Longest Palindromic Subsequence (LPS)
Last Updated :
10 Mar, 2025
Given a string s, find the length of the Longest Palindromic Subsequence in it.
Note: The Longest Palindromic Subsequence (LPS) is the maximum-length subsequence of a given string that is also a Palindrome.
Longest Palindromic SubsequenceExamples:
Input: s = "bbabcbcab"
Output: 7
Explanation: Subsequence "babcbab" is the longest subsequence which is also a palindrome.
Input: s = "abcd"
Output: 1
Explanation: "a", "b", "c" and "d" are palindromic and all have a length 1.
[Naive Approach] - Using Recursion - O(2^n) Time and O(1) Space
The idea is to recursively generate all possible subsequences of the given string s and find the longest palindromic subsequence. To do so, create two counters low and high and set them to point to first and last character of string s. Start matching the characters from both the ends of the string. For each case, there are two possibilities:
- If characters are matching, increment the value low and decrement the value high by 1 and recur to find the LPS of new substring. And return the value result + 2.
- Else make two recursive calls for (low + 1, hi) and (lo, hi-1). And return the max of 2 calls.
C++
// C++ program to find the lps
#include <bits/stdc++.h>
using namespace std;
// Returns the length of the longest
// palindromic subsequence in seq
int lps(const string& s, int low, int high) {
// Base case
if(low > high) return 0;
// If there is only 1 character
if (low == high)
return 1;
// If the first and last characters match
if (s[low] == s[high])
return lps(s, low + 1, high - 1) + 2;
// If the first and last characters do not match
return max(lps(s, low, high - 1), lps(s, low + 1, high));
}
int longestPalinSubseq(string &s) {
return lps(s, 0, s.size() - 1);
}
int main() {
string s = "bbabcbcab";
cout << longestPalinSubseq(s);
return 0;
}
// C program to find the lps
#include <stdio.h>
#include <string.h>
// Returns the length of the longest
// palindromic subsequence in seq
int lps(const char *s, int low, int high) {
// Base case
if (low > high) return 0;
// If there is only 1 character
if (low == high)
return 1;
// If the first and last characters match
if (s[low] == s[high])
return lps(s, low + 1, high - 1) + 2;
// If the first and last characters do not match
int a = lps(s, low, high - 1);
int b = lps(s, low + 1, high);
return (a > b)? a: b;
}
int longestPalinSubseq(char *s) {
int n = strlen(s);
return lps(s, 0, n - 1);
}
int main() {
char s[] = "bbabcbcab";
printf("%d", longestPalinSubseq(s));
return 0;
}
// Java program to find the lps
class GfG {
// Returns the length of the longest
// palindromic subsequence in seq
static int lps(String s, int low, int high) {
// Base case
if (low > high) return 0;
// If there is only 1 character
if (low == high)
return 1;
// If the first and last characters match
if (s.charAt(low) == s.charAt(high))
return lps(s, low + 1, high - 1) + 2;
// If the first and last characters do not match
return Math.max(lps(s, low, high - 1), lps(s, low + 1, high));
}
static int longestPalinSubseq(String s) {
return lps(s, 0, s.length() - 1);
}
public static void main(String[] args) {
String s = "bbabcbcab";
System.out.println(longestPalinSubseq(s));
}
}
# Python program to find the lps
# Returns the length of the longest
# palindromic subsequence in seq
def lps(s, low, high):
# Base case
if low > high:
return 0
# If there is only 1 character
if low == high:
return 1
# If the first and last characters match
if s[low] == s[high]:
return lps(s, low + 1, high - 1) + 2
# If the first and last characters do not match
return max(lps(s, low, high - 1), lps(s, low + 1, high))
def longestPalinSubseq(s):
return lps(s, 0, len(s) - 1)
if __name__ == "__main__":
s = "bbabcbcab"
