Printing Longest Common Subsequence
Last Updated :
23 Jul, 2025
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. The function discussed there was mainly to find the length of LCS. To find length of LCS, a 2D table L[][] was constructed. In this post, the function to construct and print LCS is discussed.
Following is detailed algorithm to print the LCS. It uses the same 2D table L[][].
- Construct L[m+1][n+1] using the steps discussed in previous post.
- The value L[m][n] contains length of LCS. Create a character array lcs[] of length equal to the length of lcs plus 1 (one extra to store \0).
- Traverse the 2D array starting from L[m][n]. Do following for every cell L[i][j]
- If characters (in X and Y) corresponding to L[i][j] are same (Or X[i-1] == Y[j-1]), then include this character as part of LCS.
- Else compare values of L[i-1][j] and L[i][j-1] and go in direction of greater value.
The following table (taken from Wiki) shows steps (highlighted) followed by the above algorithm.
Following is the implementation of the above approach.
C++14
/* Dynamic Programming implementation of LCS problem */
#include <cstdlib>
#include <cstring>
#include <iostream>
using namespace std;
/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
void lcs(char* X, char* Y, int m, int n)
{
int L[m + 1][n + 1];
/* Following steps build L[m+1][n+1] in bottom up
fashion. Note that L[i][j] contains length of LCS of
X[0..i-1] and Y[0..j-1] */
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
if (i == 0 || j == 0)
L[i][j] = 0;
else if (X[i - 1] == Y[j - 1])
L[i][j] = L[i - 1][j - 1] + 1;
else
L[i][j] = max(L[i - 1][j], L[i][j - 1]);
}
}
// Following code is used to print LCS
int index = L[m][n];
// Create a character array to store the lcs string
char lcs[index + 1];
lcs[index] = '\0'; // Set the terminating character
// Start from the right-most-bottom-most corner and
// one by one store characters in lcs[]
int i = m, j = n;
while (i > 0 && j > 0) {
// If current character in X[] and Y are same, then
// current character is part of LCS
if (X[i - 1] == Y[j - 1]) {
lcs[index - 1]
= X[i - 1]; // Put current character in result
i--;
j--;
index--; // reduce values of i, j and index
}
// If not same, then find the larger of two and
// go in the direction of larger value
else if (L[i - 1][j] > L[i][j - 1])
i--;
else
j--;
}
// Print the lcs
cout << "LCS of " << X << " and " << Y << " is " << lcs;
}
/* Driver program to test above function */
int main()
{
char X[] = "AGGTAB";
char Y[] = "GXTXAYB";
int m = strlen(X);
int n = strlen(Y);
lcs(X, Y, m, n);
return 0;
}
Java
// Dynamic Programming implementation of LCS problem in Java
import java.io.*;
class LongestCommonSubsequence {
// Returns length of LCS for X[0..m-1], Y[0..n-1]
static void lcs(String X, String Y, int m, int n)
{
int[][] L = new int[m + 1][n + 1];
// Following steps build L[m+1][n+1] in bottom up
// fashion. Note that L[i][j] contains length of LCS
// of X[0..i-1] and Y[0..j-1]
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
if (i == 0 || j == 0)
L[i][j] = 0;
else if (X.charAt(i - 1) == Y.charAt(j - 1))
L[i][j] = L[i - 1][j - 1] + 1;
else
L[i][j] = Math.max(L[i - 1][j],
L[i][j - 1]);
}
}
// Following code is used to print LCS
int index = L[m][n];
int temp = index;
// Create a character array to store the lcs string
char[] lcs = new char[index + 1];
lcs[index]
= '\u0000'; // Set the terminating character
// Start from the right-most-bottom-most corner and
// one by one store characters in lcs[]
int i = m;
int j = n;
while (i > 0 && j > 0) {
// If current character in X[] and Y are same,
// then current character is part of LCS
if (X.charAt(i - 1) == Y.charAt(j - 1)) {
// Put current character in result
lcs[index - 1] = X.charAt(i - 1);
// reduce values of i, j and index
i--;
