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

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 \(s\).

Python3JavaC++Go

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)
}

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