Given two unsorted arrays that represent two sets (elements in every array are distinct), find union and intersection of two arrays.
For example, if the input arrays are:
arr1[] = {7, 1, 5, 2, 3, 6}
arr2[] = {3, 8, 6, 20, 7}
Then your program should print Union as {1, 2, 3, 5, 6, 7, 8, 20} and Intersection as {3, 6, 7}. Note that the elements of union and intersection can be printed in any order.
Method 1 (Naive)
Union:
- Initialize union U as empty.
- Copy all elements of first array to U.
- Do following for every element x of second array:
- If x is not present in first array, then copy x to U.
- Return U.
Intersection:
- Initialize intersection I as empty.
- Do following for every element x of first array
- If x is present in second array, then copy x to I.
- Return I.
Time complexity of this method is O(mn) for both operations. Here m and n are number of elements in arr1[] and arr2[] respectively.
Method 2 (Use Sorting)
- Sort arr1[] and arr2[]. This step takes O(mLogm + nLogn) time.
- Use O(m + n) algorithms to find union and intersection of two sorted arrays.
Overall time complexity of this method is O(mLogm + nLogn).
Method 3 (Use Sorting and Searching)
Union:
- Initialize union U as empty.
- Find smaller of m and n and sort the smaller array.
- Copy the smaller array to U.
- For every element x of larger array, do following
- Binary Search x in smaller array. If x is not present, then copy it to U.
- Return U.
Intersection:
- Initialize intersection I as empty.
- Find smaller of m and n and sort the smaller array.
- For every element x of larger array, do following
- Binary Search x in smaller array. If x is present, then copy it to I.
- Return I.
Time complexity of this method is min(mLogm + nLogm, mLogn + nLogn) which can also be written as O((m+n)Logm, (m+n)Logn). This approach works much better than the previous approach when difference between sizes of two arrays is significant.
Thanks to use_the_force for suggesting this method in a comment here.
Below is the implementation of this method.
C++
// A C++ program to print union and intersection /// of two unsorted arrays #include <algorithm> #include <iostream> using namespace std; int binarySearch( int arr[], int l, int r, int x); // Prints union of arr1[0..m-1] and arr2[0..n-1] void printUnion( int arr1[], int arr2[], int m, int n) { // Before finding union, make sure arr1[0..m-1] // is smaller if (m > n) { int * tempp = arr1; arr1 = arr2; arr2 = tempp; int temp = m; m = n; n = temp; } // Now arr1[] is smaller // Sort the first array and print its elements (these // two steps can be swapped as order in output is not // important) sort(arr1, arr1 + m); for ( int i = 0; i < m; i++) cout << arr1[i] << " " ; // Search every element of bigger array in smaller array // and print the element if not found for ( int i = 0; i < n; i++) if (binarySearch(arr1, 0, m - 1, arr2[i]) == -1) cout << arr2[i] << " " ; } // Prints intersection of arr1[0..m-1] and arr2[0..n-1] void printIntersection( int arr1[], int arr2[], int m, int n) { // Before finding intersection, make sure arr1[0..m-1] // is smaller if (m > n) { int * tempp = arr1; arr1 = arr2; arr2 = tempp; int temp = m; m = n; n = temp; } // Now arr1[] is smaller // Sort smaller array arr1[0..m-1] sort(arr1, arr1 + m); // Search every element of bigger array in smaller // array and print the element if found for ( int i = 0; i < n; i++) if (binarySearch(arr1, 0, m - 1, arr2[i]) != -1) cout << arr2[i] << " " ; } // A recursive binary search function. It returns // location of x in given array arr[l..r] is present, // otherwise -1 int binarySearch( int arr[], int l, int r, int x) { if (r >= l) { int mid = l + (r - l) / 2; // If the element is present at the middle itself if (arr[mid] == x) return mid; // If element is smaller than mid, then it can only // be presen in left subarray if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x); // Else the element can only be present in right // subarray return binarySearch(arr, mid + 1, r, x); } // We reach here when element is not present in array return -1; } /* Driver program to test above function */ int main() { int arr1[] = { 7, 1, 5, 2, 3, 6 }; int arr2[] = { 3, 8, 6, 20, 7 }; int m = sizeof (arr1) / sizeof (arr1[0]); int n = sizeof (arr2) / sizeof (arr2[0]); // Function call cout << "Union of two arrays is n" ; printUnion(arr1, arr2, m, n); cout << "nIntersection of two arrays is n" ; printIntersection(arr1, arr2, m, n); return 0; } |
Java
// A Java program to print union and intersection /// of two unsorted arrays import java.util.Arrays; class UnionAndIntersection { // Prints union of arr1[0..m-1] and arr2[0..n-1] void printUnion( int arr1[], int arr2[], int m, int n) { // Before finding union, make sure arr1[0..m-1] // is smaller if (m > n) { int tempp[] = arr1; arr1 = arr2; arr2 = tempp; int temp = m; m = n; n = temp; } // Now arr1[] is smaller // Sort the first array and print its elements // (these two steps can be swapped as order in // output is not important) Arrays.sort(arr1); for ( int i = 0 ; i < m; i++) System.out.print(arr1[i] + " " ); // Search every element of bigger array in smaller // array and print the element if not found for ( int i = 0 ; i < n; i++) { if (binarySearch(arr1, 0 , m - 1 , arr2[i]) == - 1 ) System.out.print(arr2[i] + " " ); } } // Prints intersection of arr1[0..m-1] and arr2[0..n-1] void printIntersection( int arr1[], int arr2[], int m, int n) { // Before finding intersection, make sure // arr1[0..m-1] is smaller if (m > n) { int tempp[] = arr1; arr1 = arr2; arr2 = tempp; int temp = m; m = n; n = temp; } // Now arr1[] is smaller // Sort smaller array arr1[0..m-1] Arrays.sort(arr1); // Search every element of bigger array in smaller // array and print the element if found for ( int i = 0 ; i < n; i++) { if (binarySearch(arr1, 0 , m - 1 , arr2[i]) != - 1 ) System.out.print(arr2[i] + " " ); } } // A recursive binary search function. It returns // location of x in given array arr[l..r] is present, // otherwise -1 int binarySearch( int arr[], int l, int r, int x) { if (r >= l) { int mid = l + (r - l) / 2 ; // If the element is present at the middle // itself if (arr[mid] == x) return mid; // If element is smaller than mid, then it can // only be present in left subarray if (arr[mid] > x) return binarySearch(arr, l, mid - 1 , x); // Else the element can only be present in right // subarray return binarySearch(arr, mid + 1 , r, x); } // We reach here when element is not present in // array return - 1 ; } // Driver code public static void main(String[] args) { UnionAndIntersection u_i = new UnionAndIntersection(); int arr1[] = { 7 , 1 , 5 , 2 , 3 , 6 }; int arr2[] = { 3 , 8 , 6 , 20 , 7 }; int m = arr1.length; int n = arr2.length; // Function call System.out.println( "Union of two arrays is " ); u_i.printUnion(arr1, arr2, m, n); System.out.println( "" ); System.out.println( "Intersection of two arrays is " ); u_i.printIntersection(arr1, arr2, m, n); } } |
Python3
