Given two strings 's1' and 's2', find the length of the longest common substring.
Example:
Input: s1 = "GeeksforGeeks", s2 = "GeeksQuiz"
Output : 5
Explanation:
The longest common substring is "Geeks" and is of length 5.
Input: s1 = "abcdxyz", s2 = "xyzabcd"
Output : 4
Explanation:
The longest common substring is "abcd" and is of length 4.
Input: s1 = "abc", s2 = ""
Output : 0.
Naive Iterative Method
A simple solution is to try all substrings beginning with every pair of index from s1 and s2 and keep track of the longest matching substring. We run nested loops to generate all pairs of indexes and then inside this nested loop, we try all lengths. We find the maximum length beginning with every pair. And finally take the max of all maximums.
C++
#include <bits/stdc++.h>
using namespace std;
int maxCommStr(string & s1, string& s2) {
int m = s1.length();
int n = s2.length();
// Consider every pair of index and find the length
// of the longest common substring beginning with
// every pair. Finally return max of all maximums.
int res = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int curr = 0;
while ((i + curr) < m && (j + curr) < n
&& s1[i + curr] == s2[j + curr]) {
curr++;
}
res = max(res, curr);
}
}
return res;
}
int main() {
string s1 = "geeksforgeeks";
string s2 = "practicewritegeekscourses";
cout << maxCommStr(s1, s2) << endl;
return 0;
}
#include <stdio.h>
#include <string.h>
int maxCommStr(char* s1, char* s2) {
int m = strlen(s1);
int n = strlen(s2);
int res = 0;
// Consider every pair of index and find the length
// of the longest common substring beginning with
// every pair. Finally return max of all maximums.
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int curr = 0;
while ((i + curr) < m && (j + curr) < n
&& s1[i + curr] == s2[j + curr]) {
curr++;
}
if (curr > res) {
res = curr;
}
}
}
return res;
}
int main() {
char s1[] = "geeksforgeeks";
char s2[] = "practicewritegeekscourses";
printf("%d\n", maxCommStr(s1, s2));
return 0;
}
class GfG {
// Function to find the length of the longest common substring
static int maxCommStr(String s1, String s2) {
int m = s1.length();
int n = s2.length();
int res = 0;
// Consider every pair of index and find the length
// of the longest common substring beginning with
// every pair. Finally return max of all maximums.
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int curr = 0;
while ((i + curr) < m && (j + curr) < n
&& s1.charAt(i + curr) == s2.charAt(j + curr)) {
curr++;
}
res = Math.max(res, curr);
}
}
return res;
}
public static void main(String[] args) {
String s1 = "geeksforgeeks";
String s2 = "practicewritegeekscourses";
System.out.println(maxCommStr(s1, s2));
}
}
def maxCommStr(s1, s2):
m = len(s1)
n = len(s2)
res = 0
# Consider every pair of index and find the length
# of the longest common substring beginning with
# every pair. Finally return max of all maximums.
for i in range(m):
for j in range(n):
curr = 0
while (i + curr) < m and (j + curr) < n and s1[i + curr] == s2[j + curr]:
curr += 1
res = max(res, curr)
return res
s1 = "geeksforgeeks"
s2 = "practicewritegeekscourses"
print(maxCommStr(s1, s2))
using System;
class GfG {
static int MaxCommStr(string s1, string s2) {
int m = s1.Length;
int n = s2.Length;
int res = 0;
// Consider every pair of index and find the length
// of the longest common substring beginning with
// every pair. Finally return max of all maximums.