j--;
index--;
}
// If not same, then find the larger of two and
// go in the direction of larger value
else if (L[i - 1][j] > L[i][j - 1])
i--;
else
j--;
}
// Print the lcs
System.out.print("LCS of " + X + " and " + Y
+ " is ");
for (int k = 0; k <= temp; k++)
System.out.print(lcs[k]);
}
// driver program
public static void main(String[] args)
{
String X = "AGGTAB";
String Y = "GXTXAYB";
int m = X.length();
int n = Y.length();
lcs(X, Y, m, n);
}
}
// Contributed by Pramod Kumar
Python3
# Dynamic programming implementation of LCS problem
# Returns length of LCS for X[0..m-1], Y[0..n-1]
def lcs(X, Y, m, n):
L = [[0 for i in range(n+1)] for j in range(m+1)]
# Following steps build L[m+1][n+1] in bottom up fashion. Note
# that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1]
for i in range(m+1):
for j in range(n+1):
if i == 0 or j == 0:
L[i][j] = 0
elif X[i-1] == Y[j-1]:
L[i][j] = L[i-1][j-1] + 1
else:
L[i][j] = max(L[i-1][j], L[i][j-1])
# Create a string variable to store the lcs string
lcs = ""
# 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 X[i-1] == Y[j-1]:
lcs += X[i-1]
i -= 1
j -= 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
# We traversed the table in reverse order
# LCS is the reverse of what we got
lcs = lcs[::-1]
print("LCS of " + X + " and " + Y + " is " + lcs)
# Driver program
X = "AGGTAB"
Y = "GXTXAYB"
m = len(X)
n = len(Y)
lcs(X, Y, m, n)
# This code is contributed by AMAN ASATI
C#
// Dynamic Programming implementation
// of LCS problem in C#
using System;
class GFG {
// Returns length of LCS for X[0..m-1], Y[0..n-1]
static void lcs(String X, String Y, int m, int n)
{
int[, ] L = new int[m + 1, n + 1];
// Following steps build L[m+1][n+1] in
// bottom up fashion. Note that L[i][j]
// contains length of LCS of X[0..i-1]
// and Y[0..j-1]
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
if (i == 0 || j == 0)
L[i, j] = 0;
else if (X[i - 1] == Y[j - 1])
L[i, j] = L[i - 1, j - 1] + 1;
else
L[i, j] = Math.Max(L[i - 1, j],
L[i, j - 1]);
}
}
// Following code is used to print LCS
int index = L[m, n];
int temp = index;
// Create a character array
// to store the lcs string
char[] lcs = new char[index + 1];
// Set the terminating character
lcs[index] = '\0';
// Start from the right-most-bottom-most corner
// and one by one store characters in lcs[]
int k = m, l = n;
while (k > 0 && l > 0) {
// If current character in X[] and Y
// are same, then current character
// is part of LCS
if (X[k - 1] == Y[l - 1]) {
// Put current character in result
lcs[index - 1] = X[k - 1];
// reduce values of i, j and index
k--;
l--;
index--;
}
// If not same, then find the larger of two and
// go in the direction of larger value
else if (L[k - 1, l] > L[k, l - 1])
k--;
else
l--;
}
// Print the lcs
Console.Write("LCS of " + X + " and " + Y + " is ");
for (int q = 0; q <= temp; q++)
Console.Write(lcs[q]);
}
// Driver program
public static void Main()
{
String X = "AGGTAB";
String Y = "GXTXAYB";
int m = X.Length;
int n = Y.Length;
lcs(X, Y, m, n);
}
}
// This code is contributed by Sam007
JavaScript
<script>
function ReverseString(str) {
return str.split('').reverse().join('')
}
function max(a, b)
{
if (a > b)
return a;
else
return b;
}
function printLCS(str1, str2) {
var len1 = str1.length;
var len2 = str2.length;
var lcs = new Array(len1 + 1);
for (var i = 0; i <= len1; i++) {
lcs[i] = new Array(len2 + 1)
}
for (var i = 0; i <= len1; i++) {
for (var j = 0; j <= len2; j++) {
if (i == 0 || j == 0) {
lcs[i][j] = 0;
}
else {
if (str1[i - 1] == str2[j - 1]) {
lcs[i][j] = 1 + lcs[i - 1][j - 1];
}
else {
lcs[i][j] = max(lcs[i][j - 1], lcs[i - 1][j]);
}
}
}}
var n = lcs[len1][len2];
document.write("Length of common subsequence is: " +
n + "<br>" + "The subsequence is : ");
var str="";
var i = len1;
var j = len2;
while(i>0&&j>0)