# A Python3 program to print union and intersection # of two unsorted arrays # Prints union of arr1[0..m-1] and arr2[0..n-1] def printUnion(arr1, arr2, m, n): # Before finding union, make sure arr1[0..m-1] # is smaller if (m > n): tempp = arr1 arr1 = arr2 arr2 = tempp temp = m m = n n = temp # Now arr1[] is smaller # Sort the first array and print its elements (these two # steps can be swapped as order in output is not important) arr1.sort() for i in range ( 0 , m): print (arr1[i], end = " " ) # Search every element of bigger array in smaller array # and print the element if not found for i in range ( 0 , n): if (binarySearch(arr1, 0 , m - 1 , arr2[i]) = = - 1 ): print (arr2[i], end = " " ) # Prints intersection of arr1[0..m-1] and arr2[0..n-1] def printIntersection(arr1, arr2, m, n): # Before finding intersection, make sure arr1[0..m-1] # is smaller if (m > n): tempp = arr1 arr1 = arr2 arr2 = tempp temp = m m = n n = temp # Now arr1[] is smaller # Sort smaller array arr1[0..m-1] arr1.sort() # Search every element of bigger array in smaller # array and print the element if found for i in range ( 0 , n): if (binarySearch(arr1, 0 , m - 1 , arr2[i]) ! = - 1 ): print (arr2[i], end = " " ) # A recursive binary search function. It returns # location of x in given array arr[l..r] is present, # otherwise -1 def binarySearch(arr, l, r, x): if (r > = l): mid = int (l + (r - l) / 2 ) # If the element is present at the middle itself if (arr[mid] = = x): return mid # If element is smaller than mid, then it can only # be presen in left subarray if (arr[mid] > x): return binarySearch(arr, l, mid - 1 , x) # Else the element can only be present in right subarray return binarySearch(arr, mid + 1 , r, x) # We reach here when element is not present in array return - 1 # Driver code arr1 = [ 7 , 1 , 5 , 2 , 3 , 6 ] arr2 = [ 3 , 8 , 6 , 20 , 7 ] m = len (arr1) n = len (arr2) # Function call print ( "Union of two arrays is " ) printUnion(arr1, arr2, m, n) print ( "\nIntersection of two arrays is " ) printIntersection(arr1, arr2, m, n) # This code is contributed by mits |
C#
// A C# program to print union and // intersection of two unsorted arrays using System; class GFG { // Prints union of arr1[0..m-1] and arr2[0..n-1] static void printUnion( int [] arr1, int [] arr2, int m, int n) { // Before finding union, make // sure arr1[0..m-1] is smaller if (m > n) { int [] tempp = arr1; arr1 = arr2; arr2 = tempp; int temp = m; m = n; n = temp; } // Now arr1[] is smaller // Sort the first array and print // its elements (these two steps can // be swapped as order in output is // not important) Array.Sort(arr1); for ( int i = 0; i < m; i++) Console.Write(arr1[i] + " " ); // Search every element of bigger // array in smaller array and print // the element if not found for ( int i = 0; i < n; i++) { if (binarySearch(arr1, 0, m - 1, arr2[i]) == -1) Console.Write(arr2[i] + " " ); } } // Prints intersection of arr1[0..m-1] // and arr2[0..n-1] static void printIntersection( int [] arr1, int [] arr2, int m, int n) { // Before finding intersection, // make sure arr1[0..m-1] is smaller if (m > n) { int [] tempp = arr1; arr1 = arr2; arr2 = tempp; int temp = m; m = n; n = temp; } // Now arr1[] is smaller // Sort smaller array arr1[0..m-1] Array.Sort(arr1); // Search every element of bigger array in // smaller array and print the element if found for ( int i = 0; i < n; i++) { if (binarySearch(arr1, 0, m - 1, arr2[i]) != -1) Console.Write(arr2[i] + " " ); } } // A recursive binary search function. // It returns location of x in given // array arr[l..r] is present, otherwise -1 static int binarySearch( int [] arr, int l, int r, int x) { if (r >= l) { int mid = l + (r - l) / 2; // If the element is present at // the middle itself if (arr[mid] == x) return mid; // If element is smaller than mid, then it // can only be present in left subarray if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x); // Else the element can only be // present in right subarray return binarySearch(arr, mid + 1, r, x); } // We reach here when element is // not present in array return -1; } // Driver Code static public void Main() { int [] arr1 = { 7, 1, 5, 2, 3, 6 }; int [] arr2 = { 3, 8, 6, 20, 7 }; int m = arr1.Length; int n = arr2.Length; // Function call Console.WriteLine( "Union of two arrays is " ); printUnion(arr1, arr2, m, n); Console.WriteLine( "" ); Console.WriteLine( "Intersection of two arrays is " ); printIntersection(arr1, arr2, m, n); } } // This code is contributed // by Sach_Code |