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int curr = 0;
while ((i + curr) < m && (j + curr) < n
&& s1[i + curr] == s2[j + curr]) {
curr++;
}
res = Math.Max(res, curr);
}
}
return res;
}
static void Main() {
string s1 = "geeksforgeeks";
string s2 = "practicewritegeekscourses";
Console.WriteLine(MaxCommStr(s1, s2));
}
}
function maxCommStr(s1, s2) {
let m = s1.length;
let n = s2.length;
let res = 0;
// Consider every pair of index and find the length
// of the longest common substring beginning with
// every pair. Finally return max of all maximums.
for (let i = 0; i < m; i++) {
for (let j = 0; j < n; j++) {
let curr = 0;
while ((i + curr) < m && (j + curr) < n &&
s1[i + curr] === s2[j + curr]) {
curr++;
}
res = Math.max(res, curr);
}
}
return res;
}
let s1 = "geeksforgeeks";
let s2 = "practicewritegeekscourses";
console.log(maxCommStr(s1, s2));
Time Complexity: O(m x n x min(m, n))
Auxiliary Space: O(1)
Naive Recursive Method
Below is a recursive version of the above solution, we write a recursive method to find the maximum length substring ending with given pair of indexes. We mainly find longest common suffix. We can solve by finding longest common prefix for every pair (like we did in the above iterative solution). We call this recursive function inside nested loops to get the overall maximum.
C++
#include <bits/stdc++.h>
using namespace std;
// Returns length of the longest common substring
// ending with the last characters. We mainly find
// Longest common suffix.
int LCSuf(const string& s1, const string& s2, int m, int n) {
if (m == 0 || n == 0 || s1[m - 1] != s2[n - 1])
return 0;
return 1 + LCSuf(s1, s2, m - 1, n - 1);
}
int maxCommStr(const string& s1, const string& s2) {
int res = 0;
// Find the longest common substring ending
// at every pair of characters and take the
// maximum of all.
for (int i = 1; i <= s1.size(); i++) {
for (int j = 1; j <= s2.size(); j++) {
res = max(res, LCSuf(s1, s2, i, j));
}
}
return res;
}
int main() {
string s1 = "geeksforgeeks";
string s2 = "practicewritegeekscourses";
cout << maxCommStr(s1, s2) << endl;
return 0;
}
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
// Returns length of the longest common substring
// ending with the last characters. We mainly find
// Longest common suffix.
int LCSuf(const char* s1, const char* s2, int m, int n) {
if (m == 0 || n == 0 || s1[m - 1] != s2[n - 1])
return 0;
return 1 + LCSuf(s1, s2, m - 1, n - 1);
}
int maxCommStr(const char* s1, const char* s2) {
int res = 0;
int m = strlen(s1);
int n = strlen(s2);
// Find the longest common substring ending
// at every pair of characters and take the
// maximum of all.
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
res = res > LCSuf(s1, s2, i, j) ? res : LCSuf(s1, s2, i, j);
}
}
return res;
}
int main() {
char s1[] = "geeksforgeeks";
char s2[] = "practicewritegeekscourses";
printf("%d\n", maxCommStr(s1, s2));
return 0;
}
class GfG {
// Returns length of the longest common substring
// ending with the last characters. We mainly find
// Longest common suffix.
static int LCSuf(String s1, String s2, int m, int n) {
if (m == 0 || n == 0 || s1.charAt(m - 1)
!= s2.charAt(n - 1)) {
return 0;
}
return 1 + LCSuf(s1, s2, m - 1, n - 1);
}
static int maxCommStr(String s1, String s2) {
int res = 0;
int m = s1.length();
int n = s2.length();
// Find the longest common substring ending
// at every pair of characters and take the
// maximum of all.
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
res = Math.max(res, LCSuf(s1, s2, i, j));
}
}
return res;
}
public static void main(String[] args) {
String s1 = "geeksforgeeks";
String s2 = "practicewritegeekscourses";
System.out.println(maxCommStr(s1, s2));
}
}
def LCSuf(s1, s2, m, n):
if m == 0 or n == 0 or s1[m - 1] != s2[n - 1]:
return 0
return 1 + LCSuf(s1, s2, m - 1, n - 1)
def maxCommStr(s1, s2):
res = 0
m = len(s1)
n = len(s2)
# Find the longest common substring ending
# at every pair of characters and take the
# maximum of all.
for i in range(1, m + 1):
for j in range(1, n + 1):
res = max(res, LCSuf(s1, s2, i, j))
return res
s1 = "geeksforgeeks"
s2 = "practicewritegeekscourses"
print(maxCommStr(s1, s2))
using System;
class GfG {
// Returns length of the longest common substring
// ending with the last characters. We mainly find
// Longest common suffix.
static int LCSuf(string s1, string s2, int m, int n) {
if (m == 0 || n == 0 || s1[m - 1] != s2[n - 1]) {
return 0;
}
return 1 + LCSuf(s1, s2, m - 1, n - 1);
}
static int maxCommStr(string s1, string s2) {
int res = 0;
int m = s1.Length;
int n = s2.Length;
// Find the longest common substring ending
// at every pair of characters and take the
// maximum of all.
