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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)
    }
    

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