Lowest Common Ancestor in a Binary Tree | Set 3 (Using RMQ)
Last Updated :
04 Aug, 2022
Given a rooted tree, and two nodes are in the tree, find the Lowest common ancestor of both the nodes. The LCA for two nodes u and v is defined as the farthest node from the root that is the ancestor to both u and v.
Prerequisites: LCA | SET 1
Example for the above figure :
Input : 4 5
Output : 2
Input : 4 7
Output : 1
Converting LCA to RMQ(Range Minimum Query):
Take an array named E[], which stores the order of dfs traversal i.e. the order in which the nodes are covered during the dfs traversal. For example,
The tree given above has dfs traversal in the order: 1-2-4-2-5-2-1-3.
Take another array L[], in which L[i] is the level of node E[i].
And the array H[], which stores the index of the first occurrence of ith node in the array E[].
So, for the above tree,
E[] = {1, 2, 4, 2, 5, 2, 1, 3}
L[] = {1, 2, 3, 2, 3, 2, 1, 2}
H[] = {0, 1, 7, 2, 4}
Note that the arrays E and L are with one-based indexing but the array H has zero-based indexing.
Now, to find the LCA(4, 3), first, use the array H and find the indices at which 4 and 3 are found in E i.e. H[4] and H[3]. So, the indices come out to be 2 and 7. Now, look at the subarray L[2 : 7], and find the minimum in this subarray which is 1 (at the 6th index), and the corresponding element in the array E i.e. E[6] is the LCA(4, 3).
To understand why this works, take LCA(4, 3) again. The path by which one can reach node 3 from node 4 is the subarray E[2 : 7]. And, if there is a node with the lowest level in this path, then it can simply be claimed to be the LCA(4, 3).
Now, the problem is to find the minimum in the subarray E[H[u]....H[v]] (assuming that H[u] >= H[v]). And, that could be done using a segment tree or sparse table. Below is the code using the segment tree.
Implementation:
C++
// CPP code to find LCA of given
// two nodes in a tree
#include <bits/stdc++.h>
#define sz(x) x.size()
#define pb push_back
#define left 2 * i + 1
#define right 2 * i + 2
using namespace std;
const int maxn = 100005;
// the graph
vector<vector<int>> g(maxn);
// level of each node
int level[maxn];
vector<int> e;
vector<int> l;
int h[maxn];
// the segment tree
int st[5 * maxn];
// adding edges to the graph(tree)
void add_edge(int u, int v) {
g[u].pb(v);
g[v].pb(u);
}
// assigning level to nodes
void leveling(int src) {
for (int i = 0; i < sz(g[src]); i++) {
int des = g[src][i];
if (!level[des]) {
level[des] = level[src] + 1;
leveling(des);
}
}
}
bool visited[maxn];
// storing the dfs traversal
// in the array e
void dfs(int src) {
e.pb(src);
visited[src] = 1;
for (int i = 0; i < sz(g[src]); i++) {
int des = g[src][i];
if (!visited[des]) {
dfs(des);
e.pb(src);
}
}
}
// making the array l
void setting_l(int n) {
for (int i = 0; i < sz(e); i++)
l.pb(level[e[i]]);
}
// making the array h
void setting_h(int n) {
for (int i = 0; i <= n; i++)
h[i] = -1;
for (int i = 0; i < sz(e); i++) {
// if is already stored
if (h[e[i]] == -1)
h[e[i]] = i;
}
}
// Range minimum query to return the index
// of minimum in the subarray L[qs:qe]
int RMQ(int ss, int se, int qs, int qe, int i) {
if (ss > se)
return -1;
// out of range
if (se < qs || qe < ss)
return -1;
// in the range
if (qs <= ss && se <= qe)
return st[i];
int mid = (ss + se) >> 1;
int st = RMQ(ss, mid, qs, qe, left);
int en = RMQ(mid + 1, se, qs, qe, right);
if (st != -1 && en != -1) {
if (l[st] < l[en])
return st;