print(longestPalinSubseq(s))
// C# program to find the lps
using System;
class GfG {
// Returns the length of the longest
// palindromic subsequence in seq
static int lps(string s, int low, int high) {
// Base case
if (low > high) return 0;
// If there is only 1 character
if (low == high)
return 1;
// If the first and last characters match
if (s[low] == s[high])
return lps(s, low + 1, high - 1) + 2;
// If the first and last characters do not match
return Math.Max(lps(s, low, high - 1),
lps(s, low + 1, high));
}
static int longestPalinSubseq(string s) {
return lps(s, 0, s.Length - 1);
}
static void Main(string[] args) {
string s = "bbabcbcab";
Console.WriteLine(longestPalinSubseq(s));
}
}
// JavaScript program to find the lps
// Returns the length of the longest
// palindromic subsequence in seq
function lps(s, low, high) {
// Base case
if (low > high) return 0;
// If there is only 1 character
if (low === high)
return 1;
// If the first and last characters match
if (s[low] === s[high])
return lps(s, low + 1, high - 1) + 2;
// If the first and last characters do not match
return Math.max(lps(s, low, high - 1),
lps(s, low + 1, high));
}
function longestPalinSubseq(s) {
return lps(s, 0, s.length - 1);
}
const s = "bbabcbcab";
console.log(longestPalinSubseq(s));
[Better Approach 1] Using Memoization - O(n^2) Time and O(n^2) Space
In the above approach, lps() function is calculating the same substring multiple times. The idea is to use memoization to store the result of subproblems thus avoiding repetition. To do so, create a 2d array memo[][] of order n*n, where memo[i][j] stores the length of LPS of substring s[i] to s[j]. At each step, check if the substring is already calculated, if so return the stored value else operate as in above approach.
C++
// C++ program to find the lps
#include <bits/stdc++.h>
using namespace std;
// Returns the length of the longest
// palindromic subsequence in seq
int lps(const string& s, int low, int high,
vector<vector<int>> &memo) {
// Base case
if(low > high) return 0;
// If there is only 1 character
if (low == high)
return 1;
// If the value is already calculated
if(memo[low][high] != -1)
return memo[low][high];
// If the first and last characters match
if (s[low] == s[high])
return memo[low][high] =
lps(s, low + 1, high - 1, memo) + 2;
// If the first and last characters do not match
return memo[low][high] =
max(lps(s, low, high - 1, memo),
lps(s, low + 1, high, memo));
}
int longestPalinSubseq(string &s) {
// create memoization table
vector<vector<int>> memo(s.size(),
vector<int>(s.size(), -1));
return lps(s, 0, s.size() - 1, memo);
}
int main() {
string s = "bbabcbcab";
cout << longestPalinSubseq(s);
return 0;
}
// Java program to find the lps
class GfG {
// Returns the length of the longest
// palindromic subsequence in seq
static int lps(String s, int low, int high, int[][] memo) {
// Base case
if (low > high) return 0;
// If there is only 1 character
if (low == high)
return 1;
// If the value is already calculated
if (memo[low][high] != -1)
return memo[low][high];
// If the first and last characters match
if (s.charAt(low) == s.charAt(high))
return memo[low][high] =
lps(s, low + 1, high - 1, memo) + 2;
// If the first and last characters do not match
return memo[low][high] =
Math.max(lps(s, low, high - 1, memo),
lps(s, low + 1, high, memo));
}
static int longestPalinSubseq(String s) {
// create memoization table
int n = s.length();
int[][] memo = new int[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
memo[i][j] = -1;
}
}
return lps(s, 0, n - 1, memo);
}
public static void main(String[] args) {
String s = "bbabcbcab";
System.out.println(longestPalinSubseq(s));
}
}
# Python program to find the lps
# Returns the length of the longest
# palindromic subsequence in seq
def lps(s, low, high, memo):
