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, 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

    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