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
res = Math.Max(res, LCSuf(s1, s2, i, j));
}
}
return res;
}
static void Main() {
string s1 = "geeksforgeeks";
string s2 = "practicewritegeekscourses";
Console.WriteLine(maxCommStr(s1, s2));
}
}
function LCSuf(s1, s2, m, n) {
if (m === 0 || n === 0 || s1[m - 1] !== s2[n - 1]) {
return 0;
}
return 1 + LCSuf(s1, s2, m - 1, n - 1);
}
function maxCommStr(s1, s2) {
let res = 0;
let m = s1.length;
let n = s2.length;
// Find the longest common substring ending
// at every pair of characters and take the
// maximum of all.
for (let i = 1; i <= m; i++) {
for (let j = 1; j <= n; j++) {
res = Math.max(res, LCSuf(s1, s2, i, j));
}
}
return res;
}
let s1 = "geeksforgeeks";
let s2 = "practicewritegeekscourses";
console.log(maxCommStr(s1, s2));
Time Complexity: O(m x n x min(m, n))
Auxiliary Space: O(min(m, n))
Dynamic Programming Solution
If we take a look at the above recursive solution, we can notice that there are overlapping subproblems. We call lcsEnding() for same pair of lengths. With the help of DP, we can optimize the solution to O(m x n)
The longest common suffix has following optimal substructure property
If last characters match, then we reduce both lengths by 1
- LCSuff(s1, s2, m, n) = LCSuff(s1, s2, m-1, n-1) + 1 if s1[m-1] = s2[n-1]
If last characters do not match, then result is 0, i.e.,
- LCSuff(s1, s2, m, n) = 0 if (s1[m-1] != s2[n-1])
Now ,we consider suffixes of different substrings ending at different indexes.
The maximum length Longest Common Suffix is the longest common substring.
LCSubStr(s1, s2, m, n) = Max(LCSuff(s1, s2, i, j)) where 1 <= i <= m and 1 <= j <= n
C++
#include <iostream>
#include <string>
#include <vector>
using namespace std;
// Returns length of longest common substring of
// s1[0..m-1] and s2[0..n-1]
int maxCommStr(const string& s1, const string& s2) {
int m = s1.length();
int n = s2.length();
// Create a table to store lengths of longest
// common suffixes of substrings.
vector<vector<int>> LCSuf(m + 1, vector<int>(n + 1, 0));
// Build LCSuf[m+1][n+1] in bottom-up fashion.
int res = 0;
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (s1[i - 1] == s2[j - 1]) {
LCSuf[i][j] = LCSuf[i - 1][j - 1] + 1;
res = max(res, LCSuf[i][j]);
} else {
LCSuf[i][j] = 0;
}
}
}
return res;
}
int main() {
string s1 = "geeksforgeeks";
string s2 = "ggeegeeksquizpractice";
cout << maxCommStr(s1, s2) << endl;
return 0;
}
#include <stdio.h>
#include <string.h>
// Returns length of longest common substring of
// s1[0..m-1] and s2[0..n-1]
int maxCommStr(const char* s1, const char* s2) {
int m = strlen(s1);
int n = strlen(s2);
// Create a table to store lengths of longest
// common suffixes of substrings.
int LCSuf[m + 1][n + 1];
// Build LCSuf[m+1][n+1] in bottom-up fashion.
