Introduction

In this blog post, we will explore how to solve the LeetCode problem "3130" 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 3 positive integers zero, one, and limit. A binary array arr is called stable if: The number of occurrences of 0 in arr is exactly zero. The number of occurrences of 1 in arr is exactly one. Each subarray of arr with a size greater than limit must contain both 0 and 1. Return the total number of stable binary arrays. Since the answer may be very large, return it modulo 109 + 7. Example 1: Input: zero = 1, one = 1, limit = 2 Output: 2 Explanation: The two possible stable binary arrays are [1,0] and [0,1]. Example 2: Input: zero = 1, one = 2, limit = 1 Output: 1 Explanation: The only possible stable binary array is [1,0,1]. Example 3: Input: zero = 3, one = 3, limit = 2 Output: 14 Explanation: All the possible stable binary arrays are [0,0,1,0,1,1], [0,0,1,1,0,1], [0,1,0,0,1,1], [0,1,0,1,0,1], [0,1,0,1,1,0], [0,1,1,0,0,1], [0,1,1,0,1,0], [1,0,0,1,0,1], [1,0,0,1,1,0], [1,0,1,0,0,1], [1,0,1,0,1,0], [1,0,1,1,0,0], [1,1,0,0,1,0], and [1,1,0,1,0,0]. Constraints: 1 <= zero, one, limit <= 1000

Explanation

Here's the breakdown of the problem and the Python code:

• Approach:

  • We use dynamic programming to store the number of ways to form stable arrays with a given number of zeros and ones placed, considering the limit constraint.
  • The DP state dp[i][j][k] represents the number of stable arrays formed with i zeros and j ones, where k represents the number of consecutive zeros or ones at the end of the array.
  • We iterate through the possible number of zeros and ones and transition based on whether we can add a zero or a one while respecting the limit constraint.

• Complexity:

  • Runtime: O(zero * one * (limit + 1))
  • Storage: O(zero * one * (limit + 1))

Code

    def solve():
    zero = int(input())
    one = int(input())
    limit = int(input())

    MOD = 10**9 + 7
    dp = [[[0] * (limit + 1) for _ in range(one + 1)] for _ in range(zero + 1)]

    # Initialize base cases. If zero or one is 0. return 0
    if zero == 0 or one == 0:
      return 0

    # Initialize the base cases with single '0' or single '1'. Since each must appear once.
    dp[1][0][1] = 1 # array contains [0], zero =1, one =0
    dp[0][1][1] = 1 # array contains [1], zero =0, one =1

    for i in range(zero + 1):
        for j in range(one + 1):
            for k in range(1, limit + 1):

                if i == 0 and j == 0:
                  continue

                if i > 0 and k <= limit and dp[i-1][j][0] > 0:  # Add a zero
                    dp[i][j][k] = (dp[i][j][k] + dp[i-1][j][0 if i == 1 else k -1 ]) % MOD # We can't add zero if array already contains zero


                if j > 0 and k <= limit and dp[i][j-1][1] > 0:  # Add a one
                    dp[i][j][k] = (dp[i][j][k] + dp[i][j-1][1 if j == 1 else k -1]) % MOD # We can't add one if array already contains one

                #Add up zeros and ones consecutively up to the limit, and save that sum at index 0 and 1
                if k < limit:
                  if i>0 and j>0:
                      dp[i][j][0] = (dp[i][j][0] + dp[i][j][k])% MOD
                      dp[i][j][1] = (dp[i][j][1] + dp[i][j][k])% MOD

    result = 0
    for k in range(1, limit + 1):
      result = (result + dp[zero][one][k]) % MOD

    return result

print(solve())