# Base case
if low > high:
return 0
# If there is only 1 character
if low == high:
return 1
# If the value is already calculated
if memo[low][high] != -1:
return memo[low][high]
# If the first and last characters match
if s[low] == s[high]:
memo[low][high] = lps(s, low + 1, high - 1, memo) + 2
else:
# If the first and last characters do not match
memo[low][high] = max(lps(s, low, high - 1, memo),
lps(s, low + 1, high, memo))
return memo[low][high]
def longestPalinSubseq(s):
n = len(s)
memo = [[-1 for _ in range(n)] for _ in range(n)]
return lps(s, 0, n - 1, memo)
if __name__ == "__main__":
s = "bbabcbcab"
print(longestPalinSubseq(s))
// C# program to find the lps
using System;
class GfG {
// Returns the length of the longest
// palindromic subsequence in seq
static int lps(string s, int low,
int high, int[,] memo) {
// Base case
if (low > high) return 0;
// If there is only 1 character
if (low == high)
return 1;
// If the value is already calculated
if (memo[low, high] != -1)
return memo[low, high];
// If the first and last characters match
if (s[low] == s[high])
return memo[low, high] =
lps(s, low + 1, high - 1, memo) + 2;
// If the first and last characters do not match
return memo[low, high] =
Math.Max(lps(s, low, high - 1, memo),
lps(s, low + 1, high, memo));
}
static int longestPalinSubseq(string s) {
// create memoization table
int n = s.Length;
int[,] memo = new int[n, n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
memo[i, j] = -1;
}
}
return lps(s, 0, n - 1, memo);
}
static void Main(string[] args) {
string s = "bbabcbcab";
Console.WriteLine(longestPalinSubseq(s));
}
}
// JavaScript program to find the lps
// Returns the length of the longest
// palindromic subsequence in seq
function lps(s, low, high, memo) {
// Base case
if (low > high) return 0;
// If there is only 1 character
if (low === high)
return 1;
// If the value is already calculated
if (memo[low][high] !== -1)
return memo[low][high];
// If the first and last characters match
if (s[low] === s[high]) {
memo[low][high] = lps(s, low + 1, high - 1, memo) + 2;
} else {
// If the first and last characters do not match
memo[low][high] = Math.max(
lps(s, low, high - 1, memo),
lps(s, low + 1, high, memo)
);
}
return memo[low][high];
}
function longestPalinSubseq(s) {
const n = s.length;
const memo = Array.from({ length: n },
() => Array(n).fill(-1));
return lps(s, 0, n - 1, memo);
}
const s = "bbabcbcab";
console.log(longestPalinSubseq(s));
[Better Approach 2] Using Tabulation - O(n^2) Time and O(n^2) Space
The above approach can be implemented using tabulation to minimize the auxiliary space required for recursive stack. The idea is create a 2d array dp[][] of order n*n, where element dp[i][j] stores the length of LPS of substring s[i] to s[j]. Start from the smaller substring and try to build answers for longer ones. At each step there are two possibilities:
- if s[i] == s[j], then dp[i][j] = dp[i+1][j-1] + 2
- else, dp[i][j] = max(dp[i+1][j], dp[i][j-1])
dp[0][n-1] stores the length of the longest palindromic subsequence of string s.
C++
// C++ program to find the lps
#include <bits/stdc++.h>
using namespace std;
// Function to find the LPS
int longestPalinSubseq(string& s) {
int n = s.length();
// Create a DP table
vector<vector<int>> dp(n, vector<int>(n));
// Build the DP table for all the substrings
for (int i = n - 1; i >= 0; i--) {
for (int j = i; j < n; j++) {
// If there is only one character
if(i == j){
dp[i][j] = 1;
continue;
}
// If characters at position i and j are the same
if (s[i] == s[j]) {
if(i + 1 == j) dp[i][j] = 2;
else dp[i][j] = dp[i + 1][j - 1] + 2;
}
else {
// Otherwise, take the maximum length
// from either excluding i or j
dp[i][j] = max(dp[i + 1][j], dp[i][j - 1]);
}
}
}