int res = 0;
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (s1[i - 1] == s2[j - 1]) {
LCSuf[i][j] = LCSuf[i - 1][j - 1] + 1;
if (LCSuf[i][j] > res) {
res = LCSuf[i][j];
}
} else {
LCSuf[i][j] = 0;
}
}
}
return res;
}
int main() {
char s1[] = "geeksforgeeks";
char s2[] = "ggeegeeksquizpractice";
printf("%d\n", maxCommStr(s1, s2));
return 0;
}
public class Main {
// Returns length of longest common substring of
// s1[0..m-1] and s2[0..n-1]
public static int maxCommStr(String s1, String s2) {
int m = s1.length();
int n = s2.length();
// Create a table to store lengths of longest
// common suffixes of substrings. Note that LCSuf[i][j]
// is going to contain length of longest common suffix
// of s1[0..i-1] and s2[0..j-1].
int[][] LCSuf = new int[m + 1][n + 1];
int res = 0;
// Following steps build LCSuf[m+1][n+1] in bottom up fashion.
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (s1.charAt(i - 1) == s2.charAt(j - 1)) {
LCSuf[i][j] = LCSuf[i - 1][j - 1] + 1;
res = Math.max(res, LCSuf[i][j]);
} else {
LCSuf[i][j] = 0;
}
}
}
return res;
}
public static void main(String[] args) {
String s1 = "geeksforgeeks";
String s2 = "ggeegeeksquizpractice";
System.out.println(maxCommStr(s1, s2));
}
}
def maxCommStr(s1, s2):
m = len(s1)
n = len(s2)
# Create a table to store lengths of longest
# common suffixes of substrings.
LCSuf = [[0] * (n + 1) for _ in range(m + 1)]
res = 0
# Build LCSuf[m+1][n+1] in bottom-up fashion.
for i in range(1, m + 1):
for j in range(1, n + 1):
if s1[i - 1] == s2[j - 1]:
LCSuf[i][j] = LCSuf[i - 1][j - 1] + 1
res = max(res, LCSuf[i][j])
else:
LCSuf[i][j] = 0
return res
s1 = "geeksforgeeks"
s2 = "ggeegeeksquizpractice"
print(maxCommStr(s1, s2))
using System;
public class MainClass {
public static int maxCommStr(string s1, string s2) {
int m = s1.Length;
int n = s2.Length;
// Create a table to store lengths of longest
// common suffixes of substrings.
int[,] LCSuf = new int[m + 1, n + 1];
int res = 0;
// Build LCSuf[m+1][n+1] in bottom-up fashion.
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (s1[i - 1] == s2[j - 1]) {
LCSuf[i, j] = LCSuf[i - 1, j - 1] + 1;
res = Math.Max(res, LCSuf[i, j]);
} else {
LCSuf[i, j] = 0;
}
}
}
return res;
}
public static void Main(string[] args) {
string s1 = "geeksforgeeks";
string s2 = "ggeegeeksquizpractice";
Console.WriteLine(maxCommStr(s1, s2));
}
}
function maxCommStr(s1, s2) {
let m = s1.length;
let n = s2.length;
// Create a table to store lengths of longest
// common suffixes of substrings.
let LCSuf = Array.from(Array(m + 1), () => Array(n + 1).fill(0));
let res = 0;
// Build LCSuf[m+1][n+1] in bottom-up fashion.
for (let i = 1; i <= m; i++) {
for (let j = 1; j <= n; j++) {
if (s1[i - 1] === s2[j - 1]) {
LCSuf[i][j] = LCSuf[i - 1][j - 1] + 1;
res = Math.max(res, LCSuf[i][j]);
} else {
LCSuf[i][j] = 0;
}
}
}
return res;
}
let s1 = "geeksforgeeks";
let s2 = "ggeegeeksquizpractice";
console.log(maxCommStr(s1, s2));
Time Complexity: O(m*n)
Auxiliary Space: O(m*n), since m*n extra space has been taken.
Space Optimized DP
In the above approach, we are only using the last row of the 2-D array. Hence we can optimize the space by using a 1-D array.
