# 2489. Number of Substrings With Fixed Ratio

## 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

• 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;
}
};

• 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


• 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)
}