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, 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

    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]