C++
#include <iostream>
#include <vector>
using namespace std;
// Function to find the length of the longest LCS
// with space optimization
int longestCommonSubstr(const string& s1, const string& s2) {
int m = s1.length();
int n = s2.length();
// Create a 1D array to store the previous row's results
vector<int> prev(n + 1, 0);
int res = 0;
for (int i = 1; i <= m; i++) {
// Create a temporary array to store the current row
vector<int> curr(n + 1, 0);
for (int j = 1; j <= n; j++) {
if (s1[i - 1] == s2[j - 1]) {
curr[j] = prev[j - 1] + 1;
res = max(res, curr[j]);
} else {
curr[j] = 0;
}
}
// Move the current row's data to the previous row
prev = curr;
}
return res;
}
// Driver Code
int main() {
string s1 = "geeksforgeeks";
string s2 = "ggeegeeksquizpractice";
cout << longestCommonSubstr(s1, s2) << endl;
return 0;
}
#include <stdio.h>
#include <string.h>
// Function to find the length of the longest LCS
// with space optimization
int longestCommonSubstr(const char* s1, const char* s2) {
int m = strlen(s1);
int n = strlen(s2);
// Create a 1D array to store the previous row's results
int prev[n + 1];
memset(prev, 0, sizeof(prev));
int res = 0;
for (int i = 1; i <= m; i++) {
// Create a temporary array to store the current row
int curr[n + 1];
memset(curr, 0, sizeof(curr));
for (int j = 1; j <= n; j++) {
if (s1[i - 1] == s2[j - 1]) {
curr[j] = prev[j - 1] + 1;
if (curr[j] > res) {
res = curr[j];
}
} else {
curr[j] = 0;
}
}
// Move the current row's data to the previous row
memcpy(prev, curr, sizeof(curr));
}
return res;
}
int main() {
char s1[] = "geeksforgeeks";
char s2[] = "ggeegeeksquizpractice";
printf("%d\n", longestCommonSubstr(s1, s2));
return 0;
}
public class GfG {
// Function to find the length of the longest LCS
// with space optimization
public static int longestCommonSubstr(String s1, String s2) {
int m = s1.length();
int n = s2.length();
// Create a 1D array to store the previous row's results
int[] prev = new int[n + 1];
int res = 0;
for (int i = 1; i <= m; i++) {
// Create a temporary array to store the current row
int[] curr = new int[n + 1];
for (int j = 1; j <= n; j++) {
if (s1.charAt(i - 1) == s2.charAt(j - 1)) {
curr[j] = prev[j - 1] + 1;
res = Math.max(res, curr[j]);
} else {
curr[j] = 0;
}
}
// Move the current row's data to the previous row
prev = curr;
}
return res;
}
public static void main(String[] args) {
String s1 = "geeksforgeeks";
String s2 = "ggeegeeksquizpractice";
System.out.println(longestCommonSubstr(s1, s2));
}
}
def longestCommonSubstr(s1, s2):
m = len(s1)
n = len(s2)
# Create a 1D array to store the previous row's results
prev = [0] * (n + 1)
res = 0
for i in range(1, m + 1):
# Create a temporary array to store the current row
curr = [0] * (n + 1)
for j in range(1, n + 1):
if s1[i - 1] == s2[j - 1]:
curr[j] = prev[j - 1] + 1
res = max(res, curr[j])
else:
curr[j] = 0
# Move the current row's data to the previous row
prev = curr
return res
# Driver Code
s1 = "geeksforgeeks"
s2 = "ggeegeeksquizpractice"
print(longestCommonSubstr(s1, s2))
using System;
public class MainClass {
// Function to find the length of the longest LCS
// with space optimization
public static int longestCommonSubstr(string s1, string s2) {
int m = s1.Length;
int n = s2.Length;
// Create a 1D array to store the previous row's results
int[] prev = new int[n + 1];
int res = 0;
for (int i = 1; i <= m; i++) {
// Create a temporary array to store the current row
int[] curr = new int[n + 1];
for (int j = 1; j <= n; j++) {
if (s1[i - 1] == s2[j - 1]) {
curr[j] = prev[j - 1] + 1;
res = Math.Max(res, curr[j]);
} else {
curr[j] = 0;
}
}
// Move the current row's data to the previous row
prev = curr;
}
return res;
}
public static void Main(string[] args) {
string s1 = "geeksforgeeks";
string s2 = "ggeegeeksquizpractice";
Console.WriteLine(longestCommonSubstr(s1, s2));
}
}
function longestCommonSubstr(s1, s2) {
let m = s1.length;
let n = s2.length;
// Create a 1D array to store the previous row's results
let prev = new Array(n + 1).fill(0);
let res = 0;
for (let i = 1; i <= m; i++) {
// Create a temporary array to store the current row
let curr = new Array(n + 1).fill(0);
for (let j = 1; j <= n; j++) {
if (s1[i - 1] === s2[j - 1]) {
curr[j] = prev[j - 1] + 1;
res = Math.max(res, curr[j]);
} else {
curr[j] = 0;
}
}
// Move the current row's data to the previous row
