans)
}

type pair struct{ x, i int }
type hp []pair

func (h hp) Len() int           { return len(h) }
func (h hp) Less(i, j int) bool { return h[i].x < h[j].x || h[i].x == h[j].x && h[i].i < h[j].i }
func (h hp) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
func (h *hp) Push(x any)        { *h = append(*h, x.(pair)) }
func (h *hp) Pop() (x any)      { a := *h; x = a[len(a)-1]; *h = a[:len(a)-1]; return x }