return en;
} else if (st != -1)
return st;
else if (en != -1)
return en;
}
// constructs the segment tree
void SegmentTreeConstruction(int ss, int se, int i) {
if (ss > se)
return;
if (ss == se) // leaf
{
st[i] = ss;
return;
}
int mid = (ss + se) >> 1;
SegmentTreeConstruction(ss, mid, left);
SegmentTreeConstruction(mid + 1, se, right);
if (l[st[left]] < l[st[right]])
st[i] = st[left];
else
st[i] = st[right];
}
// Function to get LCA
int LCA(int x, int y) {
if (h[x] > h[y])
swap(x, y);
return e[RMQ(0, sz(l) - 1, h[x], h[y], 0)];
}
// Driver code
int main() {
// n=number of nodes in the tree
// q=number of queries to answer
int n = 15, q = 5;
// making the tree
/*
1
/ | \
2 3 4
| \
5 6
/ | \
8 7 9 (right of 5)
/ | \ | \
10 11 12 13 14
|
15
*/
add_edge(1, 2);
add_edge(1, 3);
add_edge(1, 4);
add_edge(3, 5);
add_edge(4, 6);
add_edge(5, 7);
add_edge(5, 8);
add_edge(5, 9);
add_edge(7, 10);
add_edge(7, 11);
add_edge(7, 12);
add_edge(9, 13);
add_edge(9, 14);
add_edge(12, 15);
level[1] = 1;
leveling(1);
dfs(1);
setting_l(n);
setting_h(n);
SegmentTreeConstruction(0, sz(l) - 1, 0);
cout << LCA(10, 15) << endl;
cout << LCA(11, 14) << endl;
return 0;
}
// JAVA code to find LCA of given
// two nodes in a tree
import java.util.*;
public class GFG
{
static int maxn = 100005;
static int left(int i)
{
return (2 * i + 1);
}
static int right(int i) { return 2 * i + 2;}
// the graph
static Vector<Integer> []g = new Vector[maxn];
// level of each node
static int []level = new int[maxn];
static Vector<Integer> e = new Vector<>();
static Vector<Integer> l= new Vector<>();
static int []h = new int[maxn];
// the segment tree
static int []st = new int[5 * maxn];
// adding edges to the graph(tree)
static void add_edge(int u, int v)
{
g[u].add(v);
g[v].add(u);
}
// assigning level to nodes
static void levelling(int src)
{
for (int i = 0; i < (g[src].size()); i++)
{
int des = g[src].get(i);
if (level[des] != 0)
{
level[des] = level[src] + 1;
leveling(des);
}
}
}
static boolean []visited = new boolean[maxn];
// storing the dfs traversal
// in the array e
static void dfs(int src)
{
e.add(src);
visited[src] = true;
for (int i = 0; i < (g[src]).size(); i++)
{
int des = g[src].get(i);
if (!visited[des])
{
dfs(des);
e.add(src);
}
}
}
// making the array l
static void setting_l(int n)
{
for (int i = 0; i < e.size(); i++)
l.add(level[e.get(i)]);
}
// making the array h
static void setting_h(int n)
{
for (int i = 0; i <= n; i++)
h[i] = -1;
for (int i = 0; i < e.size(); i++)
{
// if is already stored
if (h[e.get(i)] == -1)
h[e.get(i)] = i;
}
}
// Range minimum query to return the index
// of minimum in the subarray L[qs:qe]
static int RMQ(int ss, int se, int qs, int qe, int i)
{
if (ss > se)
return -1;
// out of range
if (se < qs || qe < ss)
return -1;
// in the range
if (qs <= ss && se <= qe)
return st[i];
int mid = (ss + se)/2 ;
int st = RMQ(ss, mid, qs, qe, left(i));
int en = RMQ(mid + 1, se, qs, qe, right(i));
if (st != -1 && en != -1)
{
if (l.get(st) < l.get(en))
return st;
return en;
} else if (st != -1)
return st-2;
else if (en != -1)
return en-1;
return 0;
}
// constructs the segment tree
static void SegmentTreeConstruction(int ss,
int se, int i)
{
if (ss > se)
return;
if (ss == se) // leaf
{
st[i] = ss;
return;
}
int mid = (ss + se) /2;
SegmentTreeConstruction(ss, mid, left(i));
SegmentTreeConstruction(mid + 1, se, right(i));
if (l.get(st[left(i)]) < l.get(st[right(i)]))
st[i] = st[left(i)];
