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2466. Count Ways To Build Good Strings 👍

  • Time: $O(\texttt{high})$
  • Space: $O(\texttt{high})$
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class Solution {
 public:
  int countGoodStrings(int low, int high, int zero, int one) {
    constexpr int kMod = 1'000'000'007;
    int ans = 0;
    // dp[i] := the number of good strings with length i
    vector<int> dp(high + 1);
    dp[0] = 1;

    for (int i = 1; i <= high; ++i) {
      if (i >= zero)
        dp[i] = (dp[i] + dp[i - zero]) % kMod;
      if (i >= one)
        dp[i] = (dp[i] + dp[i - one]) % kMod;
      if (i >= low)
        ans = (ans + dp[i]) % kMod;
    }

    return ans;
  }
};

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class Solution {
  public int countGoodStrings(int low, int high, int zero, int one) {
    final int kMod = 1_000_000_007;
    int ans = 0;
    // dp[i] := the number of good strings with length i
    int[] dp = new int[high + 1];
    dp[0] = 1;

    for (int i = 1; i <= high; ++i) {
      if (i >= zero)
        dp[i] = (dp[i] + dp[i - zero]) % kMod;
      if (i >= one)
        dp[i] = (dp[i] + dp[i - one]) % kMod;
      if (i >= low)
        ans = (ans + dp[i]) % kMod;
    }

    return ans;
  }
}

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class Solution:
  def countGoodStrings(self, low: int, high: int, zero: int, one: int) -> int:
    kMod = 1_000_000_007
    ans = 0
    # dp[i] := the number of good strings with length i
    dp = [1] + [0] * high

    for i in range(1, high + 1):
      if i >= zero:
        dp[i] = (dp[i] + dp[i - zero]) % kMod
      if i >= one:
        dp[i] = (dp[i] + dp[i - one]) % kMod
      if i >= low:
        ans = (ans + dp[i]) % kMod

    return ans