Solving Leetcode Interviews in Seconds with AI: Find the Maximum Factor Score of Array
Introduction
In this blog post, we will explore how to solve the LeetCode problem "3334" 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. The factor score of an array is defined as the product of the LCM and GCD of all elements of that array. Return the maximum factor score of nums after removing at most one element from it. Note that both the LCM and GCD of a single number are the number itself, and the factor score of an empty array is 0. Example 1: Input: nums = [2,4,8,16] Output: 64 Explanation: On removing 2, the GCD of the rest of the elements is 4 while the LCM is 16, which gives a maximum factor score of 4 * 16 = 64. Example 2: Input: nums = [1,2,3,4,5] Output: 60 Explanation: The maximum factor score of 60 can be obtained without removing any elements. Example 3: Input: nums = [3] Output: 9 Constraints: 1 <= nums.length <= 100 1 <= nums[i] <= 30
Explanation
Here's an efficient solution to the problem:
High-Level Approach: Iterate through the input array
nums. For each element, temporarily remove it and calculate the factor score (GCD * LCM) of the remaining elements. Keep track of the maximum factor score found so far. If the array has only one element, handle that as a special case.Complexity:
- Runtime: O(n^2 * log(max(nums))) - due to iterating through the array (O(n)) and calculating GCD/LCM within the loop (GCD/LCM operations take O(log(max(nums))) time in the worst case, performed potentially O(n) times).
- Storage: O(1) - constant extra space is used.
import math
def gcd(a, b):
if b == 0:
return a
return gcd(b, a % b)
def lcm(a, b):
return (a * b) // gcd(a, b)
def solve():
nums = [int(x) for x in input().split(",")]
n = len(nums)
if n == 0:
print(0)
return
if n == 1:
print(nums[0] * nums[0]) # Factor score of a single element is the number squared
return
max_factor_score = 0
for i in range(n):
temp_nums = nums[:i] + nums[i+1:]
if not temp_nums:
max_factor_score = max(max_factor_score, 0)
continue
current_gcd = temp_nums[0]
current_lcm = temp_nums[0]
for j in range(1, len(temp_nums)):
current_gcd = gcd(current_gcd, temp_nums[j])
current_lcm = lcm(current_lcm, temp_nums[j])
max_factor_score = max(max_factor_score, current_gcd * current_lcm)
print(max_factor_score)
# Example usage (replace input with your actual input method)
# For testing in environment where input() doesn't exist,
# you can comment out solve() and uncomment the following lines
# and call solve_test():
# nums = [2, 4, 8, 16]
# nums = [1, 2, 3, 4, 5]
# nums = [3]
# nums = [6,10,15] # Output: 90
# solve()
def solve_test():
# Test cases
test_cases = [
([2, 4, 8, 16], 64),
([1, 2, 3, 4, 5], 60),
([3], 9),
([6, 10, 15], 90),
([1,1,1], 1),
([2,2,2], 4),
([12, 8, 24], 96)
]
for nums, expected in test_cases:
n = len(nums)
if n == 0:
result = 0
elif n == 1:
result = nums[0] * nums[0]
else:
max_factor_score = 0
for i in range(n):
temp_nums = nums[:i] + nums[i+1:]
if not temp_nums:
max_factor_score = max(max_factor_score, 0)
continue
current_gcd = temp_nums[0]
current_lcm = temp_nums[0]
for j in range(1, len(temp_nums)):
current_gcd = gcd(current_gcd, temp_nums[j])
current_lcm = lcm(current_lcm, temp_nums[j])
max_factor_score = max(max_factor_score, current_gcd * current_lcm)
result = max_factor_score
if result == expected:
print(f"Test case {nums}: Passed, result = {result}")
else:
print(f"Test case {nums}: Failed, expected = {expected}, got = {result}")
# Code
```python
import math
def gcd(a, b):
if b == 0:
return a
return gcd(b, a % b)
def lcm(a, b):
return (a * b) // gcd(a, b)
def solve():
nums = [int(x) for x in input().split(",")]
n = len(nums)
if n == 0:
print(0)
return
if n == 1:
print(nums[0] * nums[0]) # Factor score of a single element is the number squared
return
max_factor_score = 0
for i in range(n):
temp_nums = nums[:i] + nums[i+1:]
if not temp_nums:
max_factor_score = max(max_factor_score, 0)
continue
current_gcd = temp_nums[0]
current_lcm = temp_nums[0]
for j in range(1, len(temp_nums)):
current_gcd = gcd(current_gcd, temp_nums[j])
current_lcm = lcm(current_lcm, temp_nums[j])
max_factor_score = max(max_factor_score, current_gcd * current_lcm)
print(max_factor_score)
# Example usage (replace input with your actual input method)
# For testing in environment where input() doesn't exist,
# you can comment out solve() and uncomment the following lines
# and call solve_test():
# nums = [2, 4, 8, 16]
# nums = [1, 2, 3, 4, 5]
# nums = [3]
# nums = [6,10,15] # Output: 90
# solve()
def solve_test():
# Test cases
test_cases = [
([2, 4, 8, 16], 64),
([1, 2, 3, 4, 5], 60),
([3], 9),
([6, 10, 15], 90),
([1,1,1], 1),
([2,2,2], 4),
([12, 8, 24], 96)
]
for nums, expected in test_cases:
n = len(nums)
if n == 0:
result = 0
elif n == 1:
result = nums[0] * nums[0]
else:
max_factor_score = 0
for i in range(n):
temp_nums = nums[:i] + nums[i+1:]
if not temp_nums:
max_factor_score = max(max_factor_score, 0)
continue
current_gcd = temp_nums[0]
current_lcm = temp_nums[0]
for j in range(1, len(temp_nums)):
current_gcd = gcd(current_gcd, temp_nums[j])
current_lcm = lcm(current_lcm, temp_nums[j])
max_factor_score = max(max_factor_score, current_gcd * current_lcm)
result = max_factor_score
if result == expected:
print(f"Test case {nums}: Passed, result = {result}")
else:
print(f"Test case {nums}: Failed, expected = {expected}, got = {result}")