Question

1573. Number of Ways to Split a String

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Description

Given a binary string s, you can split s into 3 non-empty strings s1, s2, and s3 where s1 + s2 + s3 = s.

Return the number of ways s can be split such that the number of ones is the same in s1, s2, and s3. Since the answer may be too large, return it modulo 109 + 7.

 

Example 1:

Input: s = "10101"
Output: 4
Explanation: There are four ways to split s in 3 parts where each part contain the same number of letters '1'.
"1|010|1"
"1|01|01"
"10|10|1"
"10|1|01"

Example 2:

Input: s = "1001"
Output: 0

Example 3:

Input: s = "0000"
Output: 3
Explanation: There are three ways to split s in 3 parts.
"0|0|00"
"0|00|0"
"00|0|0"

 

Constraints:

• 3 <= s.length <= 105
• s[i] is either '0' or '1'.

Solutions

Solution 1

class Solution:
  def numWays(self, s: str) -> int:
    def find(x):
      t = 0
      for i, c in enumerate(s):
        t += int(c == '1')
        if t == x:
          return i

    cnt, m = divmod(sum(c == '1' for c in s), 3)
    if m:
      return 0
    n = len(s)
    mod = 10**9 + 7
    if cnt == 0:
      return ((n - 1) * (n - 2) // 2) % mod
    i1, i2 = find(cnt), find(cnt + 1)
    j1, j2 = find(cnt * 2), find(cnt * 2 + 1)
    return (i2 - i1) * (j2 - j1) % (10**9 + 7)

class Solution {
  private String s;

  public int numWays(String s) {
    this.s = s;
    int cnt = 0;
    int n = s.length();
    for (int i = 0; i < n; ++i) {
      if (s.charAt(i) == '1') {
        ++cnt;
      }
    }
    int m = cnt % 3;
    if (m != 0) {
      return 0;
    }
    final int mod = (int) 1e9 + 7;
    if (cnt == 0) {
      return (int) (((n - 1L) * (n - 2) / 2) % mod);
    }
    cnt /= 3;
    long i1 = find(cnt), i2 = find(cnt + 1);
    long j1 = find(cnt * 2), j2 = find(cnt * 2 + 1);
    return (int) ((i2 - i1) * (j2 - j1) % mod);
  }

  private int find(int x) {
    int t = 0;
    for (int i = 0;; ++i) {
      t += s.charAt(i) == '1' ? 1 : 0;
      if (t == x) {
        return i;
      }
    }
  }
}

class Solution {
public:
  int numWays(string s) {
    int cnt = 0;
    for (char& c : s) {
      cnt += c == '1';
    }
    int m = cnt % 3;
    if (m) {
      return 0;
    }
    const int mod = 1e9 + 7;
    int n = s.size();
    if (cnt == 0) {
      return (n - 1LL) * (n - 2) / 2 % mod;
    }
    cnt /= 3;
    auto find = [&](int x) {
      int t = 0;
      for (int i = 0;; ++i) {
        t += s[i] == '1';
        if (t == x) {
          return i;
        }
      }
    };
    int i1 = find(cnt), i2 = find(cnt + 1);
    int j1 = find(cnt * 2), j2 = find(cnt * 2 + 1);
    return (1LL * (i2 - i1) * (j2 - j1)) % mod;
  }
};

func numWays(s string) int {
  cnt := 0
  for _, c := range s {
    if c == '1' {
      cnt++
    }
  }
  m := cnt % 3
  if m != 0 {
    return 0
  }
  const mod = 1e9 + 7
  n := len(s)
  if cnt == 0 {
    return (n - 1) * (n - 2) / 2 % mod
  }
  cnt /= 3
  find := func(x int) int {
    t := 0
    for i := 0; ; i++ {
      if s[i] == '1' {
        t++
        if t == x {
          return i
        }
      }
    }
  }
  i1, i2 := find(cnt), find(cnt+1)
  j1, j2 := find(cnt*2), find(cnt*2+1)
  return (i2 - i1) * (j2 - j1) % mod
}