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


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3264" 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.  Return an integer array denoting the final state of nums after performing all k operations.   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]     Example 2:  Input: nums = [1,2], k = 3, multiplier = 4 Output: [16,8] Explanation:    Operation Result   After operation 1 [4, 2]   After operation 2 [4, 8]   After operation 3 [16, 8]       Constraints:  1 <= nums.length <= 100 1 <= nums[i] <= 100 1 <= k <= 10 1 <= multiplier <= 5  

	# Explanation
	*   **Leverage `heapq`:** Use a min-heap to efficiently find the minimum element in each operation. This avoids linear scans of the array.
*   **Heapify Indices:** Store indices in the heap instead of the actual values. This allows quick updates in the original array. We store tuples of (value, index).
*   **Update and Re-heapify:** After updating the minimum element, update its value in the array and re-add its index back into the heap with the new value to maintain the min-heap property.

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

	
	# Code
	```python
	import heapq

def modify_array(nums, k, multiplier):
    """
    Modifies the input array nums by performing k operations.

    Args:
        nums: The input integer array.
        k: The number of operations to perform.
        multiplier: The multiplier to apply to the minimum value.

    Returns:
        The modified array after k operations.
    """

    heap = [(nums[i], i) for i in range(len(nums))]
    heapq.heapify(heap)

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

    return nums
	```