Solving Leetcode Interviews in Seconds with AI: Number of Ways to Separate Numbers
Introduction
In this blog post, we will explore how to solve the LeetCode problem "1977" 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 wrote down many positive integers in a string called num. However, you realized that you forgot to add commas to seperate the different numbers. You remember that the list of integers was non-decreasing and that no integer had leading zeros. Return the number of possible lists of integers that you could have written down to get the string num. Since the answer may be large, return it modulo 109 + 7. Example 1: Input: num = "327" Output: 2 Explanation: You could have written down the numbers: 3, 27 327 Example 2: Input: num = "094" Output: 0 Explanation: No numbers can have leading zeros and all numbers must be positive. Example 3: Input: num = "0" Output: 0 Explanation: No numbers can have leading zeros and all numbers must be positive. Constraints: 1 <= num.length <= 3500 num consists of digits '0' through '9'.
Explanation
Here's the solution:
- Dynamic Programming: Use dynamic programming to store the number of valid lists ending at each prefix of the input string. The state
dp[i]represents the number of non-decreasing lists that can be formed using the firsticharacters ofnum. - Iteration: Iterate through all possible prefixes and consider all possible last numbers in the list. The last number must be a valid integer (no leading zeros) and greater than or equal to the previous number in the list.
Comparison: Compare the current potential last number with the previous number (if there is one). If the current number is smaller, then we can't use it. Otherwise, add the number of ways to form a list ending with the previous number to the current count.
Runtime Complexity: O(N^2)
- Storage Complexity: O(N^2)
Code
def solve(): num = input()
n = len(num)
mod = 10**9 + 7
if n > 1 and num[0] == '0':
print(0)
return
if n == 1 and num[0] == '0':
print(0)
return
dp = [[0] * (n + 1) for _ in range(n + 1)]
dp[0][0] = 1
for i in range(1, n + 1):
for j in range(1, i + 1):
# num[i-j:i] is the current number
current_num_str = num[i-j:i]
if current_num_str[0] == '0' and len(current_num_str) > 1:
continue
current_num = int(current_num_str)
if i - j == 0:
dp[i][j] = dp[i][j] + dp[i - j][0]
dp[i][j] %= mod
continue
for k in range(1, i - j + 1):
prev_num_str = num[i - j - k:i - j]
if prev_num_str[0] == '0' and len(prev_num_str) > 1:
continue
prev_num = int(prev_num_str)
if len(current_num_str) > len(prev_num_str):
dp[i][j] = (dp[i][j] + dp[i-j][k]) % mod
elif len(current_num_str) < len(prev_num_str):
continue
else:
if current_num >= prev_num:
dp[i][j] = (dp[i][j] + dp[i-j][k]) % mod
ans = 0
for j in range(1, n + 1):
ans = (ans + dp[n][j]) % mod
print(ans)
solve()