{
if(str1[i - 1] == str2[j - 1])
{
str += str1[i - 1];
i--;
j--;
}
else{
if(lcs[i][j-1]>lcs[i-1][j])
{
j--;
}
else
{
i--;
}
}
}
return ReverseString(str);
}
var str1 = "AGGTAB";
var str2 = "GXTXAYB";
document.write(printLCS(str1, str2));
// This code is contributed by akshitsaxenaa09
</script>
PHP
<?php
// Dynamic Programming implementation of LCS problem
// Returns length of LCS for X[0..m-1], Y[0..n-1]
function lcs( $X, $Y, $m, $n )
{
$L = array_fill(0, $m + 1,
array_fill(0, $n + 1, NULL));
/* Following steps build L[m+1][n+1] in bottom
up fashion. Note that L[i][j] contains length
of LCS of X[0..i-1] and Y[0..j-1] */
for ($i = 0; $i <= $m; $i++)
{
for ($j = 0; $j <= $n; $j++)
{
if ($i == 0 || $j == 0)
$L[$i][$j] = 0;
else if ($X[$i - 1] == $Y[$j - 1])
$L[$i][$j] = $L[$i - 1][$j - 1] + 1;
else
$L[$i][$j] = max($L[$i - 1][$j],
$L[$i][$j - 1]);
}
}
// Following code is used to print LCS
$index = $L[$m][$n];
$temp = $index;
// Create a character array to store the lcs string
$lcs = array_fill(0, $index + 1, NULL);
$lcs[$index] = ''; // Set the terminating character
// Start from the right-most-bottom-most corner
// and one by one store characters in lcs[]
$i = $m;
$j = $n;
while ($i > 0 && $j > 0)
{
// If current character in X[] and Y are same,
// then current character is part of LCS
if ($X[$i - 1] == $Y[$j - 1])
{
// Put current character in result
$lcs[$index - 1] = $X[$i - 1];
$i--;
$j--;
$index--; // reduce values of i, j and index
}
// If not same, then find the larger of two
// and go in the direction of larger value
else if ($L[$i - 1][$j] > $L[$i][$j - 1])
$i--;
else
$j--;
}
// Print the lcs
echo "LCS of " . $X . " and " . $Y . " is ";
for($k = 0; $k < $temp; $k++)
echo $lcs[$k];
}
// Driver Code
$X = "AGGTAB";
$Y = "GXTXAYB";
$m = strlen($X);
$n = strlen($Y);
lcs($X, $Y, $m, $n);
// This code is contributed by ita_c
?>
OutputLCS of AGGTAB and GXTXAYB is GTAB
Time Complexity: O(m*n)
Auxiliary Space: O(m*n)
Top-down approach for printing Longest Common Subsequence:
Follow the steps below for the implementation:
- Check if one of the two strings is of size zero, then we return an empty string because the LCS, in this case, is empty (base case).
- Check if not the base case, then if we have a solution for the current a and b saved in the memory, we return it, else we calculate the solution for the current a and b and store it in memory.
- For calculation of solution, we compare the last two characters of a and b.
- Check if they are equal,
- If true, we add this character to the solution string, then we erase the last character from each string and add to the solution string the string returned from LCS(a, b) after deleting the last characters.
- Otherwise, if the last two characters don't equal each other, we call LCS(a_without_last_character, b) and LCS(a, b_without_last_character). Then we compare the two returned strings and add the bigger string to the solution string.
- Store the solution in the memory and return it.
- Reverse the final returned solution, given that our top-down approach generates a reversed string.
Here is the implementation of the above recursive approach.
C++
#include <iostream>
#include <map>
#include <string>
using namespace std;
using mpsss = map<pair<string, string>, string>;
using mpssi = map<pair<string, string>, int>;
using ss = pair<string, string>;
mpsss dp;
// For keep track of visited subproblem or not (0 = not
// visited, 1 = visited)
mpssi vs;
// utility function to reverse a string, we need it because
// our top-down approach return a reversed solution
string reverse(string str)
{
string ans = str;
int u = 0;
int v = ans.length() - 1;
while (u < v) {
swap(ans[u], ans[v]);
u++;
v--;
}
return ans;
}
// utility function that compares two strings and return the
// longer in size.