PHP
<?php // A PHP program to print union and intersection /// of two unsorted arrays // Prints union of arr1[0..m-1] and arr2[0..n-1] function printUnion( $arr1 , $arr2 , $m , $n ) { // Before finding union, make sure arr1[0..m-1] // is smaller if ( $m > $n ) { $tempp = $arr1 ; $arr1 = $arr2 ; $arr2 = $tempp ; $temp = $m ; $m = $n ; $n = $temp ; } // Now arr1[] is smaller // Sort the first array and print its elements (these two // steps can be swapped as order in output is not important) sort( $arr1 ); for ( $i = 0; $i < $m ; $i ++) echo $arr1 [ $i ]. " " ; // Search every element of bigger array in smaller array // and print the element if not found for ( $i = 0; $i < $n ; $i ++) if (binarySearch( $arr1 , 0, $m - 1, $arr2 [ $i ]) == -1) echo $arr2 [ $i ]. " " ; } // Prints intersection of arr1[0..m-1] and arr2[0..n-1] function printIntersection( $arr1 , $arr2 , $m , $n ) { // Before finding intersection, make sure arr1[0..m-1] // is smaller if ( $m > $n ) { $tempp = $arr1 ; $arr1 = $arr2 ; $arr2 = $tempp ; $temp = $m ; $m = $n ; $n = $temp ; } // Now arr1[] is smaller // Sort smaller array arr1[0..m-1] sort( $arr1 ); // Search every element of bigger array in smaller // array and print the element if found for ( $i = 0; $i < $n ; $i ++) if (binarySearch( $arr1 , 0, $m - 1, $arr2 [ $i ]) != -1) echo $arr2 [ $i ]. " " ; } // A recursive binary search function. It returns // location of x in given array arr[l..r] is present, // otherwise -1 function binarySearch( $arr , $l , $r , $x ) { if ( $r >= $l ) { $mid = (int)( $l + ( $r - $l )/2); // If the element is present at the middle itself if ( $arr [ $mid ] == $x ) return $mid ; // If element is smaller than mid, then it can only // be presen in left subarray if ( $arr [ $mid ] > $x ) return binarySearch( $arr , $l , $mid - 1, $x ); // Else the element can only be present in right subarray return binarySearch( $arr , $mid + 1, $r , $x ); } // We reach here when element is not present in array return -1; } /* Driver program to test above function */ $arr1 = array (7, 1, 5, 2, 3, 6); $arr2 = array (3, 8, 6, 20, 7); $m = count ( $arr1 ); $n = count ( $arr2 ); echo "Union of two arrays is \n" ; printUnion( $arr1 , $arr2 , $m , $n ); echo "\nIntersection of two arrays is \n" ; printIntersection( $arr1 , $arr2 , $m , $n ); // This code is contributed by mits ?> |
Union of two arrays is n3 6 7 8 20 1 5 2 nIntersection of two arrays is n7 3 6
Another Approach (When elements in the array may not be distinct) :
C++
// C code to find intersection when // elements may not be distinct #include <bits/stdc++.h> using namespace std; // Function to find intersection void intersection( int a[], int b[], int n, int m) { int i = 0, j = 0; while (i < n && j < m) { if (a[i] > b[j]) { j++; } else if (b[j] > a[i]) { i++; } else { // when both are equal cout << a[i] << " " ; i++; j++; } } } // Driver Code int main() { int a[] = { 1, 3, 2, 3, 3, 4, 5, 5, 6 }; int b[] = { 3, 3, 5 }; int n = sizeof (a) / sizeof (a[0]); int m = sizeof (b) / sizeof (b[0]); // sort sort(a, a + n); sort(b, b + m); // Function call intersection(a, b, n, m); } |
Java
