Solving Leetcode Interviews in Seconds with AI: Apply Operations to Maximize Score
Introduction
In this blog post, we will explore how to solve the LeetCode problem "2818" 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 array nums of n positive integers and an integer k. Initially, you start with a score of 1. You have to maximize your score by applying the following operation at most k times: Choose any non-empty subarray nums[l, ..., r] that you haven't chosen previously. Choose an element x of nums[l, ..., r] with the highest prime score. If multiple such elements exist, choose the one with the smallest index. Multiply your score by x. Here, nums[l, ..., r] denotes the subarray of nums starting at index l and ending at the index r, both ends being inclusive. The prime score of an integer x is equal to the number of distinct prime factors of x. For example, the prime score of 300 is 3 since 300 = 2 2 3 5 5. Return the maximum possible score after applying at most k operations. Since the answer may be large, return it modulo 109 + 7. Example 1: Input: nums = [8,3,9,3,8], k = 2 Output: 81 Explanation: To get a score of 81, we can apply the following operations: - Choose subarray nums[2, ..., 2]. nums[2] is the only element in this subarray. Hence, we multiply the score by nums[2]. The score becomes 1 9 = 9. - Choose subarray nums[2, ..., 3]. Both nums[2] and nums[3] have a prime score of 1, but nums[2] has the smaller index. Hence, we multiply the score by nums[2]. The score becomes 9 9 = 81. It can be proven that 81 is the highest score one can obtain. Example 2: Input: nums = [19,12,14,6,10,18], k = 3 Output: 4788 Explanation: To get a score of 4788, we can apply the following operations: - Choose subarray nums[0, ..., 0]. nums[0] is the only element in this subarray. Hence, we multiply the score by nums[0]. The score becomes 1 19 = 19. - Choose subarray nums[5, ..., 5]. nums[5] is the only element in this subarray. Hence, we multiply the score by nums[5]. The score becomes 19 18 = 342. - Choose subarray nums[2, ..., 3]. Both nums[2] and nums[3] have a prime score of 2, but nums[2] has the smaller index. Hence, we multipy the score by nums[2]. The score becomes 342 14 = 4788. It can be proven that 4788 is the highest score one can obtain. Constraints: 1 <= nums.length == n <= 105 1 <= nums[i] <= 105 1 <= k <= min(n (n + 1) / 2, 109)
Explanation
Here's a breakdown of the optimal approach, followed by the Python code:
- Calculate Prime Scores: Precompute the prime scores for each number from 1 to 105 using the Sieve of Eratosthenes. This is crucial for efficiency as we'll be repeatedly querying prime scores.
- Greedy Subarray Selection: Use a priority queue (max heap) to store all possible subarrays. The priority of a subarray is determined by the highest prime score within it, and in case of ties, the smallest index of the element with the highest prime score. Initially, add all possible subarrays to the priority queue.
Iterative Score Maximization: Extract the subarray with the highest priority from the priority queue k times. Multiply the current score by the element selected from the extracted subarray. To avoid re-selecting the same subarray, mark subarrays as visited (or remove them from consideration in some other way).
Time Complexity: O(N2 log(N) + k log(N)), where N is the length of
nums. The O(N2 log(N)) part comes from building the initial priority queue of all subarrays, and the k log(N) factor from extracting the maximum k times. Computing the prime scores takes O(N log log N), which is dominated by O(N2 log N).- Space Complexity: O(N2) to store all the subarrays in the priority queue.
import heapq
def solve():
n, k = map(int, input().split())
nums = list(map(int, input().split()))
MOD = 10**9 + 7
MAX_NUM = 10**5 + 1
# Precompute prime scores using Sieve of Eratosthenes
prime_scores = [0] * MAX_NUM
for i in range(2, MAX_NUM):
if prime_scores[i] == 0:
for j in range(i, MAX_NUM, i):
prime_scores[j] += 1
# Create a priority queue to store subarrays (max heap)
pq = []
subarray_selected = set()
for i in range(n):
for j in range(i, n):
subarray = (i, j)
max_prime_score = -1
best_index = -1
for l in range(i,j+1):
if prime_scores[nums[l]] > max_prime_score:
max_prime_score = prime_scores[nums[l]]
best_index = l
elif prime_scores[nums[l]] == max_prime_score and l < best_index:
best_index = l
heapq.heappush(pq, (-max_prime_score, -best_index, i, j))
score = 1
operations = 0
while operations < k and pq:
neg_max_prime_score, neg_best_index, start, end = heapq.heappop(pq)
best_index = -neg_best_index
# Check for duplicates is no longer needed since we add new subarrays
score = (score * nums[best_index]) % MOD
operations += 1
print(score)
# Example usage: (remove input prompts for coding platforms)
# n, k = map(int, input().split())
# nums = list(map(int, input().split()))
# solve(nums, k)
def solve_example(nums, k): # Helper function to call from example problems
MOD = 10**9 + 7
MAX_NUM = 10**5 + 1
# Precompute prime scores using Sieve of Eratosthenes
prime_scores = [0] * MAX_NUM
for i in range(2, MAX_NUM):
if prime_scores[i] == 0:
for j in range(i, MAX_NUM, i):
prime_scores[j] += 1
# Create a priority queue to store subarrays (max heap)
pq = []
for i in range(len(nums)):
for j in range(i, len(nums)):
subarray = (i, j)
max_prime_score = -1
best_index = -1
for l in range(i,j+1):
if prime_scores[nums[l]] > max_prime_score:
max_prime_score = prime_scores[nums[l]]
best_index = l
elif prime_scores[nums[l]] == max_prime_score and l < best_index:
best_index = l
heapq.heappush(pq, (-max_prime_score, -best_index, i, j))
score = 1
operations = 0
while operations < k and pq:
neg_max_prime_score, neg_best_index, start, end = heapq.heappop(pq)
best_index = -neg_best_index
score = (score * nums[best_index]) % MOD
operations += 1
return score
</code>
# Code
```python
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