Union of Two Sorted Arrays
Last Updated :
23 Jul, 2025
Given two sorted arrays a[] and b[], the task is to return union of both the arrays in sorted order. Union of two arrays is an array having all distinct elements that are present in either array. The input arrays may contain duplicates.
Examples:
Input: a[] = {1, 1, 2, 2, 2, 4}, b[] = {2, 2, 4, 4}
Output: {1, 2, 4}
Explanation: 1, 2 and 4 are the distinct elements present in either array.
Input: a[] = {3, 5, 10, 10, 10, 15, 15, 20}, b[] = {5, 10, 10, 15, 30}
Output: {3, 5, 10, 15, 20, 30}
Explanation: 3, 5, 10, 15, 20 and 30 are the distinct elements present in either array.
[Naive Approach] Using Nested Loops – O(n*m) Time and O(1) Space
The intuition behind this approach is to gather unique elements from two arrays by checking each element against the current result array. The idea is to traverse both the arrays and for each element, check if the element is present in the result or not. If not, then add this element to the result.
C++
// C++ program to find union of two sorted arrays
// using nested loops
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> findUnion(vector<int>& a, vector<int>& b) {
vector<int> res;
// Traverse through a[] and search every element
// a[i] in result
for(int i = 0; i < a.size(); i++){
// check if the element is already
// in the result to avoid duplicates
int j;
for (j = 0; j < res.size(); j++) {
if (res[j] == a[i])
break;
}
if (j == res.size())
res.push_back(a[i]);
}
// Traverse through b[] and search every element
// b[i] in result
for(int i = 0; i < b.size(); i++){
// check if the element is already
// in the result to avoid duplicates
int j;
for (j = 0; j < res.size(); j++) {
if (res[j] == b[i])
break;
}
if (j == res.size())
res.push_back(b[i]);
}
sort(res.begin(), res.end());
return res;
}
int main() {
vector<int> a = {1, 1, 2, 2, 2, 4};
vector<int> b = {2, 2, 4, 4};
vector<int> res = findUnion(a, b);
for (int i = 0; i < res.size(); i++)
cout << res[i] << " ";
return 0;
}
// C program to find union of two sorted arrays
// using nested loops
#include <stdio.h>
#include <stdlib.h>
// Function to compare two integers for qsort
int compare(const void* a, const void* b) {
return (*(int*)a - *(int*)b);
}
// Function to find union of two sorted arrays
int* findUnion(int a[], int n, int b[], int m, int* size) {
int* res = (int*)malloc((m + n) * sizeof(int));
int index = 0;
// Traverse through a[] and search every element
// a[i] in result
for (int i = 0; i < n; i++) {
// Check if the element is already in the result to avoid duplicates
int j;
for (j = 0; j < index; j++) {
if (res[j] == a[i])
break;
}
if (j == index)
res[index++] = a[i];
}
// Traverse through b[] and search every element
// b[i] in result
for (int i = 0; i < m; i++) {
// Check if the element is already in the result to avoid duplicates
int j;
for (j = 0; j < index; j++) {
if (res[j] == b[i])
break;
}
if (j == index)
res[index++] = b[i];
}
// Sort the result array using qsort
qsort(res, index, sizeof(int), compare);
*size = index;
return res;
}
int main() {
int a[] = {1, 1, 2, 2, 2, 4};
int b[] = {2, 2, 4, 4};
int n = sizeof(a) / sizeof(a[0]);
int m = sizeof(b) / sizeof(b[0]);
int size;
int* result = findUnion(a, n, b, m, &size);
for (int i = 0; i < size; i++)
printf("%d ", result[i]);
free(result);
return 0;
}
// Java program to find union of two arrays
// using nested loops
import java.util.*;
class GfG {
static ArrayList<Integer> findUnion(int[] a, int[] b) {
ArrayList<Integer> res = new ArrayList<>();
// Traverse through a[] and search every element
// a[i] in result
for (int i = 0; i < a.length; i++) {
// check if the element is already in the result
// to avoid duplicates
if (!res.contains(a[i])) {
