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Mar 26, 2023
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[pull] main from doocs:main #203

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165 changes: 142 additions & 23 deletions solution/1600-1699/1638.Count Substrings That Differ by One Character/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -72,9 +72,19 @@

**方法一：枚举**

枚举不同的那个字符，然后向两边扩展。
我们可以枚举字符串 $s$ 和 $t$ 中不同的那个字符位置，然后分别向两边扩展，直到遇到不同的字符为止，这样就可以得到以该位置为中心的满足条件的子串对数目。我们记左边扩展的相同字符个数为 $l$，右边扩展的相同字符个数为 $r$，那么以该位置为中心的满足条件的子串对数目为 $(l + 1) \times (r + 1)$，累加到答案中即可。

时间复杂度 $O(m \times n \times min(m, n))$，空间复杂度 $O(1)$。其中 $m$ 和 $n$ 分别是字符串 $s$ 和 $t$ 的长度。
枚举结束后，即可得到答案。

时间复杂度 $O(m \times n \times \min(m, n))$，空间复杂度 $O(1)$。其中 $m$ 和 $n$ 分别为字符串 $s$ 和 $t$ 的长度。

**方法二：预处理 + 枚举**

方法一中，我们每次需要分别往左右两边扩展，得出 $l$ 和 $r$ 的值。实际上，我们可以预处理出以每个位置 $(i, j)$ 结尾的最长相同后缀的长度，以及以每个位置 $(i, j)$ 开头的最长相同前缀的长度，分别记录在数组 $f$ 和 $g$ 中。

接下来，与方法一类似，我们枚举字符串 $s$ 和 $t$ 中不同的那个字符位置 $(i, j)$，那么以该位置为中心的满足条件的子串对数目为 $(f[i][j] + 1) \times (g[i + 1][j + 1] + 1)$，累加到答案中即可。

时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别为字符串 $s$ 和 $t$ 的长度。

<!-- tabs:start -->

Expand All @@ -86,15 +96,36 @@
class Solution:
def countSubstrings(self, s: str, t: str) -> int:
ans = 0
m, n = len(s), len(t)
for i, a in enumerate(s):
for j, b in enumerate(t):
if a != b:
l = r = 1
while i >= l and j >= l and s[i - l] == t[j - l]:
l = r = 0
while i > l and j > l and s[i - l - 1] == t[j - l - 1]:
l += 1
while i + r < len(s) and j + r < len(t) and s[i + r] == t[j + r]:
while i + r + 1 < m and j + r + 1 < n and s[i + r + 1] == t[j + r + 1]:
r += 1
ans += l * r
ans += (l + 1) * (r + 1)
return ans
```

```python
class Solution:
def countSubstrings(self, s: str, t: str) -> int:
ans = 0
m, n = len(s), len(t)
f = [[0] * (n + 1) for _ in range(m + 1)]
g = [[0] * (n + 1) for _ in range(m + 1)]
for i, a in enumerate(s, 1):
for j, b in enumerate(t, 1):
if a == b:
f[i][j] = f[i - 1][j - 1] + 1
for i in range(m - 1, -1, -1):
for j in range(n - 1, -1, -1):
if s[i] == t[j]:
g[i][j] = g[i + 1][j + 1] + 1
else:
ans += (f[i][j] + 1) * (g[i + 1][j + 1] + 1)
return ans
```

Expand All @@ -105,19 +136,47 @@ class Solution:
```java
class Solution {
public int countSubstrings(String s, String t) {
int m = s.length(), n = t.length();
int ans = 0;
int m = s.length(), n = t.length();
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (s.charAt(i) != t.charAt(j)) {
int l = 1, r = 1;
while (i - l >= 0 && j - l >= 0 && s.charAt(i - l) == t.charAt(j - l)) {
int l = 0, r = 0;
while (i - l > 0 && j - l > 0 && s.charAt(i - l - 1) == t.charAt(j - l - 1)) {
++l;
}
while (i + r < m && j + r < n && s.charAt(i + r) == t.charAt(j + r)) {
while (i + r + 1 < m && j + r + 1 < n && s.charAt(i + r + 1) == t.charAt(j + r + 1)) {
++r;
}
ans += l * r;
ans += (l + 1) * (r + 1);
}
}
}
return ans;
}
}
```

