<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>

class Solution:
    def getFinalState(self, nums: List[int], k: int, multiplier: int) -> List[int]:
        if multiplier == 1:  # 数组不变
            return nums

        MOD = 1_000_000_007
        n = len(nums)
        mn = min(nums)
        mx = max(nums)
        t = 0
        while mn < mx:
            mn *= multiplier
            t += 1

        if k < t * n:
            # 暴力模拟
            h = [(x, i) for i, x in enumerate(nums)]
            heapify(h)
            for _ in range(k):
                x, i = h[0]
                heapreplace(h, (x * multiplier, i))
            for x, j in h:
                nums[j] = x % MOD
            return nums

        # 每个数直接暴力操作到 >= mx
        for i, x in enumerate(nums):
            while x < mx:
                x *= multiplier
                k -= 1
            nums[i] = x

        # 剩余的操作可以直接用公式计算
        pow1 = pow(multiplier, k // n, MOD)
        pow2 = pow1 * multiplier % MOD
        for i, (x, j) in enumerate(sorted((x, i) for i, x in enumerate(nums))):
            nums[j] = x * (pow2 if i < k % n else pow1) % MOD
        return nums

# 作者：灵茶山艾府
# 链接：https://leetcode.cn/problems/final-array-state-after-k-multiplication-operations-ii/solutions/2892178/zui-xiao-dui-mo-ni-shu-xue-gong-shi-pyth-z4zw/
# 来源：力扣（LeetCode）
# 著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
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