Number of Substrings with Single Character Difference - Microsoft Top Interview Questions


Problem Statement :


You are given two lowercase alphabet strings s and t. Return the number of pairs of substrings across s and t such that if we replace a single character to a different character, it becomes a substring of t.

Constraints

0 ≤ n ≤ 100 where n is the length of s

0 ≤ m ≤ 100 where m is the length of t

Example 1

Input

s = "ab"

t = "db"

Output

4

Explanation

We can have the following substrings:



"a" changed to "d"

"a" changed to "b"

"b" changed to "d"

"ab" changed to "db"


Solution :



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                        Solution in C++ :

int solve(string s, string t) {
    int m = s.length(), n = t.length();

    /*
     * dp[i][j][0] -> number of same substrings ending at the ith character in s and j th chaaracter
     * in t such that no change has been made to the string s 'yet'. dp[i][j][0] -> number of same
     * substrings ending at the ith character in s and j th chaaracter in t such that exactly one
     * change has been made to the string.
     */
    vector<vector<vector<int>>> dp(m + 1, vector<vector<int>>(n + 1, vector<int>(2)));

    for (int i = 1; i <= m; ++i) {       // ending at i - 1th character in s
        for (int j = 1; j <= n; ++j) {   // ending at the j - 1 th character in t
            if (s[i - 1] == t[j - 1]) {  // if the characters are same
                /*
                * number of equal substrings would be:
                * 1. With no change -> 1 + number of equal substrings ending at s[i - 2] and t[j -
                2] with no change 1 because s[i - 1] == t[j - 1], ie substring of length 1 is also
                valid
                * 2. With change -> number of equal substrings ending at s[i - 2] and t[j - 2] with
                exatly one change.
                */
                dp[i][j][0] = 1 + dp[i - 1][j - 1][0];
                dp[i][j][1] = dp[i - 1][j - 1][1];
            } else
                dp[i][j][1] = 1 +
                              dp[i - 1][j - 1]
                                [0];  // if charactes are not same then we have only one option, ie
                                      // to change it. So, number of equal substrings ending here ->
                                      // 1 + number of equal substrings ending at s[i - 1] and t[j -
                                      // 1] with no change. 1 because when s[i - 1] is changed to
                                      // t[j - 1] we have equal substring of length 1.
        }
    }

    // count all substrings ending at some s[i - 1] and t[j - 1] and with exactly one change
    int res = 0;
    for (int i = 0; i <= m; ++i) {
        for (int j = 0; j <= n; ++j) res += dp[i][j][1];
    }

    return res;
}
                    

                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(String s, String t) {
        /*
            definitions :
                let string s = s1 s2 s3 ..... si-1 si si+1 ..... sm-2 sm-1 sm
                let string t = t1 t2 t3 ..... tj-1 tj tj+1 ..... tn-2 tn-1 tn
                LEFT[i][j] = max size equal substring ending at position i in string s and position
           j in string t when traversing s and t from left side RIGHT[i][j] = max size equal
           substring ending at position i in string s and position j in string t when traversing s
           and t from right side

            Base Cases :
                LEFT[0][0] = 0
                RIGH[M + 1][N + 1] = 0

            Transitive Equations :
                LEFT[i][j] = (1 + LEFT[i - 1][j - 1])   ; si == tj
                LEFT[i][j] = 0                          ; si != tj
                RIGHT[i][j] = (1 + RIGHT[i + 1][j + 1]) ; si == tj
                RIGHT[i][j] = 0                         ; si != tj

            Final Answer :
                count += (1 + LEFT[i - 1][j - 1] + RIGHT[i + 1][j + 1] + (LEFT[i - 1][j - 1] *
           RIGHT[i + 1][j + 1])) ; si != tj
        */
        int M = s.length(), N = t.length(), count = 0;
        int[][] LEFT = new int[M + 1][N + 1], RIGHT = new int[M + 2][N + 2];
        for (int i = 1; i <= M; i++) {
            for (int j = 1; j <= N; j++) {
                if (s.charAt(i - 1) == t.charAt(j - 1)) {
                    LEFT[i][j] = (1 + LEFT[i - 1][j - 1]);
                }
            }
        }
        for (int i = M; i >= 1; i--) {
            for (int j = N; j >= 1; j--) {
                if (s.charAt(i - 1) == t.charAt(j - 1)) {
                    RIGHT[i][j] = (1 + RIGHT[i + 1][j + 1]);
                } else {
                    count += (1 + LEFT[i - 1][j - 1] + RIGHT[i + 1][j + 1]
                        + (LEFT[i - 1][j - 1] * RIGHT[i + 1][j + 1]));
                }
            }
        }
        return count;
    }
}
                    

                        Solution in Python : 
                            
class Solution:
    def solve(self, s, t):
        ans = 0

        for i in range(len(s)):
            for j in range(len(t)):
                mismatches = 0
                for k in range(min(len(s) - i, len(t) - j)):
                    mismatches += s[i + k] != t[j + k]
                    if mismatches == 1:
                        ans += 1

        return ans
                    

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