res.add(a[i]);
}
}
// Traverse through b[] and search every element
// b[i] in result
for (int i = 0; i < b.length; i++) {
// check if the element is already in the result
// to avoid duplicates
if (!res.contains(b[i])) {
res.add(b[i]);
}
}
Collections.sort(res);
return res;
}
public static void main(String[] args) {
int[] a = {1, 1, 2, 2, 2, 4};
int[] b = {2, 2, 4, 4};
ArrayList<Integer> res = findUnion(a, b);
for (Integer num : res) {
System.out.print(num + " ");
}
}
}
# Python program to find union of two arrays
# using nested loops
def findUnion(a, b):
res = []
# Traverse through a[] and search every element
# a[i] in result
for i in range(len(a)):
# check if the element is already
# in the result to avoid duplicates
if a[i] not in res:
res.append(a[i])
# Traverse through b[] and search every element
# b[i] in result
for i in range(len(b)):
# check if the element is already
# in the result to avoid duplicates
if b[i] not in res:
res.append(b[i])
res.sort()
return res
if __name__ == "__main__":
a = [1, 1, 2, 2, 2, 4]
b = [2, 2, 4, 4]
res = findUnion(a, b)
for i in res:
print(i, end=" ")
// C# program to find union of two arrays
// using nested loops
using System;
using System.Collections.Generic;
class GfG {
static List<int> findUnion(int[] a, int[] b) {
List<int> res = new List<int>();
// Traverse through a[] and search every element
// a[i] in result
for (int i = 0; i < a.Length; i++) {
// check if the element is already
// in the result to avoid duplicates
if (!res.Contains(a[i])) {
res.Add(a[i]);
}
}
// Traverse through b[] and search every element
// b[i] in result
for (int i = 0; i < b.Length; i++) {
// check if the element is already
// in the result to avoid duplicates
if (!res.Contains(b[i])) {
res.Add(b[i]);
}
}
res.Sort();
return res;
}
static void Main() {
int[] a = {1, 1, 2, 2, 2, 4};
int[] b = {2, 2, 4, 4};
List<int> res = findUnion(a, b);
foreach (int i in res) {
Console.Write(i + " ");
}
}
}
// JavaScript program to find union of two arrays
// using nested loops
function findUnion(a, b) {
let res = [];
// Traverse through a[] and search every element
// a[i] in result
for (let i = 0; i < a.length; i++) {
// check if the element is already
// in the result to avoid duplicates
if (!res.includes(a[i])) {
res.push(a[i]);
}
}
// Traverse through b[] and search every element
// b[i] in result
for (let i = 0; i < b.length; i++) {
// check if the element is already
// in the result to avoid duplicates
if (!res.includes(b[i])) {
res.push(b[i]);
}
}
res.sort((x, y) => x - y);
return res;
}
const a = [1, 1, 2, 2, 2, 4];
const b = [2, 2, 4, 4];
const res = findUnion(a, b);
console.log(res.join(" "));
Time Complexity: O((n + m)2), where n is size of a[] and m is size of b[]
- Copying all elements from a[] to res[] takes O(n2) time.
- Now in the worst case, there will be no common elements in a[] and b[]. So, to check if the first element of b[] is present in res[], we need n comparisons. Similarly, for second element of b[], we need (n + 1) comparisons. So for m elements, total number of comparisons will be: n + (n + 1) + (n + 2) …. (n + m) = (n * m) + (m2 / 2)
- So, overall time complexity = O(n2 + n * m + m2)
Auxiliary Space: O(1)
[Better Approach] Using Set – O((n+m)*(log (n+m))) Time and O(n+m) Space
The approach is to insert all elements from both arrays, a[] and b[], into a set. Since a set automatically removes duplicates, it gives us the union of the two arrays. Also, the set keeps the elements in sorted order, so after inserting them, we can store these sorted and unique elements in a result array.
Note: In Python and JavaScript, the set data structure does not store the elements in sorted order, so we need to explicitly sort the union array.