// Java code to find intersection when // elements may not be distinct import java.io.*; import java.util.Arrays; class GFG { // Function to find intersection static void intersection( int a[], int b[], int n, int m) { int i = 0 , j = 0 ; while (i < n && j < m) { if (a[i] > b[j]) { j++; } else if (b[j] > a[i]) { i++; } else { // when both are equal System.out.print(a[i] + " " ); i++; j++; } } } // Driver Code public static void main(String[] args) { int a[] = { 1 , 3 , 2 , 3 , 4 , 5 , 5 , 6 }; int b[] = { 3 , 3 , 5 }; int n = a.length; int m = b.length; // sort Arrays.sort(a); Arrays.sort(b); // Function call intersection(a, b, n, m); } } |
Python 3
# Python 3 code to find intersection # when elements may not be distinct # Function to find intersection def intersection(a, b, n, m): i = 0 j = 0 while (i < n and j < m): if (a[i] > b[j]): j + = 1 else : if (b[j] > a[i]): i + = 1 else : # when both are equal print (a[i], end = " " ) i + = 1 j + = 1 # Driver Code if __name__ = = "__main__" : a = [ 1 , 3 , 2 , 3 , 4 , 5 , 5 , 6 ] b = [ 3 , 3 , 5 ] n = len (a) m = len (b) # sort a.sort() b.sort() # function call intersection(a, b, n, m) # This code is contributed by Ita_c |
C#
// C# code to find intersection when // elements may not be distinct using System; class GFG { // Function to find intersection static void intersection( int [] a, int [] b, int n, int m) { int i = 0, j = 0; while (i < n && j < m) { if (a[i] > b[j]) { j++; } else if (b[j] > a[i]) { i++; } else { // when both are equal Console.Write(a[i] + " " ); i++; j++; } } } // Driver Code public static void Main() { int [] a = { 1, 3, 2, 3, 4, 5, 5, 6 }; int [] b = { 3, 3, 5 }; int n = a.Length; int m = b.Length; // sort Array.Sort(a); Array.Sort(b); // Function call intersection(a, b, n, m); } } // this code is contributed by mukul singh |
PHP
<?php // PHP code to find intersection when // elements may not be distinct // Function to find intersection function intersection( $a , $b , $n , $m ) { $i = 0; $j = 0; while ( $i < $n && $j < $m ) { if ( $a [ $i ] > $b [ $j ]) { $j ++; } else if ( $b [ $j ] > $a [ $i ]) { $i ++; } else { // when both are equal echo ( $a [ $i ] . " " ); $i ++; $j ++; } } } // Driver Code $a = array (1, 3, 2, 3, 4, 5, 5, 6); $b = array (3, 3, 5); $n = sizeof( $a ); $m = sizeof( $b ); // sort sort( $a ); sort( $b ); // Function call intersection( $a , $b , $n , $m ); // This code is contributed // by Mukul Singh ?> |
3 3 5
Thanks Sanny Kumar for suggesting the above method.
Method 4 (Use Hashing)
Union
- Initialize an empty hash set hs.
- Iterate through the first array and put every element of the first array in the set S.
- Repeat the process for the second array.
- Print the set hs.
Intersection
- Initialize an empty set hs.
- Iterate through the first array and put every element of the first array in the set S.
- For every element x of the second array, do the following :
Search x in the set hs. If x is present, then print it. Time complexity of this method is ?(m+n) under the assumption that hash table search and insert operations take ?(1) time.
Below is the implementation of the above idea:
C++
// CPP program to find union and intersection // using sets #include <bits/stdc++.h> using namespace std; // Prints union of arr1[0..n1-1] and arr2[0..n2-1] void printUnion( int arr1[], int arr2[], int n1, int n2) { set< int > hs; // Inhsert the elements of arr1[] to set hs for ( int i = 0; i < n1; i++) hs.insert(arr1[i]); // Insert the elements of arr2[] to set hs for ( int i = 0; i < n2; i++) hs.insert(arr2[i]); // Print the content of set hs for ( auto it = hs.begin(); it != hs.end(); it++) cout << *it << " " ; cout << endl; } // Prints intersection of arr1[0..n1-1] and // arr2[0..n2-1] void printIntersection( int arr1[], int arr2[], int n1, int n2) { set< int > hs; // Insert the elements of arr1[] to set S for ( int i = 0; i < n1; i++) hs.insert(arr1[i]); for ( int i = 0; i < n2; i++) // If element is present in set then // push it to vector V if (hs.find(arr2[i]) != hs.end()) cout << arr2[i] << " " ; } // Driver Program int main() { int arr1[] = { 7, 1, 5, 2, 3, 6 }; int arr2[] = { 3, 8, 6, 20, 7 }; int n1 = sizeof (arr1) / sizeof (arr1[0]); int n2 = sizeof (arr2) / sizeof (arr2[0]); // Function call printUnion(arr1, arr2, n1, n2); printIntersection(arr1, arr2, n1, n2); return 0; } |