prev = curr;
}
return res;
}
// Driver Code
let s1 = "geeksforgeeks";
let s2 = "ggeegeeksquizpractice";
console.log(longestCommonSubstr(s1, s2));
Time Complexity: O(m * n)
Auxiliary Space: O(n)
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Printing Longest Common Subsequence Given two sequences, print the longest subsequence present in both of them. Examples: LCS for input Sequences “ABCDGH†and “AEDFHR†is “ADH†of length 3. LCS for input Sequences “AGGTAB†and “GXTXAYB†is “GTAB†of length 4.We have discussed Longest Common Subsequence (LCS) problem in a previous post
15+ min read
Longest Common Subsequence | DP using Memoization Given two strings s1 and s2, the task is to find the length of the longest common subsequence present in both of them. Examples: Input: s1 = “ABCDGHâ€, s2 = “AEDFHR†Output: 3 LCS for input Sequences “AGGTAB†and “GXTXAYB†is “GTAB†of length 4. Input: s1 = “striverâ€, s2 = “raj†Output: 1 The naive s
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Longest Common Increasing Subsequence (LCS + LIS) Given two arrays, a[] and b[], find the length of the longest common increasing subsequence(LCIS). LCIS refers to a subsequence that is present in both arrays and strictly increases.Prerequisites: LCS, LIS.Examples:Input: a[] = [3, 4, 9, 1], b[] = [5, 3, 8, 9, 10, 2, 1]Output: 2Explanation: The long
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LCS (Longest Common Subsequence) of three strings Given three strings s1, s2 and s3. Your task is to find the longest common sub-sequence in all three given sequences.Note: This problem is simply an extension of LCS.Examples: Input: s1 = "geeks" , s2 = "geeksfor", s3 = "geeksforgeeks"Output : 5Explanation: Longest common subsequence is "geeks" i.e.
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C++ Program for Longest Common Subsequence LCS Problem Statement: Given two sequences, find the length of longest subsequence present in both of them. A subsequence is a sequence that appears in the same relative order, but not necessarily contiguous. For example, "abc", "abg", "bdf", "aeg", '"acefg", .. etc are subsequences of "abcdefg". So
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Java Program for Longest Common Subsequence LCS Problem Statement: Given two sequences, find the length of longest subsequence present in both of them. A subsequence is a sequence that appears in the same relative order, but not necessarily contiguous. For example, "abc", "abg", "bdf", "aeg", '"acefg", .. etc are subsequences of "abcdefg". So
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Python Program for Longest Common Subsequence LCS Problem Statement: Given two sequences, find the length of longest subsequence present in both of them. A subsequence is a sequence that appears in the same relative order, but not necessarily contiguous. For example, "abc", "abg", "bdf", "aeg", '"acefg", .. etc are subsequences of "abcdefg". So
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Problems on LCS
Edit distance and LCS (Longest Common Subsequence)In standard Edit Distance where we are allowed 3 operations, insert, delete, and replace. Consider a variation of edit distance where we are allowed only two operations insert and delete, find edit distance in this variation. Examples: Input : str1 = "cat", st2 = "cut"Output : 2We are allowed to ins
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Length of longest common subsequence containing vowelsGiven two strings X and Y of length m and n respectively. The problem is to find the length of the longest common subsequence of strings X and Y which contains all vowel characters.Examples: Input : X = "aieef" Y = "klaief"Output : aieInput : X = "geeksforgeeks" Y = "feroeeks"Output : eoeeSource:Pay
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Longest Common Subsequence (LCS) by repeatedly swapping characters of a string with characters of another stringGiven two strings A and B of lengths N and M respectively, the task is to find the length of the longest common subsequence that can be two strings if any character from string A can be swapped with any other character from B any number of times. Examples: Input: A = "abdeff", B = "abbet"Output: 4Ex
