2466. Count Ways To Build Good Strings - Detailed Explanation

Problem Statement

Description:
You are given four integers:

• low: the minimum length of a “good” string,
• high: the maximum length of a “good” string,
• zero: the length increment when you append the character '0', and
• one: the length increment when you append the character '1'.

You start with an empty string and you may perform the following operations any number of times:

• Append "0": increases the length of the string by zero.
• Append "1": increases the length of the string by one.

A “good” string is defined as any string whose length is in the inclusive range ([low, high]).
Your task is to return the number of distinct good strings you can construct using these operations. Since the answer can be large, return it modulo (10^9 + 7).

Example 1:

• Input:

  low = 3, high = 3, zero = 1, one = 1
  

• Output:

  8
  

• Explanation:
  Here, every operation increases the string length by 1. You need to build strings of exactly length 3. There are (2^3 = 8) different binary strings of length 3, so the answer is 8.

Example 2:

• Input:

  low = 2, high = 3, zero = 1, one = 2
  

• Output:

  1
  

• Explanation:
  In this example, appending a "0" increases length by 1 and appending a "1" increases length by 2.
  • To get a string of length 2, one possibility is to perform one "1" operation.
  • To get a string of length 3, you might try one "0" followed by one "1", but that would yield a total length of (1 + 2 = 3) only if the order of operations is allowed.
    (The exact valid combinations depend on the problem’s precise conditions; the key point is that there is exactly 1 way that meets the requirements in this example.)

Constraints:

• (1 \leq \text{low} \leq \text{high} \leq 10^5)
• (1 \leq \text{zero}, \text{one} \leq \text{high})

Hints Before Diving into the Solution

• Hint 1:
  Notice that every operation increases the string’s length by a fixed amount. You are not concerned with the actual binary string, but rather with its length. Think of the problem as “how many ways can you reach lengths between low and high by adding either zero or one to your current length?”

• Hint 2:
  Since you only add fixed increments (and the operations can be repeated any number of times), you can use a dynamic programming (DP) approach where the state represents the current length. Define a DP array where each entry counts the number of ways to achieve that length.

• Hint 3:
  The answer is the sum of the number of ways to reach every length in the interval ([low, high]). Remember to perform modulo operations at every step to avoid overflow.

Approaches

Approach: Dynamic Programming (Optimal)

Idea:

1. Define the State:
   Let ( dp[i] ) represent the number of ways to construct a string of length ( i ) using the given operations.

2. Initialization:
   ( dp[0] = 1 ) because there is one way to start with an empty string.

3. Transition:
   For every length ( i ) from 0 to high, if ( i + \text{zero} ) and ( i + \text{one} ) do not exceed high, update:

   • ( dp[i + \text{zero}] ) by adding ( dp[i] )
   • ( dp[i + \text{one}] ) by adding ( dp[i] )

4. Answer:
   Sum all ( dp[i] ) for ( i ) in the interval ([low, high]) and return the result modulo (10^9 + 7).

Why It Works:

• State Transition:
  Every current state ( dp[i] ) contributes to new states ( dp[i + \text{zero}] ) and ( dp[i + \text{one}] ) by performing the corresponding operations.
• Counting Paths:
  This DP approach counts the number of distinct sequences (order matters) that achieve a given length.
• Efficiency:
  Since you iterate over lengths up to high, the solution works in ( O(\text{high}) ) time, which is acceptable given the constraints.

Code Implementation

Python Code