// The final answer is stored in dp[0][n-1]
return dp[0][n - 1];
}
int main() {
string s = "bbabcbcab";
cout << longestPalinSubseq(s);
return 0;
}
// Java program to find the lps
class GfG {
// Function to find the LPS
static int longestPalinSubseq(String s) {
int n = s.length();
// Create a DP table
int[][] dp = new int[n][n];
// Build the DP table for all the substrings
for (int i = n - 1; i >= 0; i--) {
for (int j = i; j < n; j++) {
// If there is only one character
if (i == j) {
dp[i][j] = 1;
continue;
}
// If characters at position i and j are the same
if (s.charAt(i) == s.charAt(j)) {
if (i + 1 == j) dp[i][j] = 2;
else dp[i][j] = dp[i + 1][j - 1] + 2;
}
else {
// Otherwise, take the maximum length
// from either excluding i or j
dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);
}
}
}
// The final answer is stored in dp[0][n-1]
return dp[0][n - 1];
}
public static void main(String[] args) {
String s = "bbabcbcab";
System.out.println(longestPalinSubseq(s));
}
}
# Python program to find the lps
# Function to find the LPS
def longestPalinSubseq(s):
n = len(s)
# Create a DP table
dp = [[0] * n for _ in range(n)]
# Build the DP table for all the substrings
for i in range(n - 1, -1, -1):
for j in range(i, n):
# If there is only one character
if i == j:
dp[i][j] = 1
continue
# If characters at position i and j are the same
if s[i] == s[j]:
if i + 1 == j:
dp[i][j] = 2
else:
dp[i][j] = dp[i + 1][j - 1] + 2
else:
# Otherwise, take the maximum length
# from either excluding i or j
dp[i][j] = max(dp[i + 1][j], dp[i][j - 1])
# The final answer is stored in dp[0][n-1]
return dp[0][n - 1]
if __name__ == "__main__":
s = "bbabcbcab"
print(longestPalinSubseq(s))
// C# program to find the lps
using System;
class GfG {
// Function to find the LPS
static int longestPalinSubseq(string s) {
int n = s.Length;
// Create a DP table
int[,] dp = new int[n, n];
// Build the DP table for all the substrings
for (int i = n - 1; i >= 0; i--) {
for (int j = i; j < n; j++) {
// If there is only one character
if (i == j) {
dp[i, j] = 1;
continue;
}
// If characters at position i and j are the same
if (s[i] == s[j]) {
if (i + 1 == j) dp[i, j] = 2;
else dp[i, j] = dp[i + 1, j - 1] + 2;
}
else {
// Otherwise, take the maximum length
// from either excluding i or j
dp[i, j] = Math.Max(dp[i + 1, j], dp[i, j - 1]);
}
}
}
// The final answer is stored in dp[0][n-1]
return dp[0, n - 1];
}
static void Main(string[] args) {
string s = "bbabcbcab";
Console.WriteLine(longestPalinSubseq(s));
}
}
// JavaScript program to find the lps
// Function to find the LPS
function longestPalinSubseq(s) {
const n = s.length;
// Create a DP table
const dp = Array.from({ length: n },
() => Array(n).fill(0));
// Build the DP table for all the substrings
for (let i = n - 1; i >= 0; i--) {
for (let j = i; j < n; j++) {
// If there is only one character
if (i === j) {
dp[i][j] = 1;
continue;
}
// If characters at position i and j are the same
if (s[i] === s[j]) {
if (i + 1 === j) dp[i][j] = 2;
else dp[i][j] = dp[i + 1][j - 1] + 2;
}
else {
// Otherwise, take the maximum length
// from either excluding i or j
dp[i][j] = Math.max(dp[i + 1][j], dp[i][j - 1]);
}
}
}
// The final answer is stored in dp[0][n-1]
return dp[0][n - 1];
}
const s = "bbabcbcab";
console.log(longestPalinSubseq(s));
[Expected Approach] Using Tabulation - O(n^2) Time and O(n) Space
In the above approach, for calculating the LPS of substrings starting from index i, only the LPS of substrings starting from index i+1 are required. Thus instead of creating 2d array, idea is to create two arrays of size, curr[] and prev[], where curr[j] stores the lps of substring from s[i] to s[j], while prev[j] stores the lps of substring from s[i+1] to s[j]. Else everything will be similar to above approach.
