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feat: add solutions to lc problem: No.1573
No.1573.Number of Ways to Split a String
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‎solution/0900-0999/0927.Three Equal Parts/README.md

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时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 表示 `arr` 的长度。
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相似题目：
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- [1573. 分割字符串的方案数](/solution/1500-1599/1573.Number%20of%20Ways%20to%20Split%20a%20String/README.md)
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<!-- tabs:start -->
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### **Python3**

‎solution/1500-1599/1573.Number of Ways to Split a String/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：计数**
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我们先遍历字符串 $s$，统计其中字符 $1$ 的个数 $cnt$，如果 $cnt$ 不能被 $3$ 整除，那么无法分割，直接返回 $0$。如果 $cnt$ 为 $0$，说明字符串中没有字符 $1$，我们可以在 $n-1$ 个位置中任意选择两个位置，将字符串分割成三个子串，那么方案数就是 $C_{n-1}^2$。
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如果 $cnt \gt 0$，我们将 $cnt$ 更新为 $\frac{cnt}{3}$，即每个子串中字符 $1$ 的个数。
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接下来我们找到第一个子字符串的右边界的最小下标，记为 $i_1$，第一个子字符串的右边界的最大下标（不包含），记为 $i_2$；第二个子字符串的右边界的最小下标，记为 $j_1$，第二个子字符串的右边界的最大下标（不包含），记为 $j_2$。那么方案数就是 $(i_2-i_1) \times (j_2-j_1)$。
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注意答案可能很大，需要对 $10^9+7$ 取模。
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时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为字符串 $s$ 的长度。
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相似题目：
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- [927. 三等分](/solution/0900-0999/0927.Three%20Equal%20Parts/README.md)
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def numWays(self, s: str) -> int:
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def find(x):
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t = 0
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for i, c in enumerate(s):
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t += int(c == '1')
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if t == x:
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return i
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cnt, m = divmod(sum(c == '1' for c in s), 3)
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if m:
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return 0
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n = len(s)
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mod = 10**9 + 7
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if cnt == 0:
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return ((n - 1) * (n - 2) // 2) % mod
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i1, i2 = find(cnt), find(cnt + 1)
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j1, j2 = find(cnt * 2), find(cnt * 2 + 1)
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return (i2 - i1) * (j2 - j1) % (10**9 + 7)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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private String s;
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public int numWays(String s) {
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this.s = s;
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int cnt = 0;
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int n = s.length();
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for (int i = 0; i < n; ++i) {
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if (s.charAt(i) == '1') {
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++cnt;
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}
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}
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int m = cnt % 3;
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if (m != 0) {
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return 0;
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}
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final int mod = (int) 1e9 + 7;
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if (cnt == 0) {
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return (int) (((n - 1L) * (n - 2) / 2) % mod);
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}
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cnt /= 3;
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long i1 = find(cnt), i2 = find(cnt + 1);
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long j1 = find(cnt * 2), j2 = find(cnt * 2 + 1);
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return (int) ((i2 - i1) * (j2 - j1) % mod);
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}
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private int find(int x) {
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int t = 0;
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for (int i = 0;; ++i) {
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t += s.charAt(i) == '1' ? 1 : 0;
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if (t == x) {
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return i;
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}
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}
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int numWays(string s) {
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int cnt = 0;
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for (char& c : s) {
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cnt += c == '1';
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}
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int m = cnt % 3;
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if (m) {
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return 0;
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}
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const int mod = 1e9 + 7;
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int n = s.size();
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if (cnt == 0) {
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return (n - 1LL) * (n - 2) / 2 % mod;
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}
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cnt /= 3;
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auto find = [&](int x) {
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int t = 0;
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for (int i = 0; ; ++i) {
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t += s[i] == '1';
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if (t == x) {
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return i;
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}
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}
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};
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int i1 = find(cnt), i2 = find(cnt + 1);
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int j1 = find(cnt * 2), j2 = find(cnt * 2 + 1);
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return (1LL * (i2 - i1) * (j2 - j1)) % mod;
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}
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};
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```
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### **Go**
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```go
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func numWays(s string) int {
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cnt := 0
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for _, c := range s {
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if c == '1' {
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cnt++
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}
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}
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m := cnt % 3
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if m != 0 {
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return 0
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}
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const mod = 1e9 + 7
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n := len(s)
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if cnt == 0 {
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return (n - 1) * (n - 2) / 2 % mod
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}
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cnt /= 3
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find := func(x int) int {
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t := 0
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for i := 0; ; i++ {
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if s[i] == '1' {
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t++
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if t == x {
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return i
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}
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}
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}
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}
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i1, i2 := find(cnt), find(cnt+1)
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j1, j2 := find(cnt*2), find(cnt*2+1)
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return (i2 - i1) * (j2 - j1) % mod
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}
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```
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### **...**

