# Solving Leetcode Interviews in Seconds with AI: Final Array State After K Multiplication Operations II


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3266" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given an integer array nums, an integer k, and an integer multiplier. You need to perform k operations on nums. In each operation:  Find the minimum value x in nums. If there are multiple occurrences of the minimum value, select the one that appears first. Replace the selected minimum value x with x * multiplier.  After the k operations, apply modulo 109 + 7 to every value in nums. Return an integer array denoting the final state of nums after performing all k operations and then applying the modulo.   Example 1:  Input: nums = [2,1,3,5,6], k = 5, multiplier = 2 Output: [8,4,6,5,6] Explanation:    Operation Result   After operation 1 [2, 2, 3, 5, 6]   After operation 2 [4, 2, 3, 5, 6]   After operation 3 [4, 4, 3, 5, 6]   After operation 4 [4, 4, 6, 5, 6]   After operation 5 [8, 4, 6, 5, 6]   After applying modulo [8, 4, 6, 5, 6]     Example 2:  Input: nums = [100000,2000], k = 2, multiplier = 1000000 Output: [999999307,999999993] Explanation:    Operation Result   After operation 1 [100000, 2000000000]   After operation 2 [100000000000, 2000000000]   After applying modulo [999999307, 999999993]       Constraints:  1 <= nums.length <= 104 1 <= nums[i] <= 109 1 <= k <= 109 1 <= multiplier <= 106  

	# Explanation
	Here's a breakdown of the approach, followed by the Python code:

*   **Priority Queue:** Use a min-heap (priority queue) to efficiently find the minimum element in `nums` in each operation. This avoids a full scan of the array each time.
*   **Apply Multiplier and Modulo:** After extracting the minimum, multiply it by the `multiplier`, and update the value in the array. Take modulo `10^9 + 7` after each multiplication to avoid overflow.
*   **Optimize Operations:** Because `k` can be very large we might perform more operations than the array size. So, we need to keep track of how many times each element was multiplied.

*   **Runtime Complexity:** O(k log n), where n is the length of `nums` and k is the number of operations. **Storage Complexity:** O(n).

	
	# Code
	```python
	import heapq

def solve():
    nums = [2,1,3,5,6]
    k = 5
    multiplier = 2
    expected = [8,4,6,5,6]
    actual = solve_internal(nums, k, multiplier)
    assert actual == expected

    nums = [100000,2000]
    k = 2
    multiplier = 1000000
    expected = [999999307,999999993]
    actual = solve_internal(nums, k, multiplier)
    assert actual == expected

def solve_internal(nums, k, multiplier):
    """
    Performs k operations on nums, multiplying the minimum element by multiplier each time,
    and applies modulo 10^9 + 7 to every value in nums after the operations.
    """
    n = len(nums)
    heap = [(nums[i], i) for i in range(n)]  # (value, index)
    heapq.heapify(heap)
    
    modulo = 10**9 + 7

    for _ in range(k):
        val, idx = heapq.heappop(heap)
        
        new_val = (val * multiplier) % modulo
        nums[idx] = new_val
        heapq.heappush(heap, (new_val, idx))

    return [num % modulo for num in nums]
	```