Rust

Rust port of @Ray's C++ answer.

Try it online!

use std::cmp::{max, min};
use std::io::{self, BufRead};

fn main() {
    let stdin = io::stdin();
    let mut lines = stdin.lock().lines();

    // Read the two input lines
    let s1 = lines.next().unwrap().unwrap();
    let s2 = lines.next().unwrap().unwrap();

    // Convert the strings to char arrays for easier indexing
    let chars1: Vec<char> = s1.chars().collect();
    let chars2: Vec<char> = s2.chars().collect();

    let n1 = chars1.len();
    let n2 = chars2.len();

    let mut max_len = 0;
    let mut max_end = -1;

    // Loop over the range [1-n2, n1)
    for i in (1 - n2 as isize)..(n1 as isize) {
        let mut f0 = 0;
        let mut f1 = 0;
        let mut max_len2 = 0;
        let mut max_end2 = -1;

        // Determine the inclusive range for j based on i
        let start_j = max(i, 0) as usize;            // ensures j won’t be negative
        let end_j = min(n1 as isize, i + n2 as isize) as usize; // ensures j won’t exceed n1

        for j in start_j..end_j {
            // Because i might be negative, and j is always ≥ 0, the subindex for s2 is (j - i)
            let index_in_s2 = (j as isize - i) as usize;

            if chars1[j] == chars2[index_in_s2] {
                f0 += 1;
                f1 += 1;
            } else {
                f1 = f0 + 1;
                f0 = 0;
            }

            if f1 > max_len2 {
                max_len2 = f1;
                max_end2 = j as isize + 1; // +1 so it's the position after the matched section
            }
        }

        if max_len2 > max_len {
            max_len = max_len2;
            max_end = max_end2;
        }
    }

    // Equivalent to assert(max_end != -1);
    assert!(max_end != -1);

    // Print the same output as the original code: "start_index end_index"
    // Note that (max_end - max_len) + 1 is the starting index in 1-based indexing.
    let start_pos = max_end - max_len as isize + 1;
    println!("{} {}", start_pos, max_end);
}

C++ time: O(n^2), extra space: O(1)

It takes 0.2s to complete the 15K data on my machine.

To compile it, use:

g++ -std=c++11 -O3 code.cpp -o code

To run it, use:

./code < INPUT_FILE_THAT_CONTAINS_TWO_LINES_SPERATED_BY_A_LINE_BREAK

Explaination

The idea is simple, for string s1 and s2, we try to offset s2 by i:

s1: abcabcabc
s2: bcabcab

When offset is 3:

s1: abcabcabc
s2:    bcabcab

Then, for each offset i, we perform a dynamic programing scan on s1[i:] and s2. For each j, let f[j, 0] be the maximum length d such that s1[j - d:j] == s2[j - i - d: j - i]. Similarly, let f[j, 1] be the maximum length d such that strings s1[j - d:j] and s2[j - i - d:j - i] differ by at most 1 character.

So for s1[j] == s2[j - i], we have:

f[j, 0] = f[j - 1, 0] + 1  // concat solution in f[j - 1, 0] and s1[j]
f[j, 1] = f[j - 1, 1] + 1  // concat solution in f[j - 1, 1] and s1[j]

Otherwise:

f[j, 0] = 0  // the only choice is empty string
f[j, 1] = f[j - 1, 0] + 1  // concat solution in f[j - 1, 0] and s1[j] (or s2[j - i])

And:

f[-1, 0] = f[-1, 1] = 0 

Since we only need f[j - 1, :] to calculate f[j, :], only O(1) extra space is used.

Finally, the max length will be:

max(f[j, 1] for all valid j and all i)

Code

#include <string>
#include <cassert>
#include <iostream>

using namespace std;

int main() {
    string s1, s2;
    getline(cin, s1);
    getline(cin, s2);
    int n1, n2;
    n1 = s1.size();
    n2 = s2.size();
    int max_len = 0;
    int max_end = -1;
    for(int i = 1 - n2; i < n1; i++) {
        int f0, f1;
        int max_len2 = 0;
        int max_end2 = -1;
        f0 = f1 = 0;
        for(int j = max(i, 0), j_end = min(n1, i + n2); j < j_end; j++) {
            if(s1[j] == s2[j - i]) {
                f0 += 1;
                f1 += 1;
            } else {
                f1 = f0 + 1;
                f0 = 0;
            }
            if(f1 > max_len2) {
                max_len2 = f1;
                max_end2 = j + 1;
            }
        }
        if(max_len2 > max_len) {
            max_len = max_len2;
            max_end = max_end2;
        }
    }
    assert(max_end != -1);
    // cout << max_len << endl;
    cout << max_end - max_len + 1 << " " << max_end << endl;
}

R

Seems I was over complicating the problem with the previous solution. This is about 50% quicker (23 secs on 15k strings) than the previous one and pretty simple.

rm(list=ls(all=TRUE))
a="xxxappleyyyyyyy"
b="zapllezzz"
s=proc.time()
matchLen=1
matchIndex=1
indexA = 1
repeat {    
    i = 0
    repeat {
        srch = substring(a,indexA,indexA+matchLen+i)
        if (agrepl(srch,b,max.distance=list(insertions=0,deletions=0,substitutions=1)))
            i = i + 1
        else {
            if (i > 0) {
                matchLen = matchLen + i - 1
                matchIndex = indexA
            }
            break
        }
    }
    indexA=indexA+1
    if (indexA + matchLen > nchar(a)) break
}
c(matchIndex, matchLen + matchIndex)
print (substring(a,matchIndex, matchLen + matchIndex))
print(proc.time()-s)

