README.md

2466. Count Ways To Build Good Strings (Medium)

Given the integers zero, one, low, and high, we can construct a string by starting with an empty string, and then at each step perform either of the following:

• Append the character '0' zero times.
• Append the character '1' one times.

This can be performed any number of times.

A good string is a string constructed by the above process having a length between low and high (inclusive).

Return the number of different good strings that can be constructed satisfying these properties. Since the answer can be large, return it modulo 109 + 7.

 

Example 1:

Input: low = 3, high = 3, zero = 1, one = 1
Output: 8
Explanation: 
One possible valid good string is "011". 
It can be constructed as follows: "" -> "0" -> "01" -> "011". 
All binary strings from "000" to "111" are good strings in this example.

Example 2:

Input: low = 2, high = 3, zero = 1, one = 2
Output: 5
Explanation: The good strings are "00", "11", "000", "110", and "011".

 

Constraints:

• 1 <= low <= high <= 105
• 1 <= zero, one <= low

Companies: Citadel, Oracle

Related Topics:
Dynamic Programming

Similar Questions:

• Climbing Stairs (Easy)

Hints:

• Calculate the number of good strings with length less or equal to some constant x.
• Apply dynamic programming using the group size of consecutive zeros and ones.

Solution 1.

Let dp[len] be the number of ways to form good strings. The anwer is SUM( dp[len] | low <= len <= high )

dp[len] = 0 if len < 0
dp[0] = 1
dp[len] = dp[len-zero] + dp[len-one]

// OJ: https://leetcode.com/problems/count-ways-to-build-good-strings
// Author: github.com/lzl124631x
// Time: O(H)
// Space: O(H)
class Solution {
public:
    int countGoodStrings(int low, int high, int zero, int one) {
        long dp[100001] = {}, ans = 0, mod = 1e9 + 7;
        dp[0] = 1;
        for (int i = 1; i <= high; ++i) {
            if (i - zero >= 0) dp[i] = dp[i - zero];
            if (i - one >= 0) dp[i] = (dp[i] + dp[i - one]) % mod;
            if (i >= low) ans = (ans + dp[i]) % mod;
        }
        return ans;
    }
};