Solving Leetcode Interviews in Seconds with AI: Subsequences with a Unique Middle Mode I
Introduction
In this blog post, we will explore how to solve the LeetCode problem "3395" 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
Given an integer array nums, find the number of subsequences of size 5 of nums with a unique middle mode. Since the answer may be very large, return it modulo 109 + 7. A mode of a sequence of numbers is defined as the element that appears the maximum number of times in the sequence. A sequence of numbers contains a unique mode if it has only one mode. A sequence of numbers seq of size 5 contains a unique middle mode if the middle element (seq[2]) is a unique mode. Example 1: Input: nums = [1,1,1,1,1,1] Output: 6 Explanation: [1, 1, 1, 1, 1] is the only subsequence of size 5 that can be formed, and it has a unique middle mode of 1. This subsequence can be formed in 6 different ways, so the output is 6. Example 2: Input: nums = [1,2,2,3,3,4] Output: 4 Explanation: [1, 2, 2, 3, 4] and [1, 2, 3, 3, 4] each have a unique middle mode because the number at index 2 has the greatest frequency in the subsequence. [1, 2, 2, 3, 3] does not have a unique middle mode because 2 and 3 appear twice. Example 3: Input: nums = [0,1,2,3,4,5,6,7,8] Output: 0 Explanation: There is no subsequence of length 5 with a unique middle mode. Constraints: 5 <= nums.length <= 1000 -109 <= nums[i] <= 109
Explanation
Here's the breakdown of the solution:
- Iterate and Count: The core idea is to iterate through each element of the input array
numsand consider it as the middle element of a potential subsequence of length 5. - Count Combinations: For each middle element, we count the number of ways to choose two elements from the left side (smaller indices) and two elements from the right side (larger indices), ensuring that the middle element remains the unique mode in the formed subsequence. This involves counting elements smaller, equal, and greater than the middle element.
Modular Arithmetic: Since the answer can be large, we perform all calculations modulo 109 + 7 to prevent overflow.
Runtime Complexity: O(n3), Storage Complexity: O(1)
def solve():
def count_subsequences(nums):
n = len(nums)
MOD = 10**9 + 7
count = 0
for i in range(2, n - 2):
mid_val = nums[i]
# Count elements to the left
less_left = 0
equal_left = 0
for j in range(i):
if nums[j] < mid_val:
less_left += 1
elif nums[j] == mid_val:
equal_left += 1
# Count elements to the right
less_right = 0
equal_right = 0
for j in range(i + 1, n):
if nums[j] < mid_val:
less_right += 1
elif nums[j] == mid_val:
equal_right += 1
# Calculate combinations for left side
left_combinations = 0
if less_left >= 2:
left_combinations = (left_combinations + (less_left * (less_left - 1) // 2) % MOD) % MOD
if less_left >= 1 and equal_left >= 1:
left_combinations = (left_combinations + (less_left * equal_left) % MOD) % MOD
if equal_left >= 2:
left_combinations = (left_combinations + (equal_left * (equal_left - 1) // 2) % MOD) % MOD
# Calculate combinations for right side
right_combinations = 0
if less_right >= 2:
right_combinations = (right_combinations + (less_right * (less_right - 1) // 2) % MOD) % MOD
if less_right >= 1 and equal_right >= 1:
right_combinations = (right_combinations + (less_right * equal_right) % MOD) % MOD
if equal_right >= 2:
right_combinations = (right_combinations + (equal_right * (equal_right - 1) // 2) % MOD) % MOD
# Consider cases where middle element appears more than once
# and is still the unique mode
combinations = 0
if equal_left + equal_right == 0:
if less_left >= 2 and less_right >= 2:
combinations = (less_left * (less_left - 1) // 2) % MOD
combinations = (combinations * (less_right * (less_right - 1) // 2) % MOD) % MOD
elif equal_left + equal_right == 1:
if less_left >= 2 and less_right >= 1:
combinations = (less_left * (less_left - 1) // 2) % MOD
combinations = (combinations * less_right) % MOD
if less_left >= 1 and less_right >= 2:
combinations = (less_right * (less_right - 1) // 2) % MOD
combinations = (combinations * less_left) % MOD
elif equal_left + equal_right == 2:
if less_left >= 2 and less_right >= 0:
combinations = (less_left * (less_left - 1) // 2) % MOD
if less_left >= 0 and less_right >= 2:
combinations = (less_right * (less_right-1) // 2) % MOD
if less_left >= 1 and less_right >= 1:
combinations = (less_left * less_right) % MOD
count = (count + combinations) % MOD
return count
nums = [int(x) for x in input().split()]
print(count_subsequences(nums))
# Code
```python
def solve(): def count_subsequences(nums):
n = len(nums)
MOD = 10**9 + 7
count = 0
for i in range(2, n - 2):
mid_val = nums[i]
# Count elements to the left
less_left = 0
equal_left = 0
for j in range(i):
if nums[j] < mid_val:
less_left += 1
elif nums[j] == mid_val:
equal_left += 1
# Count elements to the right
less_right = 0
equal_right = 0
for j in range(i + 1, n):
if nums[j] < mid_val:
less_right += 1
elif nums[j] == mid_val:
equal_right += 1
# Calculate combinations for left side
left_combinations = 0
if less_left >= 2:
left_combinations = (left_combinations + (less_left * (less_left - 1) // 2) % MOD) % MOD
if less_left >= 1 and equal_left >= 1:
left_combinations = (left_combinations + (less_left * equal_left) % MOD) % MOD
if equal_left >= 2:
left_combinations = (left_combinations + (equal_left * (equal_left - 1) // 2) % MOD) % MOD
# Calculate combinations for right side
right_combinations = 0
if less_right >= 2:
right_combinations = (right_combinations + (less_right * (less_right - 1) // 2) % MOD) % MOD
if less_right >= 1 and equal_right >= 1:
right_combinations = (right_combinations + (less_right * equal_right) % MOD) % MOD
if equal_right >= 2:
right_combinations = (right_combinations + (equal_right * (equal_right - 1) // 2) % MOD) % MOD
# Consider cases where middle element appears more than once
# and is still the unique mode
combinations = 0
if equal_left + equal_right == 0:
if less_left >= 2 and less_right >= 2:
combinations = (less_left * (less_left - 1) // 2) % MOD
combinations = (combinations * (less_right * (less_right - 1) // 2) % MOD) % MOD
elif equal_left + equal_right == 1:
if less_left >= 2 and less_right >= 1:
combinations = (less_left * (less_left - 1) // 2) % MOD
combinations = (combinations * less_right) % MOD
if less_left >= 1 and less_right >= 2:
combinations = (less_right * (less_right - 1) // 2) % MOD
combinations = (combinations * less_left) % MOD
elif equal_left + equal_right == 2:
if less_left >= 2 and less_right >= 0:
combinations = (less_left * (less_left - 1) // 2) % MOD
if less_left >= 0 and less_right >= 2:
combinations = (less_right * (less_right-1) // 2) % MOD
if less_left >= 1 and less_right >= 1:
combinations = (less_left * less_right) % MOD
count = (count + combinations) % MOD
return count
nums = [int(x) for x in input().split()]
print(count_subsequences(nums))
solve()