‎solution/1500-1599/1573.Number of Ways to Split a String/README_EN.md

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### **Python3**
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```python
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class Solution:
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def numWays(self, s: str) -> int:
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def find(x):
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t = 0
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for i, c in enumerate(s):
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t += int(c == '1')
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if t == x:
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return i
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cnt, m = divmod(sum(c == '1' for c in s), 3)
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if m:
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return 0
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n = len(s)
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mod = 10**9 + 7
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if cnt == 0:
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return ((n - 1) * (n - 2) // 2) % mod
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i1, i2 = find(cnt), find(cnt + 1)
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j1, j2 = find(cnt * 2), find(cnt * 2 + 1)
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return (i2 - i1) * (j2 - j1) % (10**9 + 7)
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```
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### **Java**
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```java
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class Solution {
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private String s;
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public int numWays(String s) {
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this.s = s;
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int cnt = 0;
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int n = s.length();
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for (int i = 0; i < n; ++i) {
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if (s.charAt(i) == '1') {
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++cnt;
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}
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}
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int m = cnt % 3;
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if (m != 0) {
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return 0;
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}
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final int mod = (int) 1e9 + 7;
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if (cnt == 0) {
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return (int) (((n - 1L) * (n - 2) / 2) % mod);
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}
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cnt /= 3;
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long i1 = find(cnt), i2 = find(cnt + 1);
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long j1 = find(cnt * 2), j2 = find(cnt * 2 + 1);
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return (int) ((i2 - i1) * (j2 - j1) % mod);
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}
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private int find(int x) {
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int t = 0;
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for (int i = 0;; ++i) {
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t += s.charAt(i) == '1' ? 1 : 0;
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if (t == x) {
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return i;
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}
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}
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int numWays(string s) {
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int cnt = 0;
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for (char& c : s) {
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cnt += c == '1';
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}
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int m = cnt % 3;
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if (m) {
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return 0;
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}
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const int mod = 1e9 + 7;
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int n = s.size();
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if (cnt == 0) {
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return (n - 1LL) * (n - 2) / 2 % mod;
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}
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cnt /= 3;
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auto find = [&](int x) {
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int t = 0;
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for (int i = 0; ; ++i) {
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t += s[i] == '1';
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if (t == x) {
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return i;
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}
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}
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};
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int i1 = find(cnt), i2 = find(cnt + 1);
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int j1 = find(cnt * 2), j2 = find(cnt * 2 + 1);
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return (1LL * (i2 - i1) * (j2 - j1)) % mod;
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}
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};
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```
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### **Go**
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```go
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func numWays(s string) int {
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cnt := 0
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for _, c := range s {
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if c == '1' {
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cnt++
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}
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}
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m := cnt % 3
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if m != 0 {
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return 0
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}
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const mod = 1e9 + 7
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n := len(s)
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if cnt == 0 {
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return (n - 1) * (n - 2) / 2 % mod
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}
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cnt /= 3
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find := func(x int) int {
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t := 0
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for i := 0; ; i++ {
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if s[i] == '1' {
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t++
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if t == x {
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return i
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}
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}
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}
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}
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i1, i2 := find(cnt), find(cnt+1)
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j1, j2 := find(cnt*2), find(cnt*2+1)
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return (i2 - i1) * (j2 - j1) % mod
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}
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```
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### **...**

‎solution/1500-1599/1573.Number of Ways to Split a String/Solution.cpp

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

‎solution/1500-1599/1573.Number of Ways to Split a String/Solution.go

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

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