C++
// C++ program to find longest
// palindromic subsequence
#include <bits/stdc++.h>
using namespace std;
// Function to find the length of the lps
int longestPalinSubseq(const string &s) {
int n = s.size();
// Create two vectors: one for the current state (dp)
// and one for the previous state (dpPrev)
vector<int> curr(n), prev(n);
// Loop through the string in reverse (starting from the end)
for (int i = n - 1; i >= 0; --i){
// Initialize the current state of dp
curr[i] = 1;
// Loop through the characters ahead of i
for (int j = i + 1; j < n; ++j){
// If the characters at i and j are the same
if (s[i] == s[j]){
// Add 2 to the length of the palindrome between them
curr[j] = prev[j - 1] + 2;
}
else{
// Take the maximum between excluding either i or j
curr[j] = max(prev[j], curr[j - 1]);
}
}
// Update previous to the current state of dp
prev = curr;
}
return curr[n-1];
}
int main() {
string s = "bbabcbcab";
cout << longestPalinSubseq(s);
return 0;
}
// Java program to find longest
// palindromic subsequence
import java.util.*;
// Java program to find the length of the lps
class GfG {
// Function to find the length of the lps
static int longestPalinSubseq(String s) {
int n = s.length();
// Create two arrays: one for the current state (dp)
// and one for the previous state (dpPrev)
int[] curr = new int[n];
int[] prev = new int[n];
// Loop through the string in reverse (starting from the end)
for (int i = n - 1; i >= 0; --i) {
// Initialize the current state of dp
curr[i] = 1;
// Loop through the characters ahead of i
for (int j = i + 1; j < n; ++j) {
// If the characters at i and j are the same
if (s.charAt(i) == s.charAt(j)) {
// Add 2 to the length of the palindrome between them
curr[j] = prev[j - 1] + 2;
} else {
// Take the maximum between excluding either i or j
curr[j] = Math.max(prev[j], curr[j - 1]);
}
}
// Update previous to the current state of dp
prev = curr.clone();
}
return curr[n - 1];
}
public static void main(String[] args) {
String s = "bbabcbcab";
System.out.println(longestPalinSubseq(s));
}
}
# Python program to find the length of the lps
# Function to find the length of the lps
def longestPalinSubseq(s):
n = len(s)
# Create two arrays: one for the current state (dp)
# and one for the previous state (dpPrev)
curr = [0] * n
prev = [0] * n
# Loop through the string in reverse (starting from the end)
for i in range(n - 1, -1, -1):
# Initialize the current state of dp
curr[i] = 1
# Loop through the characters ahead of i
for j in range(i + 1, n):
# If the characters at i and j are the same
if s[i] == s[j]:
# Add 2 to the length of the palindrome between them
curr[j] = prev[j - 1] + 2
else:
# Take the maximum between excluding either i or j
curr[j] = max(prev[j], curr[j - 1])
# Update previous to the current state of dp
prev = curr[:]
return curr[n - 1]
if __name__ == "__main__":
s = "bbabcbcab"
print(longestPalinSubseq(s))
// C# program to find longest
// palindromic subsequence
using System;
// C# program to find the length of the lps
class GfG {
// Function to find the length of the lps
static int longestPalinSubseq(string s) {
int n = s.Length;
// Create two arrays: one for the current state (dp)
// and one for the previous state (dpPrev)
int[] curr = new int[n];
int[] prev = new int[n];
// Loop through the string in reverse (starting from the end)
for (int i = n - 1; i >= 0; --i) {
// Initialize the current state of dp
curr[i] = 1;
// Loop through the characters ahead of i
for (int j = i + 1; j < n; ++j) {
// If the characters at i and j are the same
if (s[i] == s[j]) {
// Add 2 to the length of the palindrome between them
curr[j] = prev[j - 1] + 2;
} else {
// Take the maximum between excluding either i or j
curr[j] = Math.Max(prev[j], curr[j - 1]);
}
}
// Update previous to the current state of dp
Array.Copy(curr, prev, n);
}
return curr[n - 1];
}
static void Main(string[] args) {
string s = "bbabcbcab";
Console.WriteLine(longestPalinSubseq(s));
}
}
// JavaScript program to find the length of the lps
// Function to find the length of the lps
function longestPalinSubseq(s) {
const n = s.length;
// Create two arrays: one for the current state (dp)
// and one for the previous state (dpPrev)
let curr = new Array(n).fill(0);
let prev = new Array(n).fill(0);
// Loop through the string in reverse (starting from the end)
for (let i = n - 1; i >= 0; --i) {
// Initialize the current state of dp
curr[i] = 1;
// Loop through the characters ahead of i
for (let j = i + 1; j < n; ++j) {
// If the characters at i and j are the same
if (s[i] === s[j]) {
// Add 2 to the length of the palindrome between them
curr[j] = prev[j - 1] + 2;
} else {
// Take the maximum between excluding either i or j
curr[j] = Math.max(prev[j], curr[j - 1]);
}
}
// Update previous to the current state of dp
prev = [...curr];
}
return curr[n - 1];
}
const s = "bbabcbcab";
console.log(longestPalinSubseq(s));
[Alternate Approach] Using Longest Common Subsequence - O(n^2) Time and O(n) Space
The idea is to reverse the given string s and find the length of the longest common subsequence of original and reversed string.