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Longest Common Subsequence with at most k changes allowedGiven two sequence P and Q of numbers. The task is to find Longest Common Subsequence of two sequences if we are allowed to change at most k element in first sequence to any value. Examples: Input : P = { 8, 3 } Q = { 1, 3 } K = 1 Output : 2 If we change first element of first sequence from 8 to 1,
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Minimum cost to make Longest Common Subsequence of length kGiven two string X, Y and an integer k. Now the task is to convert string X with the minimum cost such that the Longest Common Subsequence of X and Y after conversion is of length k. The cost of conversion is calculated as XOR of old character value and new character value. The character value of 'a
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Longest Common SubstringGiven two strings 's1' and 's2', find the length of the longest common substring. Example: Input: s1 = "GeeksforGeeks", s2 = "GeeksQuiz" Output : 5 Explanation:The longest common substring is "Geeks" and is of length 5.Input: s1 = "abcdxyz", s2 = "xyzabcd" Output : 4Explanation:The longest common su
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Longest Common Subsequence of two arrays out of which one array consists of distinct elements onlyGiven two arrays firstArr[], consisting of distinct elements only, and secondArr[], the task is to find the length of LCS between these 2 arrays. Examples: Input: firstArr[] = {3, 5, 1, 8}, secondArr[] = {3, 3, 5, 3, 8}Output: 3.Explanation: LCS between these two arrays is {3, 5, 8}. Input : firstAr
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Longest Repeating SubsequenceGiven a string s, the task is to find the length of the longest repeating subsequence, such that the two subsequences don't have the same string character at the same position, i.e. any ith character in the two subsequences shouldn't have the same index in the original string. Examples:Input: s= "ab
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Longest Common Anagram SubsequenceGiven two strings str1 and str2 of length n1 and n2 respectively. The problem is to find the length of the longest subsequence which is present in both the strings in the form of anagrams. Note: The strings contain only lowercase letters. Examples: Input : str1 = "abdacp", str2 = "ckamb" Output : 3
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Length of Longest Common Subsequence with given sum KGiven two arrays a[] and b[] and an integer K, the task is to find the length of the longest common subsequence such that sum of elements is equal to K. Examples: Input: a[] = { 9, 11, 2, 1, 6, 2, 7}, b[] = {1, 2, 6, 9, 2, 3, 11, 7}, K = 18Output: 3Explanation: Subsequence { 11, 7 } and { 9, 2, 7 }
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Longest Common Subsequence with no repeating characterGiven two strings s1 and s2, the task is to find the length of the longest common subsequence with no repeating character. Examples: Input: s1= "aabbcc", s2= "aabc"Output: 3Explanation: "aabc" is longest common subsequence but it has two repeating character 'a'.So the required longest common subsequ
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Find the Longest Common Subsequence (LCS) in given K permutationsGiven K permutations of numbers from 1 to N in a 2D array arr[][]. The task is to find the longest common subsequence of these K permutations. Examples: Input: N = 4, K = 3arr[][] = {{1, 4, 2, 3}, {4, 1, 2, 3}, {1, 2, 4, 3}}Output: 3Explanation: Longest common subsequence is {1, 2, 3} which has leng
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Find length of longest subsequence of one string which is substring of another stringGiven two strings X and Y. The task is to find the length of the longest subsequence of string X which is a substring in sequence Y.Examples: Input : X = "ABCD", Y = "BACDBDCD"Output : 3Explanation: "ACD" is longest subsequence of X which is substring of Y.Input : X = "A", Y = "A"Output : 1Perquisit
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Length of longest common prime subsequence from two given arraysGiven two arrays arr1[] and arr2[] of length N and M respectively, the task is to find the length of the longest common prime subsequence that can be obtained from the two given arrays. Examples: Input: arr1[] = {1, 2, 3, 4, 5, 6, 7, 8, 9}, arr2[] = {2, 5, 6, 3, 7, 9, 8} Output: 4 Explanation: The l
