Introduction

In this blog post, we will explore how to solve the LeetCode problem "2801" 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 two positive integers low and high represented as strings, find the count of stepping numbers in the inclusive range [low, high]. A stepping number is an integer such that all of its adjacent digits have an absolute difference of exactly 1. Return an integer denoting the count of stepping numbers in the inclusive range [low, high]. Since the answer may be very large, return it modulo 109 + 7. Note: A stepping number should not have a leading zero. Example 1: Input: low = "1", high = "11" Output: 10 Explanation: The stepping numbers in the range [1,11] are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. There are a total of 10 stepping numbers in the range. Hence, the output is 10. Example 2: Input: low = "90", high = "101" Output: 2 Explanation: The stepping numbers in the range [90,101] are 98 and 101. There are a total of 2 stepping numbers in the range. Hence, the output is 2. Constraints: 1 <= int(low) <= int(high) < 10100 1 <= low.length, high.length <= 100 low and high consist of only digits. low and high don't have any leading zeros.

Explanation

Here's the breakdown of the solution:

• Dynamic Programming: Use dynamic programming to count stepping numbers of a given length. The state dp[i][j] represents the number of stepping numbers of length i ending with digit j.

• Counting within Range: Define a helper function count_stepping_numbers(s) to count the number of stepping numbers less than or equal to a given string s. This involves iterating through the digits of s and using the DP table. Handle the edge cases carefully.

• Calculate and Return: Calculate the number of stepping numbers less than or equal to high and subtract the number of stepping numbers less than low. Add 1 if low itself is a stepping number. Take the modulo at each step to prevent integer overflow.

• Time & Space Complexity: O(10 N) where N is the maximum length of the input strings. Both runtime and storage are dominated by the DP table of size O(10 N).

def solve():
    MOD = 10**9 + 7

    def is_stepping(n):
        s = str(n)
        for i in range(len(s) - 1):
            if abs(int(s[i]) - int(s[i+1])) != 1:
                return False
        return True

    def count_stepping_numbers(s):
        n = len(s)
        dp = [[0] * 10 for _ in range(n + 1)]

        for j in range(1, 10):
            dp[1][j] = 1

        for i in range(2, n + 1):
            for j in range(10):
                if j > 0:
                    dp[i][j] = (dp[i][j] + dp[i-1][j-1]) % MOD
                if j < 9:
                    dp[i][j] = (dp[i][j] + dp[i-1][j+1]) % MOD

        ans = 0
        for i in range(1, n):
            ans = (ans + sum(dp[i])) % MOD

        # Handle numbers with same length as s
        for i in range(n):
            digit = int(s[i])
            for j in range( (1 if i == 0 else 0), digit):
                if i == 0 and j == 0: continue
                if i == 0:
                  ans = (ans + dp[n - i][j]) % MOD
                elif abs(j - int(s[i-1])) == 1:
                    ans = (ans + dp[n - i][j]) % MOD

            if i > 0 and abs(digit - int(s[i-1])) != 1:
                break
            if i == n - 1:
                ans = (ans + 1) % MOD
        return ans


    low = input()
    high = input()

    low_count = count_stepping_numbers(low)
    high_count = count_stepping_numbers(high)

    ans = (high_count - low_count + MOD) % MOD

    if is_stepping(int(low)):
        ans = (ans + 1) % MOD

    print(ans)


    # Code
    ```python
    def solve():    MOD = 10**9 + 7

    def is_stepping(n):
        s = str(n)
        for i in range(len(s) - 1):
            if abs(int(s[i]) - int(s[i+1])) != 1:
                return False
        return True

    def count_stepping_numbers(s):
        n = len(s)
        dp = [[0] * 10 for _ in range(n + 1)]

        for j in range(1, 10):
            dp[1][j] = 1

        for i in range(2, n + 1):
            for j in range(10):
                if j > 0:
                    dp[i][j] = (dp[i][j] + dp[i-1][j-1]) % MOD
                if j < 9:
                    dp[i][j] = (dp[i][j] + dp[i-1][j+1]) % MOD

        ans = 0
        for i in range(1, n):
            ans = (ans + sum(dp[i])) % MOD

        # Handle numbers with same length as s
        for i in range(n):
            digit = int(s[i])
            for j in range( (1 if i == 0 else 0), digit):
                if i == 0 and j == 0: continue
                if i == 0:
                  ans = (ans + dp[n - i][j]) % MOD
                elif abs(j - int(s[i-1])) == 1:
                    ans = (ans + dp[n - i][j]) % MOD

            if i > 0 and abs(digit - int(s[i-1])) != 1:
                break
            if i == n - 1:
                ans = (ans + 1) % MOD
        return ans


    low = input()
    high = input()

    low_count = count_stepping_numbers(low)
    high_count = count_stepping_numbers(high)

    ans = (high_count - low_count + MOD) % MOD

    if is_stepping(int(low)):
        ans = (ans + 1) % MOD

    print(ans)

solve()