string max_str(string a, string b)
{
if (a.length() > b.length())
return a;
else
return b;
}
string LCS_core(string a, string b)
{
// size of string a
int n_a = a.length();
// size of string b
int n_b = b.length();
// Base case
if (n_a == 0 || n_b == 0)
return "";
// dp index to access the dp structure
ss dp_i = make_pair(a, b);
// ans points to the memory location in the dp
// structure in which the solution string will be stored
string& ans = dp[dp_i];
// if visited return solution from memory.
if (vs[dp_i] == 1)
return dp[dp_i];
// if not visited, set the visit value to be one
// (meaning its now visited)
else
vs[dp_i] = 1;
// if the last two character match
if (a[n_a - 1] == b[n_b - 1]) {
// Add this character to the solution string
ans += a[n_a - 1];
// Erase last character from a
a.erase(n_a - 1, 1);
// Erase last character from b
b.erase(n_b - 1, 1);
// add to the solution string the value of
// LCS_core(a, b) (the remaining strings after
// deleting last characters)
ans += LCS_core(a, b);
return ans;
}
// Return longest string
ans += max_str(LCS_core(a.substr(0, n_a - 1), b),
LCS_core(a, b.substr(0, n_b - 1)));
return ans;
}
string LCS(string a, string b)
{
;
// Reverse obtained result
return reverse(LCS_core(a, b));
}
int main()
{
string a = "AGGTAB";
string b = "GXTXAYB";
cout << LCS(a, b);
return 0;
}
Java
import java.util.HashMap;
import java.util.Map;
class Main {
public static void main(String[] args) {
String a = "AGGTAB";
String b = "GXTXAYB";
System.out.println(LCS(a, b));
}
private static Map<String, Map<String, String>> dp = new HashMap<>();
private static Map<String, Map<String, Integer>> vs = new HashMap<>();
private static String reverse(String str) {
String ans = "";
for (int i = str.length() - 1; i >= 0; i--) {
ans += str.charAt(i);
}
return ans;
}
private static String max_str(String a, String b) {
return a.length() > b.length() ? a : b;
}
private static String LCS_core(String a, String b) {
if (a.isEmpty() || b.isEmpty()) {
return "";
}
if (vs.containsKey(a) && vs.get(a).containsKey(b)) {
return dp.get(a).get(b);
}
String ans = "";
if (a.charAt(a.length() - 1) == b.charAt(b.length() - 1)) {
ans += a.charAt(a.length() - 1);
a = a.substring(0, a.length() - 1);
b = b.substring(0, b.length() - 1);
ans += LCS_core(a, b);
} else {
ans += max_str(LCS_core(a.substring(0, a.length() - 1), b),
LCS_core(a, b.substring(0, b.length() - 1)));
}
if (!dp.containsKey(a)) {
dp.put(a, new HashMap<>());
}
if (!vs.containsKey(a)) {
vs.put(a, new HashMap<>());
}
dp.get(a).put(b, ans);
vs.get(a).put(b, 1);
return ans;
}
private static String LCS(String a, String b) {
return reverse(LCS_core(a, b));
}
}
// This code is contributed by Susobhan Akhuli
Python3
from typing import Dict, Tuple
# Initialize an empty dictionary to store the solutions of subproblems
dp: Dict[Tuple[str, str], str] = {}
# Initialize an empty dictionary to keep track of visited subproblems
vs: Dict[Tuple[str, str], int] = {}
# Utility function to reverse a string, we need it because our top-down approach
# return a reversed solution
def reverse(s: str) -> str:
ans = list(s)
u, v = 0, len(ans) - 1
while u < v:
ans[u], ans[v] = ans[v], ans[u] #swap operation
u += 1
v -= 1
return "".join(ans)