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A Space Optimized Solution of LCSGiven two strings, s1 and s2, the task is to find the length of the Longest Common Subsequence. If there is no common subsequence, return 0.Examples:Input: s1 = “ABCDGHâ€, s2 = “AEDFHRâ€Output: 3Explanation: The longest subsequence present in both strings is "ADH".Input: s1 = “AGGTABâ€, s2 = “GXTXAYBâ€O
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Longest common subarray in the given two arraysGiven two arrays A[] and B[] of N and M integers respectively, the task is to find the maximum length of an equal subarray or the longest common subarray between the two given array. Examples: Input: A[] = {1, 2, 8, 2, 1}, B[] = {8, 2, 1, 4, 7} Output: 3 Explanation: The subarray that is common to b
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Number of ways to insert a character to increase the LCS by oneGiven two strings A and B. The task is to count the number of ways to insert a character in string A to increase the length of the Longest Common Subsequence between string A and string B by 1. Examples: Input : A = "aa", B = "baaa" Output : 4 The longest common subsequence shared by string A and st
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Longest common subsequence with permutations allowedGiven two strings in lowercase, find the longest string whose permutations are subsequences of given two strings. The output longest string must be sorted. Examples: Input : str1 = "pink", str2 = "kite" Output : "ik" The string "ik" is the longest sorted string whose one permutation "ik" is subseque
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Longest subsequence such that adjacent elements have at least one common digitGiven an array arr[], the task is to find the length of the longest sub-sequence such that adjacent elements of the subsequence have at least one digit in common.Examples: Input: arr[] = [1, 12, 44, 29, 33, 96, 89] Output: 5 Explanation: The longest sub-sequence is [1 12 29 96 89]Input: arr[] = [12,
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Longest subsequence with different adjacent charactersGiven string str. The task is to find the longest subsequence of str such that all the characters adjacent to each other in the subsequence are different. Examples:Â Â Input: str = "ababa"Â Output: 5Â Explanation:Â "ababa" is the subsequence satisfying the condition Input: str = "xxxxy"Â Output: 2Â Explan
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Longest subsequence such that difference between adjacents is oneGiven an array arr[] of size n, the task is to find the longest subsequence such that the absolute difference between adjacent elements is 1.Examples: Input: arr[] = [10, 9, 4, 5, 4, 8, 6]Output: 3Explanation: The three possible subsequences of length 3 are [10, 9, 8], [4, 5, 4], and [4, 5, 6], wher
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Longest Uncommon SubsequenceGiven two strings, find the length of longest uncommon subsequence of the two strings. The longest uncommon subsequence is defined as the longest subsequence of one of these strings which is not a subsequence of other strings. Examples: Input : "abcd", "abc"Output : 4The longest subsequence is 4 bec
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LCS formed by consecutive segments of at least length KGiven two strings s1, s2 and K, find the length of the longest subsequence formed by consecutive segments of at least length K. Examples: Input : s1 = aggayxysdfa s2 = aggajxaaasdfa k = 4 Output : 8 Explanation: aggasdfa is the longest subsequence that can be formed by taking consecutive segments, m
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Longest Increasing Subsequence using Longest Common Subsequence AlgorithmGiven an array arr[] of N integers, the task is to find and print the Longest Increasing Subsequence.Examples: Input: arr[] = {12, 34, 1, 5, 40, 80} Output: 4 {12, 34, 40, 80} and {1, 5, 40, 80} are the longest increasing subsequences.Input: arr[] = {10, 22, 9, 33, 21, 50, 41, 60, 80} Output: 6 Prer
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