source: https://leetcode.com/problems/count-ways-to-build-good-strings/description/
Count Ways To Build Good Strings
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 10^9 + 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
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class Solution { public int countGoodStrings(int low, int high, int zero, int one) { Integer[] memo = new Integer[high+1]; return dfs(0, low, high, zero, one, memo); } private static final int MOD = 1_000_000_007; public int dfs(int length, int low, int high, int zero, int one, Integer[] memo){ if(length>high){ return 0; } if(memo[length]!=null) return memo[length]; long count = 0; if(length >= low && length <= high){ count = count+1; } //either append zero count = (count + dfs(length+zero, low, high, zero, one, memo))%MOD; //or append one count = (count + dfs(length+one, low, high, zero, one, memo))%MOD; memo[length] = (int)count; return (int)count; } } |