C++
// C++ program to find union of two sorted arrays using Set
#include <iostream>
#include <set>
#include <vector>
using namespace std;
vector<int> findUnion(vector<int> &a, vector<int> &b) {
set<int> st;
// Put all elements of a[] in st
for (int i = 0; i < a.size(); i++)
st.insert(a[i]);
// Put all elements of b[] in st
for (int i = 0; i < b.size(); i++)
st.insert(b[i]);
vector<int> res;
// iterate through the set to fill the result array
for (auto it : st)
res.push_back(it);
return res;
}
int main() {
vector<int> a = {1, 1, 2, 2, 2, 4};
vector<int> b = {2, 2, 4, 4};
vector<int> res = findUnion(a, b);
for (int i = 0; i < res.size(); i++)
cout << res[i] << " ";
return 0;
}
// Java program to find union of two sorted arrays using Set
import java.util.*;
class GfG {
static ArrayList<Integer> findUnion(int a[],
int b[]) {
Set<Integer> st = new TreeSet<>();
// Put all elements of a[] in st
for (int i = 0; i < a.length; i++)
st.add(a[i]);
// Put all elements of b[] in st
for (int i = 0; i < b.length; i++)
st.add(b[i]);
ArrayList<Integer> res = new ArrayList<>(st);
return res;
}
public static void main(String[] args) {
int a[] = { 1, 1, 2, 2, 2, 4 };
int b[] = { 2, 2, 4, 4 };
ArrayList<Integer> res = findUnion(a, b);
for (int i = 0; i < res.size(); i++)
System.out.print(res.get(i) + " ");
}
}
# Python program to find union of two sorted arrays using Set
def findUnion(a, b):
st = set()
# Put all elements of a[] in st
for i in range(len(a)):
st.add(a[i])
# Put all elements of b[] in st
for i in range(len(b)):
st.add(b[i])
res = []
# iterate through the set to fill the result array
for it in st:
res.append(it)
res.sort()
return res
if __name__ == "__main__":
a = [1, 1, 2, 2, 2, 4]
b = [2, 2, 4, 4]
res = findUnion(a, b)
for i in res:
print(i, end=" ")
// C# Program to find union of two sorted arrays using Set
using System;
using System.Collections.Generic;
class GfG {
static List<int> FindUnion(int[] a, int[] b) {
SortedSet<int> st = new SortedSet<int>();
// Put all elements of a[] in st
for (int i = 0; i < a.Length; i++)
st.Add(a[i]);
// Put all elements of b[] in st
for (int i = 0; i < b.Length; i++)
st.Add(b[i]);
List<int> res = new List<int>(st);
return res;
}
static void Main() {
int[] a = {1, 1, 2, 2, 2, 4};
int[] b = {2, 2, 4, 4};
List<int> res = FindUnion(a, b);
for (int i = 0; i < res.Count; i++)
Console.Write(res[i] + " ");
}
}
// JavaScript program to find union of two sorted arrays using Set
function findUnion(a, b) {
let st = new Set();
// Put all elements of a[] in st
for (let i = 0; i < a.length; i++)
st.add(a[i]);
// Put all elements of b[] in st
for (let i = 0; i < b.length; i++)
st.add(b[i]);
let res = Array.from(st);
res.sort((x, y) => x - y);
return res;
}
let a = [1, 1, 2, 2, 2, 4];
let b = [2, 2, 4, 4];
let res = findUnion(a, b);
console.log(res.join(" "));
Time Complexity: O((n + m) * (log (n + m))) , where n is the size of array a[] and m is the size of array b[]
Auxiliary Space: O(n + m)
[Expected Approach] Using Merge Step of Merge Sort - O(n+m) Time and O(1) Space
The idea is to find the union of two sorted arrays using merge step in merge sort. We maintain two pointers to traverse both arrays simultaneously.
- If the element in first array is smaller, add it to the result and move the pointer of first array forward.
- If the element in second array is smaller, add it to the result and move the pointer of second array forward.
- If both elements are equal, add one of them and move both the pointers forward.
Also, while traversing both the arrays, we will compare the current element with its previous element and skip the current element if it is same as previous element. This will ensure that the union will contain only distinct elements.