# Utility function that compares two strings and return the longer in size.
def max_str(a: str, b: str) -> str:
return a if len(a) > len(b) else b
# Recursive function that takes two strings as input, and returns the LCS of them
def LCS_core(a: str, b: str) -> str:
# Base case
if not a or not b:
return ""
# dp index to access the dp structure
dp_i = (a, b)
# if visited return solution from memory
if dp_i in vs:
return dp[dp_i]
else:
vs[dp_i] = 1
# if the last two character match
if a[-1] == b[-1]:
ans = a[-1] + LCS_core(a[:-1], b[:-1])
dp[dp_i] = ans
return ans
# Return longest string
ans = max_str(LCS_core(a[:-1], b), LCS_core(a, b[:-1]))
dp[dp_i] = ans
return ans
# Final wrapper function to call the recursive function and reverse the result
def LCS(a: str, b: str) -> str:
return reverse(LCS_core(a, b))
a = "AGGTAB"
b = "GXTXAYB"
print(LCS(a, b))
# This code is contributed by Shivam Tiwari
C#
using System;
using System.Collections.Generic
namespace LongestCommonSubsequence
{
public class GFG{
static void Main(string[] args)
{
string a = "AGGTAB";
string b = "GXTXAYB";
Console.WriteLine(LCS(a, b));
}
// Utility function to reverse a string
static string Reverse(string str)
{
char[] charArray = str.ToCharArray();
Array.Reverse(charArray);
return new string(charArray);
}
// Utility function to return the larger string
static string MaxStr(string a, string b)
{
return a.Length > b.Length ? a : b;
}
// Main function that returns the LCS
static string LCS_Core(string a, string b, Dictionary<string, string> dp, Dictionary<string, int> vs)
{
int n_a = a.Length;
int n_b = b.Length;
// Base case
if (n_a == 0 || n_b == 0)
{
return "";
}
// dp index to access the dp dictionary
string dp_i = a + "," + b;
// ans points to the memory location in the dp
// dictionary in which the solution string will be stored
string ans;
if (dp.TryGetValue(dp_i, out ans))
{
// If visited return solution from memory
return ans;
}
// If not visited, set the visit value to be one
// (meaning it's now visited)
vs[dp_i] = 1;
// If the last two characters match
if (a[n_a - 1] == b[n_b - 1])
{
// Add this character to the solution string
ans = a[n_a - 1] + LCS_Core(a.Substring(0, n_a - 1), b.Substring(0, n_b - 1), dp, vs);
dp[dp_i] = ans;
return ans;
}
// Return longest string
ans = MaxStr(LCS_Core(a.Substring(0, n_a - 1), b, dp, vs), LCS_Core(a, b.Substring(0, n_b - 1), dp, vs));
dp[dp_i] = ans;
return ans;
}
static string LCS(string a, string b)
{
Dictionary<string, string> dp = new Dictionary<string, string>();
Dictionary<string, int> vs = new Dictionary<string, int>();
string ans = LCS_Core(a, b, dp, vs);
// Reverse the obtained result
return Reverse(ans);
}
}
}
JavaScript
// Initialize an empty dictionary to store the solutions of subproblems
let dp = {};
// Initialize an empty dictionary to keep track of visited subproblems
let vs = {};
// Utility function to reverse a string, we need it because our top-down approach
// return a reversed solution
function reverse(s) {
let ans = s.split("");
let u = 0;
let v = ans.length - 1;
while (u < v) {
[ans[u], ans[v]] = [ans[v], ans[u]]; // swap operation
u++;
v--;
}
return ans.join("");
}
// Utility function that compares two strings and return the longer in size.
function maxStr(a, b) {
return a.length > b.length ? a : b;
}
// Recursive function that takes two strings as input,
// and returns the LCS of them
function LCS_core(a, b) {
// Base case
if (!a || !b) return "";
// dp index to access the dp structure
let dp_i = `${a},${b}`;
// if visited return solution from memory
if (dp_i in vs) return dp[dp_i];
else vs[dp_i] = 1;
// if the last two character match
if (a[a.length - 1] === b[b.length - 1]) {
let ans = a[a.length - 1] + LCS_core(a.slice(0, -1), b.slice(0, -1));
dp[dp_i] = ans;
return ans;
}
// Return longest string
let ans = maxStr(LCS_core(a.slice(0, -1), b), LCS_core(a, b.slice(0, -1)));
dp[dp_i] = ans;
return ans;
}
// Final wrapper function to call the
// recursive function and reverse the result
function LCS(a, b) {
return reverse(LCS_core(a, b));
}
// Driver Code
const a = "AGGTAB";
const b = "GXTXAYB";
console.log(LCS(a, b));
Time complexity: O(m*n) where m is the length of the first string and n is the length of the second string. This is because, for each character in both strings, we need to check whether they are equal and then recursively call the LCS_core() function.
Auxiliary space: O(m*n) as well. This is because we are using a map structure to keep track of the visited subproblems and the result of each subproblem. This structure is having a size of O(m*n).
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String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
2 min read
Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
2 min read
Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Hereâs the comparison of Linked List vs Arrays Linked List:
2 min read
Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
2 min read
Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
2 min read
Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
4 min read
Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
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Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
15+ min read
Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
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Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
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Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
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