```java
class Solution {
public int countSubstrings(String s, String t) {
int ans = 0;
int m = s.length(), n = t.length();
int[][] f = new int[m + 1][n + 1];
int[][] g = new int[m + 1][n + 1];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (s.charAt(i) == t.charAt(j)) {
f[i + 1][j + 1] = f[i][j] + 1;
}
}
}
for (int i = m - 1; i >= 0; --i) {
for (int j = n - 1; j >= 0; --j) {
if (s.charAt(i) == t.charAt(j)) {
g[i][j] = g[i + 1][j + 1] + 1;
} else {
ans += (f[i][j] + 1) * (g[i + 1][j + 1] + 1);
}
}
}
Expand All @@ -132,21 +191,51 @@ class Solution {
class Solution {
public:
int countSubstrings(string s, string t) {
int ans = 0;
int m = s.size(), n = t.size();
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (s[i] != t[j]) {
int l = 0, r = 0;
while (i - l > 0 && j - l > 0 && s[i - l - 1] == t[j - l - 1]) {
++l;
}
while (i + r + 1 < m && j + r + 1 < n && s[i + r + 1] == t[j + r + 1]) {
++r;
}
ans += (l + 1) * (r + 1);
}
}
}
return ans;
}
};
```

```cpp
class Solution {
public:
int countSubstrings(string s, string t) {
int ans = 0;
int m = s.length(), n = t.length();
int f[m + 1][n + 1];
int g[m + 1][n + 1];
memset(f, 0, sizeof(f));
memset(g, 0, sizeof(g));
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (s[i] == t[j]) {
continue;
f[i + 1][j + 1] = f[i][j] + 1;
}
int l = 1, r = 1;
while (i - l >= 0 && j - l >= 0 && s[i - l] == t[j - l]) {
++l;
}
while (i + r < m && j + r < n && s[i + r] == t[j + r]) {
++r;
}
}
for (int i = m - 1; i >= 0; --i) {
for (int j = n - 1; j >= 0; --j) {
if (s[i] == t[j]) {
g[i][j] = g[i + 1][j + 1] + 1;
} else {
ans += (f[i][j] + 1) * (g[i + 1][j + 1] + 1);
}
ans += l * r;
}
}
return ans;
Expand All @@ -158,17 +247,47 @@ public:

```go
func countSubstrings(s string, t string) (ans int) {
m, n := len(s), len(t)
for i, a := range s {
for j, b := range t {
if a != b {
l, r := 1, 1
for i >= l && j >= l && s[i-l] == t[j-l] {
l, r := 0, 0
for i > l && j > l && s[i-l-1] == t[j-l-1] {
l++
}
for i+r < len(s) && j+r < len(t) && s[i+r] == t[j+r] {
for i+r+1 < m && j+r+1 < n && s[i+r+1] == t[j+r+1] {
r++
}
ans += l * r
ans += (l + 1) * (r + 1)
}
}
}
return
}
```

```go
func countSubstrings(s string, t string) (ans int) {
m, n := len(s), len(t)
f := make([][]int, m+1)
g := make([][]int, m+1)
for i := range f {
f[i] = make([]int, n+1)
g[i] = make([]int, n+1)
}
for i, a := range s {
for j, b := range t {
if a == b {
f[i+1][j+1] = f[i][j] + 1
}
}
}
for i := m - 1; i >= 0; i-- {
for j := n - 1; j >= 0; j-- {
if s[i] == t[j] {
g[i][j] = g[i+1][j+1] + 1
} else {
ans += (f[i][j] + 1) * (g[i+1][j+1] + 1)
}
}
}
Expand Down
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