C++
// C++ program to find union of two sorted arrays using
// merge step of merge sort
#include <bits/stdc++.h>
using namespace std;
vector<int> findUnion(vector<int>& a, vector<int>& b) {
vector<int> res;
int n = a.size();
int m = b.size();
// This is similar to merge of merge sort
int i = 0, j = 0;
while(i < n && j < m) {
// Skip duplicate elements in the first array
if(i > 0 && a[i - 1] == a[i]) {
i++;
continue;
}
// Skip duplicate elements in the second array
if(j > 0 && b[j - 1] == b[j]) {
j++;
continue;
}
// select and add the smaller element and move
if(a[i] < b[j]) {
res.push_back(a[i]);
i++;
}
else if(a[i] > b[j]) {
res.push_back(b[j]);
j++;
}
// If equal, then add to result and move both
else {
res.push_back(a[i]);
i++;
j++;
}
}
// Add the remaining elements of a[]
while (i < n) {
// Skip duplicate elements in the first array
if(i > 0 && a[i - 1] == a[i]) {
i++;
continue;
}
res.push_back(a[i]);
i++;
}
// Add the remaining elements of b[]
while (j < m) {
// Skip duplicate elements in the second array
if(j > 0 && b[j - 1] == b[j]) {
j++;
continue;
}
res.push_back(b[j]);
j++;
}
return res;
}
int main() {
vector<int> a = {1, 1, 2, 2, 2, 4};
vector<int> b = {2, 2, 4, 4};
vector<int> res = findUnion(a, b);
for (int x : res) {
cout << x << " ";
}
}
// C program to find union of two sorted arrays using
// merge step of merge sort
#include <stdio.h>
// Function to find union of two sorted arrays
int* findUnion(int a[], int n, int b[], int m, int* size) {
int* res = (int*)malloc((m + n) * sizeof(int));
int i = 0, j = 0;
int index = 0;
// This is similar to merge of merge sort
while (i < n && j < m) {
// Skip duplicate elements in the first array
if (i > 0 && a[i - 1] == a[i]) {
i++;
continue;
}
// Skip duplicate elements in the second array
if (j > 0 && b[j - 1] == b[j]) {
j++;
continue;
}
// select and add the smaller element and move
if (a[i] < b[j]) {
res[index++] = a[i];
i++;
}
else if (a[i] > b[j]) {
res[index++] = b[j];
j++;
}
// If equal, then add to result and move both
else {
res[index++] = a[i];
i++;
j++;
}
}
// Add the remaining elements of a[]
while (i < n) {
// Skip duplicate elements in the first array
if (i > 0 && a[i - 1] == a[i]) {
i++;
continue;
}
res[index++] = a[i];
i++;
}
// Add the remaining elements of b[]
while (j < m) {
// Skip duplicate elements in the second array
if (j > 0 && b[j - 1] == b[j]) {
j++;
continue;
}
res[index++] = b[j];
j++;
}
// Update the size of the result
*size = index;
return res;
}
int main() {
int a[] = {1, 1, 2, 2, 2, 4};
int b[] = {2, 2, 4, 4};
int n = sizeof(a) / sizeof(a[0]);
int m = sizeof(b) / sizeof(b[0]);
int size;
int* result = findUnion(a, n, b, m, &size);
// Print the result
for (int i = 0; i < size; i++) {
printf("%d ", result[i]);
}
free(result);
return 0;
}
// Java program to find union of two sorted arrays using
// merge step of merge sort
import java.util.*;
class GfG {
static ArrayList<Integer> findUnion(int[] a, int[] b) {
ArrayList<Integer> res = new ArrayList<>();
int n = a.length;
int m = b.length;
// This is similar to merge of merge sort
int i = 0, j = 0;
while(i < n && j < m) {
// Skip duplicate elements in the first array
if(i > 0 && a[i - 1] == a[i]) {
i++;
continue;
}
// Skip duplicate elements in the second array
if(j > 0 && b[j - 1] == b[j]) {
j++;
continue;
}
// select and add the smaller element and move
if(a[i] < b[j]) {
res.add(a[i]);
i++;
}
else if(a[i] > b[j]) {
res.add(b[j]);
j++;
}
// If equal, then add to result and move both
else {
res.add(a[i]);
i++;
j++;
}
}
// Add the remaining elements of a[]
while (i < n) {
// Skip duplicate elements in the first array
if(i > 0 && a[i - 1] == a[i]) {
i++;
continue;
}
res.add(a[i]);
i++;
}
// Add the remaining elements of b[]
while (j < m) {
// Skip duplicate elements in the second array
if(j > 0 && b[j - 1] == b[j]) {
j++;
continue;
}
res.add(b[j]);
j++;
}
return res;
}
public static void main(String[] args) {
int[] a = {1, 1, 2, 2, 2, 4};
int[] b = {2, 2, 4, 4};
List<Integer> res = findUnion(a, b);
for (int x : res) {
System.out.print(x + " ");