This will never be a contender due to the language, but I did have a bit of fun doing it.
Not sure of the complexity of it, but over a couple of ~15k strings it takes 43 secs using a single thread. The largest portion of that was the sorting of the arrays. I tried some other libraries, but without significant improvement.

a="xxxappleyyyyyyy"
b="zapllezzz"
s=proc.time()
N=nchar
S=substring
U=unlist
V=strsplit
A=N(a)
B=N(b)
a=S(a,1:A)
b=S(b,1:B)
a=sort(a,method="quick")
b=sort(b,method="quick")
print(proc.time()-s)
C=D=1
E=X=Y=I=0
repeat{
    if(N(a[C])>E && N(b[D])>E){
        for(i in E:min(N(a[C]),N(b[D]))){
            if (sum(U(V(S(a[C],1,i),''))==U(V(S(b[D],1,i),'')))>i-2){
                F=i
            } else break
        }
        if (F>E) {
            X=A-N(a[C])+1
            Y=X+F-1
            E=F
        }
        if (a[C]<b[D])
            C=C+1
            else
            D=D+1
    } else
        if(S(a[C],1,1)<S(b[D],1,1))C=C+1 else D=D+1
    if(C>A||D>B)break
}
c(X,Y)
print(proc.time()-s)

Method:

• Create a suffix array for each string
• Order the suffix arrays
• Step through each of the arrays in a staggered sort of way comparing the beginning of each

C++

I tried thinking of a good algorithm to do this, but I'm a bit distracted today and couldn't think of anything that would work well. This runs at O(n^3) time, so it's veeeery slow. The other option I thought of could have been theoretically faster but would have taken O(n^2) space, and with inputs of 1M, it would have been worse even.

It's shameful, it takes 190 seconds for inputs of 15K. I'll try to improve it. Edit: Added multiprocessing. Now takes 37 seconds for 15K input on 8 threads.

#include <string>
#include <vector>
#include <sstream>
#include <chrono>
#include <thread>
#include <atomic>
#undef cin
#undef cout
#include <iostream>

using namespace std;

typedef pair<int, int> range;

int main(int argc, char ** argv)
{
    string a = "xxxappleyyyyyyy";
    string b = "zapllezzz";

    getline(cin, a);
    getline(cin, b);

    range longestA;
    range longestB;

    using namespace std::chrono;

    high_resolution_clock::time_point t1 = high_resolution_clock::now();

    unsigned cores = thread::hardware_concurrency(); cores = cores > 0 ? cores : 1;

    cout << "Processing on " << cores << " cores." << endl;

    atomic<int> processedCount(0);

    vector<thread> threads;

    range* longestAs = new range[cores];
    range* longestBs = new range[cores];
    for (int t = 0; t < cores; ++t)
    {
        threads.push_back(thread([&processedCount, cores, t, &a, &b, &longestBs, &longestAs]()
        {
            int la = a.length();
            int l = la / cores + (t==cores-1? la % cores : 0);
            int lb = b.length();
            int aS = t*(la/cores);

            for (int i = aS; i < aS + l; ++i)
            {
                int count = processedCount.fetch_add(1);
                if ((count+1) * 100 / la > count * 100 / la)
                {
                    cout << (count+1) * 100 / la << "%" << endl;
                }
                for (int j = 0; j < lb; ++j)
                {
                    range currentB = make_pair(j, j);
                    bool letterChanged = false;
                    for (int k = 0; k + j < lb && k + i < la; ++k)
                    {
                        if (a[i + k] == b[j + k])
                        {
                            currentB = make_pair(j, j + k);
                        }
                        else if (!letterChanged)
                        {
                            letterChanged = true;
                            currentB = make_pair(j, j + k);
                        }
                        else
                        {
                            break;
                        }
                    }
                    if (currentB.second - currentB.first > longestBs[t].second - longestBs[t].first)
                    {
                        longestBs[t] = currentB;
                        longestAs[t] = make_pair(i, i + currentB.second - currentB.first);
                    }
                }
            }
        }));
    }

    longestA = make_pair(0,0);
    for(int t = 0; t < cores; ++t)
    {
        threads[t].join();

        if (longestAs[t].second - longestAs[t].first > longestA.second - longestA.first)
        {
            longestA = longestAs[t];
            longestB = longestBs[t];
        }
    }

    high_resolution_clock::time_point t2 = high_resolution_clock::now();

    duration<double> time_span = duration_cast<duration<double>>(t2 - t1);

    cout << "First substring at range (" << longestA.first << ", " << longestA.second << "):" << endl;
    cout << a.substr(longestA.first, longestA.second - longestA.first + 1) << endl;
    cout << "Second substring at range (" << longestB.first << ", " << longestB.second << "):" << endl;
    cout << b.substr(longestB.first, longestB.second - longestB.first + 1) << endl;
    cout << "It took me " << time_span.count() << " seconds for input lengths " << a.length() << " and " << b.length() <<"." << endl;

    char c;
    cin >> c;
    return 0;
}