Description#
Given two strings s and t, find the number of ways you can choose a non-empty substring of s and replace a single character by a different character such that the resulting substring is a substring of t. In other words, find the number of substrings in s that differ from some substring in t by exactly one character.
For example, the underlined substrings in "computer" and "computation" only differ by the 'e'/'a', so this is a valid way.
Return the number of substrings that satisfy the condition above.
A substring is a contiguous sequence of characters within a string.
Example 1:
Input: s = "aba", t = "baba"
Output: 6
Explanation: The following are the pairs of substrings from s and t that differ by exactly 1 character:
("aba", "baba")
("aba", "baba")
("aba", "baba")
("aba", "baba")
("aba", "baba")
("aba", "baba")
The underlined portions are the substrings that are chosen from s and t.
Example 2:
Input: s = "ab", t = "bb"
Output: 3
Explanation: The following are the pairs of substrings from s and t that differ by 1 character:
("ab", "bb")
("ab", "bb")
("ab", "bb")
The underlined portions are the substrings that are chosen from s and t.
Constraints:
1 <= s.length, t.length <= 100s and t consist of lowercase English letters only.
Solutions#
Solution 1#
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| 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 = 0
while i > l and j > l and s[i - l - 1] == t[j - l - 1]:
l += 1
while (
i + r + 1 < m and j + r + 1 < n and s[i + r + 1] == t[j + r + 1]
):
r += 1
ans += (l + 1) * (r + 1)
return ans
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| class Solution {
public int countSubstrings(String s, String t) {
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 = 0, r = 0;
while (i - l > 0 && j - l > 0 && s.charAt(i - l - 1) == t.charAt(j - l - 1)) {
++l;
}
while (i + r + 1 < m && j + r + 1 < n
&& s.charAt(i + r + 1) == t.charAt(j + r + 1)) {
++r;
}
ans += (l + 1) * (r + 1);
}
}
}
return ans;
}
}
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| 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;
}
};
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| 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 := 0, 0
for i > l && j > l && s[i-l-1] == t[j-l-1] {
l++
}
for i+r+1 < m && j+r+1 < n && s[i+r+1] == t[j+r+1] {
r++
}
ans += (l + 1) * (r + 1)
}
}
}
return
}
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Solution 2#
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| 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
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| 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);
}
}
}
return ans;
}
}
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| 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]) {
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[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;
}
};
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| 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)
}
}
}
return
}
|