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Tabulation vs Memoization Tabulation and memoization are two techniques used to implement dynamic programming. Both techniques are used when there are overlapping subproblems (the same subproblem is executed multiple times). Below is an overview of two approaches.Memoization:Top-down approachStores the results of function ca
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Optimal Substructure Property in Dynamic Programming | DP-2 The following are the two main properties of a problem that suggest that the given problem can be solved using Dynamic programming: 1) Overlapping Subproblems 2) Optimal Substructure We have already discussed the Overlapping Subproblem property. Let us discuss the Optimal Substructure property here.
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Overlapping Subproblems Property in Dynamic Programming | DP-1 Dynamic Programming is an algorithmic paradigm that solves a given complex problem by breaking it into subproblems using recursion and storing the results of subproblems to avoid computing the same results again. Following are the two main properties of a problem that suggests that the given problem
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Steps to solve a Dynamic Programming Problem Steps to solve a Dynamic programming problem:Identify if it is a Dynamic programming problem.Decide a state expression with the Least parameters.Formulate state and transition relationship.Apply tabulation or memorization.Step 1: How to classify a problem as a Dynamic Programming Problem? Typically,
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Advanced Topics
Count Ways To Assign Unique Cap To Every PersonGiven n people and 100 types of caps labelled from 1 to 100, along with a 2D integer array caps where caps[i] represents the list of caps preferred by the i-th person, the task is to determine the number of ways the n people can wear different caps.Example:Input: caps = [[3, 4], [4, 5], [5]] Output:
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Digit DP | IntroductionPrerequisite : How to solve a Dynamic Programming Problem ?There are many types of problems that ask to count the number of integers 'x' between two integers say 'a' and 'b' such that x satisfies a specific property that can be related to its digits.So, if we say G(x) tells the number of such intege
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Sum over Subsets | Dynamic ProgrammingPrerequisite: Basic Dynamic Programming, Bitmasks Consider the following problem where we will use Sum over subset Dynamic Programming to solve it. Given an array of 2n integers, we need to calculate function F(x) = ?Ai such that x&i==i for all x. i.e, i is a bitwise subset of x. i will be a bit
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Easy problems in Dynamic programming
Coin Change - Count Ways to Make SumGiven an integer array of coins[] of size n representing different types of denominations and an integer sum, the task is to count all combinations of coins to make a given value sum. Note: Assume that you have an infinite supply of each type of coin. Examples: Input: sum = 4, coins[] = [1, 2, 3]Out
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Subset Sum ProblemGiven an array arr[] of non-negative integers and a value sum, the task is to check if there is a subset of the given array whose sum is equal to the given sum. Examples: Input: arr[] = [3, 34, 4, 12, 5, 2], sum = 9Output: TrueExplanation: There is a subset (4, 5) with sum 9.Input: arr[] = [3, 34, 4
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Introduction and Dynamic Programming solution to compute nCr%pGiven three numbers n, r and p, compute value of nCr mod p. Example: Input: n = 10, r = 2, p = 13 Output: 6 Explanation: 10C2 is 45 and 45 % 13 is 6.We strongly recommend that you click here and practice it, before moving on to the solution.METHOD 1: (Using Dynamic Programming) A Simple Solution is
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Rod CuttingGiven a rod of length n inches and an array price[]. price[i] denotes the value of a piece of length i. The task is to determine the maximum value obtainable by cutting up the rod and selling the pieces.Note: price[] is 1-indexed array.Input: price[] = [1, 5, 8, 9, 10, 17, 17, 20]Output: 22Explanati
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Painting Fence AlgorithmGiven a fence with n posts and k colors, the task is to find out the number of ways of painting the fence so that not more than two consecutive posts have the same color.Examples:Input: n = 2, k = 4Output: 16Explanation: We have 4 colors and 2 posts.Ways when both posts have same color: 4 Ways when
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Longest Common Subsequence (LCS)Given 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. A subsequence is a string generated from the original string by deleting 0 or more characters, without changing the relative order of the remaining characters.
