Question -- Given two strings, write a program to find the longest string that is a substring of both the given strings.
input: str1 = "abcdefg", str2 = "xyabcz"
output: "abc"
input: str1 = "XYXYXYZ", str2 = "XYZYX"
output: "XYZ"
Illustration from Wikipedia
The longest common substring of the strings "ABABC", "BABCA" and "ABCBA" is string "ABC" of length 3. Other common substrings are "A", "AB", "B", "BA", "BC" and "C".
ABABC
|||
BABCA
|||
ABCBA
Reference -- Wikipedia
/**
* @author MadhavBahl
* @date 03/01/2018
* Referance: https://en.wikipedia.org/wiki/Longest_common_substring_problem
*/
function longestSubstring (str1, str2) {
// initialize the longest substring matrix with all zeroes
let strMat = [],
len1 = str1.length,
len2 = str2.length;
for (let i=0; i<=len2; i++) {
let row = [];
for (let j=0; j<=len1; j++) {
row.push(0);
}
strMat.push(row);
}
// Fill the longest substring matrix and find the maximum element simultaneously
let maxi = 0, maxj = 0, max = strMat[0][0];
for (let i=1; i<=len2; i++) {
for (let j=1; j<=len1; j++) {
if (str2[i-1] === str1[j-1]) {
strMat[i][j] = strMat[i-1][j-1] + 1;
if (strMat[i][j] > max) {
max = strMat[i][j];
maxi = i;
maxj = j;
}
}
}
}
// Find the longest substring
let maxSubStr = '';
for (i=maxi, j=maxj; i>=0; i--, j--) {
if (!(i<=0 || j<=0) && strMat[i][j] !== 0) {
maxSubStr += str2[i-1];
} else {
break;
}
}
// Return the reverse of maxSubStr
return maxSubStr.split('').reverse().join('');
}
console.log(longestSubstring ("abcdefg", "xyabcz"));
console.log(longestSubstring ("XYXYXYZ", "XYZYX"));/**
* @author:divyakhetan
* @date: 3/1/2019
*/
#include <bits/stdc++.h>
using namespace std;
int main() {
string s1,s2;
cin >> s1 >> s2;
int n = s1.length();
int m = s2.length();
int lcs[n + 1][m + 1];
int ma = 0;
int mi = 0;
int mj = 0;
for(int i =0; i <=n; i++){
for(int j = 0; j <=m;j++){
if(i == 0 || j == 0) lcs[i][j] = 0;
else if(s1[i -1] == s2[j - 1]){
lcs[i][j] = 1 + lcs[i - 1][j - 1];
if(lcs[i][j] > ma){
mi = i;
mj = j;
ma = lcs[i][j];
}
}
else{
lcs[i][j]= 0;
}
}
}
int i = 0;
int j = 0;
string ans= "";
for (i=mi, j=mj; i>=0; i--, j--) {
if (!(i<=0 || j<=0) && lcs[i][j] != 0) {
ans += s1[i-1];
} else {
break;
}
}
reverse(ans.begin(), ans.end());
cout << "len : " << ma << " is " << ans;
}/*
* @author : imkaka
* @date : 04/01/2019
*/
#include<iostream>
#include<string>
using namespace std;
void longestCommonSubstrig(string, string);
int main(){
string str1 = "bisect", str2 = "secret";
longestCommonSubstrig(str1, str2); // sec
longestCommonSubstrig("abcxyzey", "cxyazeal"); //cxy
return 0;
}
void longestCommonSubstrig(string str1, string str2){
int m = str1.size();
int n = str2.size();
int LCW[m+1][n+1];
//Track Len and location.
int maxLCW = 0;
int loc = m+1;
for(int i = 0; i <= m; ++i){
LCW[i][n] = 0;
}
for(int i = 0; i <= n; ++i){
LCW[m][i] = 0;
}
// Calculating the max length of common subword
for(int c = n-1; c >= 0; --c){
for(int r = m-1; r >= 0; --r){
if(str1[r] == str2[c]){
LCW[r][c] = 1 + LCW[r+1][c+1];
}
else{
LCW[r][c] = 0;
}
if(LCW[r][c] > maxLCW){
maxLCW = LCW[r][c];
loc = r;
}
}
}
// Finding the Common SubWord
string re = "";
for(int r = loc; r < loc + maxLCW; ++r){
re += char(str1[r]);
}
cout << re << endl;
}
/**
* @date 03/01/1998
* @author SPREEHA DUTTA
*/
import java.util.*;
public class longestCommonSubstring {
static int max=0;static String sm;
public static void generate(String s,String st)
{
String str;
for(int i=0;i<=s.length()-1;i++)
{
for(int j=i+1;j<=s.length();j++)
{
str=s.substring(i,j);
if(st.contains(str))
{
if(max<str.length())
{
max=str.length();
sm=str;
}
}
}
}
}
public static void main(String []args)
{
Scanner sc=new Scanner(System.in);
String s1,s2;
System.out.println("Enter the two strings ");
s1=sc.next();
s2=sc.next();
generate(s1,s2);
generate(s2,s1);
System.out.println("Longest common substring is "+sm);
}
}"""
@author : vishalshirke7
@date : 03/01/2019
"""
def lcs(str1, str2):
m, n = len(str1), len(str2)
L = [[0 for y in range(n + 1)] for x in range(m + 1)]
for i in range(m + 1):
for j in range(n + 1):
if i == 0 or j == 0:
L[i][j] = 0
elif str1[i - 1] == str2[j - 1]:
L[i][j] = L[i - 1][j - 1] + 1
else:
L[i][j] = max(L[i - 1][j], L[i][j - 1])
index = L[m][n]
# Create a character array to store the lcs string
lcs_str = [""] * (index + 1)
lcs_str[index] = ""
# Start from the right-most-bottom-most corner and
# one by one store characters in lcs[]
i = m
j = n
while i > 0 and j > 0:
# If current character in X[] and Y are same, then
# current character is part of LCS
if str1[i - 1] == str2[j - 1]:
lcs_str[index - 1] = str1[i - 1]
i -= 1
j -= 1
index -= 1
# If not same, then find the larger of two and
# go in the direction of larger value
elif L[i - 1][j] > L[i][j - 1]:
i -= 1
else:
j -= 1
print("LCS of " + str1 + " and " + str2 + " is -" + "".join(lcs_str))
lcs(*(input().split()))'''
@author: aaditkamat
@date: 03/01/2019
'''
def print_output(str1, str2):
print(f'The longest common substring in {str1} and {str2} is: {longest_common_substring(str1, str2)}')
def substrings(string):
lst = []
n = len(string)
for i in range(1, n + 1):
for j in range(n + 1 - i):
lst.append(string[j : j + i])
print(set(lst))
return set(lst)
def longest_common_substring(str1, str2):
str1_substrings = substrings(str1)
str2_substrings = substrings(str2)
common_substrings = str1_substrings & str2_substrings
if len(common_substrings) == 0:
common_substrings.add("")
return max(common_substrings, key = lambda x: len(x))
print_output("abcdefg", "xyabcz")
print_output("XYXYXYZ", "XYZYX")
print_output("abcd", "efgh")