Java
// Java program to find union and intersection // using Hashing import java.util.HashSet; class Test { // Prints union of arr1[0..m-1] and arr2[0..n-1] static void printUnion( int arr1[], int arr2[]) { HashSet<Integer> hs = new HashSet<>(); for ( int i = 0 ; i < arr1.length; i++) hs.add(arr1[i]); for ( int i = 0 ; i < arr2.length; i++) hs.add(arr2[i]); System.out.println(hs); } // Prints intersection of arr1[0..m-1] and arr2[0..n-1] static void printIntersection( int arr1[], int arr2[]) { HashSet<Integer> hs = new HashSet<>(); HashSet<Integer> hs1 = new HashSet<>(); for ( int i = 0 ; i < arr1.length; i++) hs.add(arr1[i]); for ( int i = 0 ; i < arr2.length; i++) if (hs.contains(arr2[i])) System.out.print(arr2[i] + " " ); } // Driver code public static void main(String[] args) { int arr1[] = { 7 , 1 , 5 , 2 , 3 , 6 }; int arr2[] = { 3 , 8 , 6 , 20 , 7 }; // Function call System.out.println( "Union of two arrays is : " ); printUnion(arr1, arr2); System.out.println( "Intersection of two arrays is : " ); printIntersection(arr1, arr2); } } |
Python
# Python program to find union and intersection # using sets def printUnion(arr1, arr2, n1, n2): hs = set () # Inhsert the elements of arr1[] to set hs for i in range ( 0 , n1): hs.add(arr1[i]) # Inhsert the elements of arr1[] to set hs for i in range ( 0 , n2): hs.add(arr2[i]) print ( "Union:" ) for i in hs: print (i, end = " " ) print ( "\n" ) # Prints intersection of arr1[0..n1-1] and # arr2[0..n2-1] def printIntersection(arr1, arr2, n1, n2): hs = set () # Insert the elements of arr1[] to set S for i in range ( 0 , n1): hs.add(arr1[i]) print ( "Intersection:" ) for i in range ( 0 , n2): # If element is present in set then # push it to vector V if arr2[i] in hs: print (arr2[i], end = " " ) # Driver Program arr1 = [ 7 , 1 , 5 , 2 , 3 , 6 ] arr2 = [ 3 , 8 , 6 , 20 , 7 ] n1 = len (arr1) n2 = len (arr2) # Function call printUnion(arr1, arr2, n1, n2) printIntersection(arr1, arr2, n1, n2) # This artice is contributed by Kumar Suman . |
C#
// C# program to find union and intersection // using Hashing using System; using System.Linq; using System.Collections.Generic; class GFG { // Prints union of arr1[0..m-1] and arr2[0..n-1] static void printUnion( int []arr1, int []arr2) { HashSet< int > hs = new HashSet< int >(); for ( int i = 0; i < arr1.Length; i++) hs.Add(arr1[i]); for ( int i = 0; i < arr2.Length; i++) hs.Add(arr2[i]); Console.WriteLine( string .Join( ", " , hs)); } // Prints intersection of arr1[0..m-1] and arr2[0..n-1] static void printIntersection( int []arr1, int []arr2) { HashSet< int > hs = new HashSet< int >(); for ( int i = 0; i < arr1.Length; i++) hs.Add(arr1[i]); for ( int i = 0; i < arr2.Length; i++) if (hs.Contains(arr2[i])) Console.Write(arr2[i] + " " ); } // Driver Code static void Main() { int []arr1 = {7, 1, 5, 2, 3, 6}; int []arr2 = {3, 8, 6, 20, 7}; Console.WriteLine( "Union of two arrays is : " ); printUnion(arr1, arr2); Console.WriteLine( "\nIntersection of two arrays is : " ); printIntersection(arr1, arr2); } } // This code is contributed by mits |
1 2 3 5 6 7 8 20 3 6 7
This method is contributed by Ankur Singh.