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Longest Increasing Subsequence (LIS)Given an array arr[] of size n, the task is to find the length of the Longest Increasing Subsequence (LIS) i.e., the longest possible subsequence in which the elements of the subsequence are sorted in increasing order.Examples: Input: arr[] = [3, 10, 2, 1, 20]Output: 3Explanation: The longest increa
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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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Maximum size square sub-matrix with all 1sGiven a binary matrix mat of size n * m, the task is to find out the maximum length of a side of a square sub-matrix with all 1s.Example:Input: mat = [ [0, 1, 1, 0, 1], [1, 1, 0, 1, 0], [0, 1, 1, 1, 0], [1, 1, 1, 1, 0], [1, 1, 1, 1, 1], [0, 0, 0, 0, 0] ]Output: 3Explanation: The maximum length of a
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Min Cost PathYou are given a 2D matrix cost[][] of dimensions m × n, where each cell represents the cost of traversing through that position. Your goal is to determine the minimum cost required to reach the bottom-right cell (m-1, n-1) starting from the top-left cell (0,0).The total cost of a path is the sum of
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Longest Common Substring (Space optimized DP solution)Given two strings ‘s1‘ and ‘s2‘, find the length of the longest common substring. Example: Input: s1 = “GeeksforGeeksâ€, s2 = “GeeksQuiz†Output : 5 Explanation:The longest common substring is “Geeks†and is of length 5.Input: s1 = “abcdxyzâ€, s2 = “xyzabcd†Output : 4Explanation:The longest common su
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Count ways to reach the nth stair using step 1, 2 or 3A child is running up a staircase with n steps and can hop either 1 step, 2 steps, or 3 steps at a time. The task is to implement a method to count how many possible ways the child can run up the stairs.Examples: Input: 4Output: 7Explanation: There are seven ways: {1, 1, 1, 1}, {1, 2, 1}, {2, 1, 1},
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Grid Unique Paths - Count Paths in matrixGiven an matrix of size m x n, the task is to find the count of all unique possible paths from top left to the bottom right with the constraints that from each cell we can either move only to the right or down.Examples: Input: m = 2, n = 2Output: 2Explanation: There are two paths(0, 0) -> (0, 1)
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Unique paths in a Grid with ObstaclesGiven a matrix mat[][] of size n * m, where mat[i][j] = 1 indicates an obstacle and mat[i][j] = 0 indicates an empty space. The task is to find the number of unique paths to reach (n-1, m-1) starting from (0, 0). You are allowed to move in the right or downward direction. Note: In the grid, cells ma
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Medium problems on Dynamic programming
0/1 Knapsack ProblemGiven n items where each item has some weight and profit associated with it and also given a bag with capacity W, [i.e., the bag can hold at most W weight in it]. The task is to put the items into the bag such that the sum of profits associated with them is the maximum possible. Note: The constraint
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Printing Items in 0/1 KnapsackGiven weights and values of n items, put these items in a knapsack of capacity W to get the maximum total value in the knapsack. In other words, given two integer arrays, val[0..n-1] and wt[0..n-1] represent values and weights associated with n items respectively. Also given an integer W which repre
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Unbounded Knapsack (Repetition of items allowed)Given a knapsack weight, say capacity and a set of n items with certain value vali and weight wti, The task is to fill the knapsack in such a way that we can get the maximum profit. This is different from the classical Knapsack problem, here we are allowed to use an unlimited number of instances of
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Egg Dropping Puzzle | DP-11You are given n identical eggs and you have access to a k-floored building from 1 to k.There exists a floor f where 0 <= f <= k such that any egg dropped from a floor higher than f will break, and any egg dropped from or below floor f will not break. There are a few rules given below:An egg th
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Word BreakGiven a string s and y a dictionary of n words dictionary, check if s can be segmented into a sequence of valid words from the dictionary, separated by spaces.Examples:Input: s = "ilike", dictionary[] = ["i", "like", "gfg"]Output: trueExplanation: The string can be segmented as "i like".Input: s = "
