README_EN.md

comments	difficulty	edit_url	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/2400-2499/2489.Number%20of%20Substrings%20With%20Fixed%20Ratio/README_EN.md	Hash Table Math String Prefix Sum

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 $s$.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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)
}