Method 5 (Kind of hashing technique without using any predefined Java Collections)
- Intialize array with size of m+n
- Fill first array value in resultant array by doing hashing(to find appropriate position)
- Repeat for second array
- While doing hashing if collision happens increment the position in recursive way
Below is the implementation of the above code:
Java
// Java program to find union and intersection // using similar Hashing Technique // without using any predefined Java Collections // Time Complexity best case & avg case = O(m+n) // Worst case = O(nlogn) // package com.arrays.math; public class UnsortedIntersectionUnion { // Prints intersection of arr1[0..n1-1] and // arr2[0..n2-1] public void findPosition( int a[], int b[]) { int v = (a.length + b.length); int ans[] = new int [v]; int zero1 = 0 ; int zero2 = 0 ; System.out.print( "Intersection : " ); // Iterate first array for ( int i = 0 ; i < a.length; i++) zero1 = iterateArray(a, v, ans, i); // Iterate second array for ( int j = 0 ; j < b.length; j++) zero2 = iterateArray(b, v, ans, j); int zero = zero1 + zero2; placeZeros(v, ans, zero); printUnion(v, ans, zero); } // Prints union of arr1[0..n1-1] and arr2[0..n2-1] private void printUnion( int v, int [] ans, int zero) { int zero1 = 0 ; System.out.print( "\nUnion : " ); for ( int i = 0 ; i < v; i++) { if ((zero == 0 && ans[i] == 0 ) || (ans[i] == 0 && zero1 > 0 )) continue ; if (ans[i] == 0 ) zero1++; System.out.print(ans[i] + "," ); } } private void placeZeros( int v, int [] ans, int zero) { if (zero == 2 ) { System.out.println( "0" ); int d[] = { 0 }; placeValue(d, ans, 0 , 0 , v); } if (zero == 1 ) { int d[] = { 0 }; placeValue(d, ans, 0 , 0 , v); } } // Function to itreate array private int iterateArray( int [] a, int v, int [] ans, int i) { if (a[i] != 0 ) { int p = a[i] % v; placeValue(a, ans, i, p, v); } else return 1 ; return 0 ; } private void placeValue( int [] a, int [] ans, int i, int p, int v) { p = p % v; if (ans[p] == 0 ) ans[p] = a[i]; else { if (ans[p] == a[i]) System.out.print(a[i] + "," ); else { // Hashing collision happened increment // position and do recursive call p = p + 1 ; placeValue(a, ans, i, p, v); } } } // Driver code public static void main(String args[]) { int a[] = { 7 , 1 , 5 , 2 , 3 , 6 }; int b[] = { 3 , 8 , 6 , 20 , 7 }; // Function call UnsortedIntersectionUnion uiu = new UnsortedIntersectionUnion(); uiu.findPosition(a, b); } } // This code is contributed by Mohanakrishnan S. |
Python3
# Python3 program to find union and intersection # using similar Hashing Technique # without using any predefined Java Collections # Time Complexity best case & avg case = O(m+n) # Worst case = O(nlogn) # Prints intersection of arr1[0..n1-1] and # arr2[0..n2-1] def findPosition(a, b): v = len (a) + len (b); ans = [ 0 ] * v; zero1 = zero2 = 0 ; print ( "Intersection :" ,end = " " ); # Iterate first array for i in range ( len (a)): zero1 = iterateArray(a, v, ans, i); # Iterate second array for j in range ( len (b)): zero2 = iterateArray(b, v, ans, j); zero = zero1 + zero2; placeZeros(v, ans, zero); printUnion(v, ans, zero); # Prints union of arr1[0..n1-1] and arr2[0..n2-1] def printUnion(v, ans,zero): zero1 = 0 ; print ( "\nUnion :" ,end = " " ); for i in range (v): if ((zero = = 0 and ans[i] = = 0 ) or (ans[i] = = 0 and zero1 > 0 )): continue ; if (ans[i] = = 0 ): zero1 + = 1 ; print (ans[i],end = "," ); def placeZeros(v, ans, zero): if (zero = = 2 ): print ( "0" ); d = [ 0 ]; placeValue(d, ans, 0 , 0 , v); if (zero = = 1 ): d = [ 0 ]; placeValue(d, ans, 0 , 0 , v); # Function to itreate array def iterateArray(a,v,ans,i): if (a[i] ! = 0 ): p = a[i] % v; placeValue(a, ans, i, p, v); else : return 1 ; return 0 ; def placeValue(a,ans,i,p,v): p = p % v; if (ans[p] = = 0 ): ans[p] = a[i]; else : if (ans[p] = = a[i]): print (a[i],end = "," ); else : # Hashing collision happened increment # position and do recursive call p = p + 1 ; placeValue(a, ans, i, p, v); # Driver code a = [ 7 , 1 , 5 , 2 , 3 , 6 ]; b = [ 3 , 8 , 6 , 20 , 7 ]; findPosition(a, b); # This code is contributed by mits |
C#
