Question

2489. Number of Substrings With Fixed Ratio

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Description

You are given a binary string s, and two integers num1 and num2. num1 and num2 are coprime numbers.

A ratio substring is a substring of s where the ratio between the number of 0's and the number of 1's in the substring is exactly num1 : num2.

• For example, if num1 = 2 and num2 = 3, then "01011" and "1110000111" are ratio substrings, while "11000" is not.

Return _the number of non-empty ratio substrings of _s.

Note that:

• A substring is a contiguous sequence of characters within a string.
• Two values x and y are coprime if gcd(x, y) == 1 where gcd(x, y) is the greatest common divisor of x and y.

 

Example 1:

Input: s = "0110011", num1 = 1, num2 = 2
Output: 4
Explanation: There exist 4 non-empty ratio substrings.
- The substring s[0..2]: "0110011". It contains one 0 and two 1's. The ratio is 1 : 2.
- The substring s[1..4]: "0110011". It contains one 0 and two 1's. The ratio is 1 : 2.
- The substring s[4..6]: "0110011". It contains one 0 and two 1's. The ratio is 1 : 2.
- The substring s[1..6]: "0110011". It contains two 0's and four 1's. The ratio is 2 : 4 == 1 : 2.
It can be shown that there are no more ratio substrings.

Example 2:

Input: s = "10101", num1 = 3, num2 = 1
Output: 0
Explanation: There is no ratio substrings of s. We return 0.

 

Constraints:

• 1 <= s.length <= 105
• 1 <= num1, num2 <= s.length
• num1 and num2 are coprime integers.

Solutions

Solution 1: Prefix Sum + Counting

We use $one[i]$ to represent the number of $1$s in the substring $s[0,..i]$, and $zero[i]$ to represent the number of $0$s in the substring $s[0,..i]$. A substring meets the condition if

$$ \frac{zero[j] - zero[i]}{one[j] - one[i]} = \frac{num1}{num2} $$

where $i < j$. We can transform the above equation into

$$ one[j] \times num1 - zero[j] \times num2 = one[i] \times num1 - zero[i] \times num2 $$

When we iterate to index $j$, we only need to count how many indices $i$ satisfy the above equation. Therefore, we can use a hash table to record the number of occurrences of $one[i] \times num1 - zero[i] \times num2$, and when we iterate to index $j$, we only need to count the number of occurrences of $one[j] \times num1 - zero[j] \times num2$.

The hash table initially only has one key-value pair $(0, 1)$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the string $s$.

class Solution:
  def fixedRatio(self, s: str, num1: int, num2: int) -> int:
    n0 = n1 = 0
    ans = 0
    cnt = Counter({0: 1})
    for c in s:
      n0 += c == '0'
      n1 += c == '1'
      x = n1 * num1 - n0 * num2
      ans += cnt[x]
      cnt[x] += 1
    return ans

class Solution {
  public long fixedRatio(String s, int num1, int num2) {
    long n0 = 0, n1 = 0;
    long ans = 0;
    Map<Long, Long> cnt = new HashMap<>();
    cnt.put(0L, 1L);
    for (char c : s.toCharArray()) {
      n0 += c == '0' ? 1 : 0;
      n1 += c == '1' ? 1 : 0;
      long x = n1 * num1 - n0 * num2;
      ans += cnt.getOrDefault(x, 0L);
      cnt.put(x, cnt.getOrDefault(x, 0L) + 1);
    }
    return ans;
  }
}

using ll = long long;

class Solution {
public:
  long long fixedRatio(string s, int num1, int num2) {
    ll n0 = 0, n1 = 0;
    ll ans = 0;
    unordered_map<ll, ll> cnt;
    cnt[0] = 1;
    for (char& c : s) {
      n0 += c == '0';
      n1 += c == '1';
      ll x = n1 * num1 - n0 * num2;
      ans += cnt[x];
      ++cnt[x];
    }
    return ans;
  }
};

func fixedRatio(s string, num1 int, num2 int) int64 {
  n0, n1 := 0, 0
  ans := 0
  cnt := map[int]int{0: 1}
  for _, c := range s {
    if c == '0' {
      n0++
    } else {
      n1++
    }
    x := n1*num1 - n0*num2
    ans += cnt[x]
    cnt[x]++
  }
  return int64(ans)
}