两个字符串之间的子字符串差异

Question

Given two strings of length n,P = p1...pn and Q = q1...qn, we define M(i, j, k) as the number of mismatches between pi...pi+k-1 and qj..qj+k-1. That is in set notation, M(i, j, k) refers to the size of the set { 0<=x<k | pi+x not equal to qj+x| } { 0<=x<k | pi+x not equal to qj+x| } { 0<=x<k | pi+x not equal to qj+x| } .

Given an integer K, your task is to find the maximum length L such that there exists pair of indices (i,j) for which we have M(i, j, L) <= K . Of course, we should also have i+L-1 <=n and j+L-1 <=n . Input

First line of input contains a single integer T (1 <=T <=10). T test cases follow. Each test case consists of an integer K and two strings P and Q separated by a single space. Output

For each test case output a single integer L which is the maximum value for which there exists pair of indices (i,j) such that M(i, j, L) <=K.

Constraints

0 <= K <= length of the string P Both P & Q would have the same length The size of each of the string would be at the max 1500 All characters in P & Q are lower-case English letters.

Sample Input

3
2 tabriz torino
0 abacba abcaba
3 helloworld yellomarin

Sample Output

4
3
8 

Explanation: First test-case: If we take "briz" from the first string, and "orin" from the second string, then the number of mismatches between these two substrings is equal to 2, and the length of these substrings are 4. That's we have chosen i=3, j=2, L=4, and we have M(3,2,4) = 2.

Second test-case: Since K=0, we should find the longest common substring for the given input strings. We can choose "aba" as the result, and we don't have longer common substring between two strings. So, the answer is 3 for this test-case. That's we have chosen i=1, j=4, and L=3, and we have M(1,4,3)=0.

Third test-case: We can choose "hellowor" from first string and "yellomar" from the second string. So, we have chosen i=1, j=1, and L=8, and we have M(1,1,8)=3. Of course we can also choose i=2, j=2, and L=8 and we still have M(2,2,8)=3.

here is my implementation

import java.io.*;
import java.util.*;

class Solution {

    public static int mismatch(String a, String b, int ii, int jj, int xx) {
        int i, j = 0;
        for (i = 0; i < xx; i++) {
            if (a.charAt(ii) != b.charAt(jj)) {
                j++;
            }
            ii++;
            jj++;
        }
        return j;
    }

    public static boolean find(int x, String a, String b, int kx) {
        int nn = a.length();
        for (int i = 0; i <= (nn - x); i++) {
            for (int j = 0; j <= (nn - x); j++) {
                int k;
                k = mismatch(a, b, i, j, x);
                if (k == kx) {
                    return true;
                }
            }
        }
        return false;
    }

    public static void main(String args[]) throws IOException {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        while (t > 0) {
            int k, n;
            String a, b;
            k = scanner.nextInt();
            a = scanner.next();
            b = scanner.next();
            n = a.length();
            int i = (n + k) / 2;
            int st = k, en = n
                while (i != k || i != n) {
                boolean ch = false, chh = false;
                ch = find(i, a, b, k);
                if (i != n) {
                    chh = find(i + 1, a, b, k);
                }
                if (i == n && ch == true) {
                    System.out.println(i);
                    break;
                }
                if (ch == true && chh == false) {
                    System.out.println(i);
                    break;
                }
                if (ch) {
                    st = i;
                    i = (i + en + 1) / 2;
                } else {
                    en = i;
                    i = (st + i) / 2;
                }
            }
            t--;
        }
    }
}

the above implementation is taking 5.1 sec for input 0f 1500 string length.But maximum time limit in java is 5sec.if any one can improve this code,please kindly share yor thougths

Answer 1

Your code doesn't take 5.1s on the site. They stop running your code as soon as it exceeds the time limit. Your code might be taking even minutes. So, even if you optimize it with this algorithm you will again get 5.1s in details section. So work on your algo, not optimization!

Answer 2

You could make a boolean array compare[n,n] , for which compare[i,j]=(a[i]==b[j]). Later use it instead of making repeating comparisons. You'll have incomparably less comparisons and addressing.

public static int mismatch(String a, String b, int ii, int jj, int xx) {
    int i, j = 0;
    for (i = 0; i < xx; i++) {
        if (! compare[ii,jj]) {
            j++;
        }
        ii++;
        jj++;
    }
    return j;
}