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1638. Count Substrings That Differ by One Character

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 <= 100
  • s and t consist of lowercase English letters only.

Solutions

Solution 1

Thinking

Both strings have length at most \(100\). A valid pair differs in exactly one position; extending from that mismatch, the runs of equal characters on each side determine how many cuts work.

Enumerate \((i,j)\) with \(s[i]\ne t[j]\), measure equal spans \(l\) and \(r\), and add \((l+1)(r+1)\).

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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
}

Solution 2

Thinking

Solution 1 extends from every mismatch and repeats comparisons. Precompute the longest equal suffix \(f\) ending at \((i,j)\) and the longest equal prefix \(g\) starting there.

Each mismatch then contributes \((f[i][j]+1)(g[i+1][j+1]+1)\) in \(O(mn)\) total time.

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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
}

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