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Vertex Cover Problem (Dynamic Programming Solution for Tree)A vertex cover of an undirected graph is a subset of its vertices such that for every edge (u, v) of the graph, either ‘u’ or ‘v’ is in vertex cover. Although the name is Vertex Cover, the set covers all edges of the given graph. The problem to find minimum size vertex cover of a graph is NP complet
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Tile Stacking ProblemGiven integers n (the height of the tower), m (the maximum size of tiles available), and k (the maximum number of times each tile size can be used), the task is to calculate the number of distinct stable towers of height n that can be built. Note:A stable tower consists of exactly n tiles, each stac
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Box Stacking ProblemGiven three arrays height[], width[], and length[] of size n, where height[i], width[i], and length[i] represent the dimensions of a box. The task is to create a stack of boxes that is as tall as possible, but we can only stack a box on top of another box if the dimensions of the 2-D base of the low
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Partition a Set into Two Subsets of Equal SumGiven an array arr[], the task is to check if it can be partitioned into two parts such that the sum of elements in both parts is the same.Note: Each element is present in either the first subset or the second subset, but not in both.Examples: Input: arr[] = [1, 5, 11, 5]Output: true Explanation: Th
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Travelling Salesman Problem using Dynamic ProgrammingGiven a 2d matrix cost[][] of size n where cost[i][j] denotes the cost of moving from city i to city j. The task is to complete a tour from city 0 (0-based index) to all other cities such that we visit each city exactly once and then at the end come back to city 0 at minimum cost.Note the difference
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Longest Palindromic Subsequence (LPS)Given a string s, find the length of the Longest Palindromic Subsequence in it. Note: The Longest Palindromic Subsequence (LPS) is the maximum-length subsequence of a given string that is also a Palindrome. Longest Palindromic SubsequenceExamples:Input: s = "bbabcbcab"Output: 7Explanation: Subsequen
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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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Find all distinct subset (or subsequence) sums of an arrayGiven an array arr[] of size n, the task is to find a distinct sum that can be generated from the subsets of the given sets and return them in increasing order. It is given that the sum of array elements is small.Examples: Input: arr[] = [1, 2]Output: [0, 1, 2, 3]Explanation: Four distinct sums can
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Weighted Job SchedulingGiven a 2D array jobs[][] of order n*3, where each element jobs[i] defines start time, end time, and the profit associated with the job. The task is to find the maximum profit you can take such that there are no two jobs with overlapping time ranges.Note: If the job ends at time X, it is allowed to
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Count Derangements (Permutation such that no element appears in its original position)A Derangement is a permutation of n elements, such that no element appears in its original position. For example, a derangement of [0, 1, 2, 3] is [2, 3, 1, 0].Given a number n, find the total number of Derangements of a set of n elements.Examples : Input: n = 2Output: 1Explanation: For two balls [1
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Minimum insertions to form a palindromeGiven a string s, the task is to find the minimum number of characters to be inserted to convert it to a palindrome.Examples:Input: s = "geeks"Output: 3Explanation: "skgeegks" is a palindromic string, which requires 3 insertions.Input: s= "abcd"Output: 3Explanation: "abcdcba" is a palindromic string
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Ways to arrange Balls such that adjacent balls are of different typesThere are 'p' balls of type P, 'q' balls of type Q and 'r' balls of type R. Using the balls we want to create a straight line such that no two balls of the same type are adjacent.Examples : Input: p = 1, q = 1, r = 0Output: 2Explanation: There are only two arrangements PQ and QPInput: p = 1, q = 1,
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