}
}
}
# Python program to find union of two sorted arrays using
# merge step of merge sort
def findUnion(a, b):
res = []
n, m = len(a), len(b)
i, j = 0, 0
# This is similar to merge of merge sort
while i < n and j < m:
# Skip duplicate elements in the first array
if i > 0 and a[i - 1] == a[i]:
i += 1
continue
# Skip duplicate elements in the second array
if j > 0 and b[j - 1] == b[j]:
j += 1
continue
# select and add the smaller element and move
if a[i] < b[j]:
res.append(a[i])
i += 1
elif a[i] > b[j]:
res.append(b[j])
j += 1
# If equal, then add to result and move both
else:
res.append(a[i])
i += 1
j += 1
# Add the remaining elements of a[]
while i < n:
# Skip duplicate elements in the first array
if i > 0 and a[i - 1] == a[i]:
i += 1
continue
res.append(a[i])
i += 1
# Add the remaining elements of b[]
while j < m:
# Skip duplicate elements in the second array
if j > 0 and b[j - 1] == b[j]:
j += 1
continue
res.append(b[j])
j += 1
return res
if __name__ == "__main__":
a = [1, 1, 2, 2, 2, 4]
b = [2, 2, 4, 4]
res = findUnion(a, b)
for x in res:
print(x ,end = " ");
// C# program to find union of two sorted arrays using
// merge step of merge sort
using System;
using System.Collections.Generic;
class GfG {
static List<int> findUnion(int[] a, int[] b) {
List<int> res = new List<int>();
int n = a.Length;
int m = b.Length;
// This is similar to merge of merge sort
int i = 0, j = 0;
while (i < n && j < m) {
// Skip duplicate elements in the first array
if (i > 0 && a[i - 1] == a[i]) {
i++;
continue;
}
// Skip duplicate elements in the second array
if (j > 0 && b[j - 1] == b[j]) {
j++;
continue;
}
// select and add the smaller element and move
if (a[i] < b[j]) {
res.Add(a[i]);
i++;
}
else if (a[i] > b[j]) {
res.Add(b[j]);
j++;
}
// If equal, then add to result and move both
else {
res.Add(a[i]);
i++;
j++;
}
}
// Add the remaining elements of a[]
while (i < n) {
// Skip duplicate elements in the first array
if (i > 0 && a[i - 1] == a[i]) {
i++;
continue;
}
res.Add(a[i]);
i++;
}
// Add the remaining elements of b[]
while (j < m) {
// Skip duplicate elements in the second array
if (j > 0 && b[j - 1] == b[j]) {
j++;
continue;
}
res.Add(b[j]);
j++;
}
return res;
}
static void Main() {
int[] a = { 1, 1, 2, 2, 2, 4 };
int[] b = { 2, 2, 4, 4 };
List<int> res = findUnion(a, b);
foreach (int x in res) {
Console.Write(x + " ");
}
}
}
// JavaScript program to find union of two sorted arrays using
// merge step of merge sort
function findUnion(a, b) {
let res = [];
let n = a.length, m = b.length;
let i = 0, j = 0;
// This is similar to merge of merge sort
while(i < n && j < m) {
// Skip duplicate elements in the first array
if(i > 0 && a[i - 1] === a[i]) {
i++;
continue;
}
// Skip duplicate elements in the second array
if(j > 0 && b[j - 1] === b[j]) {
j++;
continue;
}
// select and add the smaller element and move
if(a[i] < b[j]) {
res.push(a[i]);
i++;
} else if(a[i] > b[j]) {
res.push(b[j]);
j++;
}
// If equal, then add to result and move both
else {
res.push(a[i]);
i++;
j++;
}
}
// Add the remaining elements of a[]
while(i < n) {
// Skip duplicate elements in the first array
if(i > 0 && a[i - 1] === a[i]) {
i++;
continue;
}
res.push(a[i]);
i++;
}
// Add the remaining elements of b[]
while(j < m) {
// Skip duplicate elements in the second array
if(j > 0 && b[j - 1] === b[j]) {
j++;
continue;
}
res.push(b[j]);
j++;
}
return res;
}
let a = [1, 1, 2, 2, 2, 4];
let b = [2, 2, 4, 4];
console.log(findUnion(a, b).join(" "));
Time Complexity: O(n + m), Where n is the size of a[] and m is the size of b[]
Auxiliary Space: O(1)
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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
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
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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
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
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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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