else
st[i] = st[right(i)];
}
// Function to get LCA
static int LCA(int x, int y)
{
if (h[x] > h[y])
{
int t = x;
x = y;
y = t;
}
return e.get(RMQ(0, l.size() - 1, h[x], h[y], 0));
}
// Driver code
public static void main(String[] args)
{
// n=number of nodes in the tree
// q=number of queries to answer
int n = 15, q = 5;
for (int i = 0; i < g.length; i++)
g[i] = new Vector<Integer>();
// making the tree
/*
1
/ | \
2 3 4
| \
5 6
/ | \
8 7 9 (right of 5)
/ | \ | \
10 11 12 13 14
|
15
*/
add_edge(1, 2);
add_edge(1, 3);
add_edge(1, 4);
add_edge(3, 5);
add_edge(4, 6);
add_edge(5, 7);
add_edge(5, 8);
add_edge(5, 9);
add_edge(7, 10);
add_edge(7, 11);
add_edge(7, 12);
add_edge(9, 13);
add_edge(9, 14);
add_edge(12, 15);
level[1] = 1;
leveling(1);
dfs(1);
setting_l(n);
setting_h(n);
SegmentTreeConstruction(0, l.size() - 1, 0);
System.out.print(LCA(10, 15) +"\n");
System.out.print(LCA(11, 14) +"\n");
}
}
// This code is contributed by Rajput-Ji
# Python code to find LCA of given
# two nodes in a tree
maxn = 100005
# the graph
g = [[] for i in range(maxn)]
# level of each node
level = [0] * maxn
e = []
l = []
h = [0] * maxn
# the segment tree
st = [0] * (5 * maxn)
# adding edges to the graph(tree)
def add_edge(u: int, v: int):
g[u].append(v)
g[v].append(u)
# assigning level to nodes
def levelling(src: int):
for i in range(len(g[src])):
des = g[src][i]
if not level[des]:
level[des] = level[src] + 1
leveling(des)
visited = [False] * maxn
# storing the dfs traversal
# in the array e
def dfs(src: int):
e.append(src)
visited[src] = True
for i in range(len(g[src])):
des = g[src][i]
if not visited[des]:
dfs(des)
e.append(src)
# making the array l
def setting_l(n: int):
for i in range(len(e)):
l.append(level[e[i]])
# making the array h
def setting_h(n: int):
for i in range(n + 1):
h[i] = -1
for i in range(len(e)):
# if is already stored
if h[e[i]] == -1:
h[e[i]] = i
# Range minimum query to return the index
# of minimum in the subarray L[qs:qe]
def RMQ(ss: int, se: int, qs: int, qe: int, i: int) -> int:
global st
if ss > se:
return -1
# out of range
if se < qs or qe < ss:
return -1
# in the range
if qs <= ss and se <= qe:
return st[i]
mid = (se + ss) >> 1
stt = RMQ(ss, mid, qs, qe, 2 * i + 1)
en = RMQ(mid + 1, se, qs, qe, 2 * i + 2)
if stt != -1 and en != -1:
if l[stt] < l[en]:
return stt
return en
elif stt != -1:
return stt
elif en != -1:
return en
# constructs the segment tree
def segmentTreeConstruction(ss: int, se: int, i: int):
if ss > se:
return
if ss == se: # leaf
st[i] = ss
return
mid = (ss + se) >> 1
segmentTreeConstruction(ss, mid, 2 * i + 1)
segmentTreeConstruction(mid + 1, se, 2 * i + 2)
if l[st[2 * i + 1]] < l[st[2 * i + 2]]:
st[i] = st[2 * i + 1]
else:
st[i] = st[2 * i + 2]
# Function to get LCA
def LCA(x: int, y: int) -> int:
if h[x] > h[y]:
x, y = y, x
return e[RMQ(0, len(l) - 1, h[x], h[y], 0)]
# Driver Code
if __name__ == "__main__":
# n=number of nodes in the tree
# q=number of queries to answer
n = 15
q = 5
# making the tree
# /*
# 1
# / | \
# 2 3 4
# | \
# 5 6
# / | \
# 8 7 9 (right of 5)
# / | \ | \
# 10 11 12 13 14
# |
# 15
# */
add_edge(1, 2)
add_edge(1, 3)
add_edge(1, 4)
add_edge(3, 5)
add_edge(4, 6)
add_edge(5, 7)
add_edge(5, 8)
add_edge(5, 9)
add_edge(7, 10)
add_edge(7, 11)
add_edge(7, 12)
add_edge(9, 13)
add_edge(9, 14)
add_edge(12, 15)
level[1] = 1
leveling(1)
dfs(1)
setting_l(n)
setting_h(n)
segmentTreeConstruction(0, len(l) - 1, 0)
print(LCA(10, 15))
