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2466. Count Ways To Build Good Strings

Description

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

Solutions

Solution 1: Memoization Search

We design a function $dfs(i)$ to represent the number of good strings constructed starting from the $i$-th position. The answer is $dfs(0)$.

The computation process of the function $dfs(i)$ is as follows:

  • If $i > high$, return $0$;
  • If $low \leq i \leq high$, increment the answer by $1$, then after $i$, we can add either zero number of $0$s or one number of $1$s. Therefore, the answer is incremented by $dfs(i + zero) + dfs(i + one)$.

During the process, we need to take the modulus of the answer, and we can use memoization search to reduce redundant computations.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n = high$.

  • class Solution {
    private static final int MOD = (int) 1e9 + 7;
    private int[] f;
    private int lo;
    private int hi;
    private int zero;
    private int one;

    public int countGoodStrings(int low, int high, int zero, int one) {
        f = new int[high + 1];
        Arrays.fill(f, -1);
        lo = low;
        hi = high;
        this.zero = zero;
        this.one = one;
        return dfs(0);
    }

    private int dfs(int i) {
        if (i > hi) {
            return 0;
        }
        if (f[i] != -1) {
            return f[i];
        }
        long ans = 0;
        if (i >= lo && i <= hi) {
            ++ans;
        }
        ans += dfs(i + zero) + dfs(i + one);
        ans %= MOD;
        f[i] = (int) ans;
        return f[i];
    }
}

  • class Solution {
public:
    const int mod = 1e9 + 7;

    int countGoodStrings(int low, int high, int zero, int one) {
        vector<int> f(high + 1, -1);
        function<int(int)> dfs = [&](int i) -> int {
            if (i > high) return 0;
            if (f[i] != -1) return f[i];
            long ans = i >= low && i <= high;
            ans += dfs(i + zero) + dfs(i + one);
            ans %= mod;
            f[i] = ans;
            return ans;
        };
        return dfs(0);
    }
};

  • class Solution:
    def countGoodStrings(self, low: int, high: int, zero: int, one: int) -> int:
        @cache
        def dfs(i):
            if i > high:
                return 0
            ans = 0
            if low <= i <= high:
                ans += 1
            ans += dfs(i + zero) + dfs(i + one)
            return ans % mod

        mod = 10**9 + 7
        return dfs(0)


  • func countGoodStrings(low int, high int, zero int, one int) int {
	f := make([]int, high+1)
	for i := range f {
		f[i] = -1
	}
	const mod int = 1e9 + 7
	var dfs func(i int) int
	dfs = func(i int) int {
		if i > high {
			return 0
		}
		if f[i] != -1 {
			return f[i]
		}
		ans := 0
		if i >= low && i <= high {
			ans++
		}
		ans += dfs(i+zero) + dfs(i+one)
		ans %= mod
		f[i] = ans
		return ans
	}
	return dfs(0)
}

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