// C# program to find union and intersection // using similar Hashing Technique // without using any predefined Java Collections // Time Complexity best case & avg case = O(m+n) // Worst case = O(nlogn) //package com.arrays.math; using System; class UnsortedIntersectionUnion { // Prints intersection of arr1[0..n1-1] and // arr2[0..n2-1] public void findPosition( int []a, int []b) { int v = (a.Length + b.Length); int []ans = new int [v]; int zero1 = 0; int zero2 = 0; Console.Write( "Intersection : " ); // Iterate first array for ( int i = 0; i < a.Length; i++) zero1 = iterateArray(a, v, ans, i); // Iterate second array for ( int j = 0; j < b.Length; j++) zero2 = iterateArray(b, v, ans, j); int zero = zero1 + zero2; placeZeros(v, ans, zero); printUnion(v, ans, zero); } // Prints union of arr1[0..n1-1] // and arr2[0..n2-1] private void printUnion( int v, int [] ans, int zero) { int zero1 = 0; Console.Write( "\nUnion : " ); for ( int i = 0; i < v; i++) { if ((zero == 0 && ans[i] == 0) || (ans[i] == 0 && zero1 > 0)) continue ; if (ans[i] == 0) zero1++; Console.Write(ans[i] + "," ); } } private void placeZeros( int v, int [] ans, int zero) { if (zero == 2) { Console.WriteLine( "0" ); int []d = { 0 }; placeValue(d, ans, 0, 0, v); } if (zero == 1) { int []d = { 0 }; placeValue(d, ans, 0, 0, v); } } // Function to itreate array private int iterateArray( int [] a, int v, int [] ans, int i) { if (a[i] != 0) { int p = a[i] % v; placeValue(a, ans, i, p, v); } else return 1; return 0; } private void placeValue( int [] a, int [] ans, int i, int p, int v) { p = p % v; if (ans[p] == 0) ans[p] = a[i]; else { if (ans[p] == a[i]) Console.Write(a[i] + "," ); else { //Hashing collision happened increment // position and do recursive call p = p + 1; placeValue(a, ans, i, p, v); } } } // Driver code public static void Main() { int []a = { 7, 1, 5, 2, 3, 6 }; int []b = { 3, 8, 6, 20, 7 }; UnsortedIntersectionUnion uiu = new UnsortedIntersectionUnion(); uiu.findPosition(a, b); } } // This code is contributed by PrinciRaj1992 |
Intersection : 3,6,7, Union : 1,2,3,5,6,7,8,20,
Python3
# Python program to find union and intersection # using sets def printUnion(arr1, arr2, n1, n2): hs = set () # Inhsert the elements of arr1[] to set hs for i in range ( 0 , n1): hs.add(arr1[i]) # Inhsert the elements of arr1[] to set hs for i in range ( 0 , n2): hs.add(arr2[i]) print ( "Union:" ) for i in hs: print (i, end = " " ) print ( "\n" ) # Prints intersection of arr1[0..n1-1] and # arr2[0..n2-1] def printIntersection(arr1, arr2, n1, n2): hs = set () # Insert the elements of arr1[] to set S for i in range ( 0 , n1): hs.add(arr1[i]) print ( "Intersection:" ) for i in range ( 0 , n2): # If element is present in set then # push it to vector V if arr2[i] in hs: print (arr2[i], end = " " ) # Driver Program arr1 = [ 7 , 1 , 5 , 2 , 3 , 6 ] arr2 = [ 3 , 8 , 6 , 20 , 7 ] n1 = len (arr1) n2 = len (arr2) # Function call printUnion(arr1, arr2, n1, n2) printIntersection(arr1, arr2, n1, n2) # This artice is contributed by Kumar Suman . |
See following post for sorted arrays.
Find Union and Intersection of two sorted arrays
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