print(LCA(11, 14))
# This code is contributed by
# sanjeev2552
// C# code to find LCA of given
// two nodes in a tree
using System;
using System.Collections.Generic;
public class GFG
{
static int maxn = 100005;
static int left(int i)
{
return (2 * i + 1);
}
static int right(int i) { return 2 * i + 2;}
// the graph
static List<int> []g = new List<int>[maxn];
// level of each node
static int []level = new int[maxn];
static List<int> e = new List<int>();
static List<int> l= new List<int>();
static int []h = new int[maxn];
// the segment tree
static int []st;
// adding edges to the graph(tree)
static void add_edge(int u, int v)
{
g[u].Add(v);
g[v].Add(u);
}
// assigning level to nodes
static void leveling(int src)
{
for (int i = 0; i < (g[src].Count); i++)
{
int des = g[src][i];
if (level[des] != 0)
{
level[des] = level[src] + 1;
leveling(des);
}
}
}
static bool []visited = new bool[maxn];
// storing the dfs traversal
// in the array e
static void dfs(int src)
{
e.Add(src);
visited[src] = true;
for (int i = 0; i < (g[src]).Count; i++)
{
int des = g[src][i];
if (!visited[des])
{
dfs(des);
e.Add(src);
}
}
}
// making the array l
static void setting_l(int n)
{
for (int i = 0; i < e.Count; i++)
l.Add(level[e[i]]);
}
// making the array h
static void setting_h(int n)
{
for (int i = 0; i <= n; i++)
h[i] = -1;
for (int i = 0; i < e.Count; i++)
{
// if is already stored
if (h[e[i]] == -1)
h[e[i]] = i;
}
}
// Range minimum query to return the index
// of minimum in the subarray L[qs:qe]
static int RMQ(int ss, int se, int qs, int qe, int i)
{
if (ss > se)
return -1;
// out of range
if (se < qs || qe < ss)
return -1;
// in the range
if (qs <= ss && se <= qe)
return st[i];
int mid = (ss + se)/2 ;
int sti = RMQ(ss, mid, qs, qe, left(i));
int en = RMQ(mid + 1, se, qs, qe, right(i));
if (sti != -1 && en != -1)
{
if (l[sti] < l[en])
return sti;
return en;
} else if (sti != -1)
return sti-2;
else if (en != -1)
return en-1;
return 0;
}
// constructs the segment tree
static void SegmentTreeConstruction(int ss,
int se, int i)
{
if (ss > se)
return;
if (ss == se) // leaf
{
st[i] = ss;
return;
}
int mid = (ss + se) /2;
SegmentTreeConstruction(ss, mid, left(i));
SegmentTreeConstruction(mid + 1, se, right(i));
if (l[st[left(i)]] < l[st[right(i)]])
st[i] = st[left(i)];
else
st[i] = st[right(i)];
}
// Function to get LCA
static int LCA(int x, int y)
{
if (h[x] > h[y])
{
int t = x;
x = y;
y = t;
}
return e[RMQ(0, l.Count - 1, h[x], h[y], 0)];
}
// Driver code
public static void Main(String[] args)
{
st = new int[5 * maxn];
// n=number of nodes in the tree
// q=number of queries to answer
int n = 15;
for (int i = 0; i < g.Length; i++)
g[i] = new List<int>();
// making the tree
/*
1
/ | \
2 3 4
| \
5 6
/ | \
8 7 9 (right of 5)
/ | \ | \
10 11 12 13 14
|
15
*/
add_edge(1, 2);
add_edge(1, 3);
add_edge(1, 4);
add_edge(3, 5);
add_edge(4, 6);
add_edge(5, 7);
add_edge(5, 8);
add_edge(5, 9);
add_edge(7, 10);
add_edge(7, 11);
add_edge(7, 12);
add_edge(9, 13);
add_edge(9, 14);
add_edge(12, 15);
level[1] = 1;
leveling(1);
dfs(1);
setting_l(n);
setting_h(n);
SegmentTreeConstruction(0, l.Count - 1, 0);
Console.Write(LCA(10, 15) +"\n");
Console.Write(LCA(11, 14) +"\n");
}
}
// This code is contributed by gauravrajput1
<script>
// JavaScript code to find LCA of given
// two nodes in a tree
let maxn = 100005;
function left(i)
{
return (2 * i + 1);
}
function right(i) { return 2 * i + 2;}
// the graph
let g = new Array(maxn);
// level of each node
let level = new Array(maxn);
level.fill(0);
let e = [];
let l= [];
let h = new Array(maxn);
h.fill(0);
// the segment tree
let st = new Array(5 * maxn);
st.fill(0);
// adding edges to the graph(tree)
function add_edge(u, v)
{
g[u].push(v);
g[v].push(u);
}
// assigning level to nodes
function levelling(src)
{
for (let i = 0; i < (g[src].length); i++)
{
let des = g[src][i];
if (level[des] != 0)
{
level[des] = level[src] + 1;
levelling(des);
}
}
}
let visited = new Array(maxn);
visited.fill(false);
// storing the dfs traversal
// in the array e
function dfs(src)
{
e.push(src);
visited[src] = true;
for (let i = 0; i < (g[src]).length; i++)
{
let des = g[src][i];
if (!visited[des])
{
dfs(des);
e.push(src);
}
}
}
// making the array l
function setting_l(n)
{
for (let i = 0; i < e.length; i++)
l.push(level[e[i]]);
}
// making the array h
function setting_h(n)
{
for (let i = 0; i <= n; i++)
h[i] = -1;
for (let i = 0; i < e.length; i++)
{
// if is already stored
if (h[e[i]] == -1)
h[e[i]] = i;
}
}
// Range minimum query to return the index
// of minimum in the subarray L[qs:qe]
function RMQ(ss, se, qs, qe, i)
{
if (ss > se)
return -1;
// out of range
if (se < qs || qe < ss)
return -1;
// in the range
if (qs <= ss && se <= qe)
return st[i];
let mid = parseInt((ss + se)/2 , 10);
let St = RMQ(ss, mid, qs, qe, left(i));
let en = RMQ(mid + 1, se, qs, qe, right(i));
if (St != -1 && en != -1)
{
if (l[St] < l[en])
return St;
return en;
} else if (St != -1)
return St-2;
else if (en != -1)
return en-1;
return 0;
}
// constructs the segment tree
function SegmentTreeConstruction(ss, se, i)
{
if (ss > se)
return;
if (ss == se) // leaf
{
st[i] = ss;
return;
}
let mid = parseInt((ss + se) /2, 10);
SegmentTreeConstruction(ss, mid, left(i));
SegmentTreeConstruction(mid + 1, se, right(i));
if (l[st[left(i)]] < l[st[right(i)]])
st[i] = st[left(i)];
else
st[i] = st[right(i)];
}
// Function to get LCA
function LCA(x, y)
{
if (h[x] > h[y])
{
let t = x;
x = y;
y = t;
}
return e[RMQ(0, l.length - 1, h[x], h[y], 0)];
}
// n=number of nodes in the tree
// q=number of queries to answer
let n = 15, q = 5;
for (let i = 0; i < g.length; i++)
g[i] = [];
// making the tree
/*
1
/ | \
2 3 4
| \
5 6
/ | \
8 7 9 (right of 5)
/ | \ | \
10 11 12 13 14
|
15
*/
add_edge(1, 2);
add_edge(1, 3);
add_edge(1, 4);
add_edge(3, 5);
add_edge(4, 6);
add_edge(5, 7);
add_edge(5, 8);
add_edge(5, 9);
add_edge(7, 10);
add_edge(7, 11);
add_edge(7, 12);
add_edge(9, 13);
add_edge(9, 14);
add_edge(12, 15);
level[1] = 1;
levelling(1);
dfs(1);
setting_l(n);
setting_h(n);
SegmentTreeConstruction(0, l.length - 1, 0);
document.write(LCA(10, 15) +"</br>");
document.write(LCA(11, 14) +"</br>");
</script>
Time Complexity: The arrays defined are stored in O(n). The segment tree construction also takes O(n) time. The LCA function calls the function RMQ which takes O(logn) per query (as it uses the segment tree). So overall time complexity is O(n + q * logn).
Auxiliary Space: O(n)
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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
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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. Here’s the comparison of Linked List vs Arrays Linked List:
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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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