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Partition an array into two subsets with equal count of unique elements

• Last Updated : 16 Aug, 2022

Given an array arr[] consisting of N integers, the task is to partition the array into two subsets such that the count of unique elements in both the subsets is the same and for each element, print 1 if that element belongs to the first subset. Otherwise, print 2. If it is not possible to do such a partition, then print “-1”.

Examples:

Input: arr[] = {1, 1, 2, 3, 4, 4}
Output: 1 1 1 2 1 1
Explanation: Consider the first and the second subset of the partition of the array as {1, 1, 2, 4 4} and {3}. Now, the above partition of subsets has equal count of distinct elements..

Input: arr[] = {1, 1, 2, 2, 3}
Output: -1

Naive Approach: The given problem can be solved by generating all possible partition of the array elements into two subsets and if there exists any such partition whose count of distinct elements are the same, then print 1 and 2 according to the respective set array elements belongs to. Otherwise, print “-1” as there doesn’t exist any such possible partition of the array.

Time Complexity: O(N*2N)
Auxiliary Space: O(N)

Efficient Approach: The above approach can also be optimized by finding the frequency of unique array elements if the frequency is odd, then there no such possible partition. Otherwise, print the partition of the subset accordingly. Follow the steps below to solve the problem:

• Initialize a Map, say M to store the frequency of all array elements.
• Initialize the array, say ans[] that stores the subset number where each array element belongs to.
• Find the count of the unique elements in the array by count the element having frequency 1 in the map M. Let this count be C.
• If the value of C is even, then move half of these elements into the first subset by marking those elements as 1 and rest all elements as 2 in the array ans[].
• Otherwise, check if there exist any element having a frequency greater than 2, then shift one instance of that element to the second subset. Otherwise, print “-1”.

Below is the implementation of the above approach:

C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to partition the array into` `// two subsets such that count of unique` `// elements in both subsets is the same` `void` `arrayPartition(``int` `a[], ``int` `n)` `{` `    ``// Stores the subset number for` `    ``// each array elements` `    ``int` `ans[n];`   `    ``// Stores the count of unique` `    ``// array elements` `    ``int` `cnt = 0;` `    ``int` `ind, flag = 0;`   `    ``// Stores the frequency of elements` `    ``map<``int``, ``int``> mp;`   `    ``// Traverse the array` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``mp[a[i]]++;` `    ``}`   `    ``// Count of elements having a` `    ``// frequency of 1` `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``if` `(mp[a[i]] == 1)` `            ``cnt++;`   `        ``// Check if there exists any` `        ``// element with frequency > 2` `        ``if` `(mp[a[i]] > 2 && flag == 0) {` `            ``flag = 1;` `            ``ind = i;` `        ``}` `    ``}`   `    ``// Count of elements needed to` `    ``// have frequency exactly 1 in` `    ``// each subset` `    ``int` `p = (cnt + 1) / 2;` `    ``int` `ans1 = 0;`   `    ``// Initialize all values in the` `    ``/// array ans[] as 1` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``ans[i] = 1;`   `    ``// Traverse the array ans[]` `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``// This array element is a` `        ``// part of first subset` `        ``if` `(mp[a[i]] == 1 && ans1 < p) {` `            ``ans[i] = 1;` `            ``ans1++;` `        ``}`   `        ``// Half array elements with` `        ``// frequency 1 are part of` `        ``// the second subset` `        ``else` `if` `(mp[a[i]] == 1) {` `            ``ans[i] = 2;` `        ``}` `    ``}`   `    ``// If count of elements is exactly` `    ``// 1 are odd and has no element` `    ``// with frequency > 2` `    ``if` `(cnt % 2 == 1 && flag == 0) {` `        ``cout << -1 << endl;` `        ``return``;` `    ``}`   `    ``// If count of elements that occurs` `    ``// exactly once are even` `    ``if` `(cnt % 2 == 0) {`   `        ``// Print the result` `        ``for` `(``int` `i = 0; i < n; i++) {` `            ``cout << ans[i] << ``" "``;` `        ``}` `    ``}`   `    ``// If the count of elements has` `    ``// exactly 1 frequency are odd` `    ``// and there is an element with` `    ``// frequency greater than 2` `    ``else` `{`   `        ``// Print the result` `        ``for` `(``int` `i = 0; i < n; i++) {` `            ``if` `(ind == i)` `                ``cout << 2 << ``" "``;` `            ``else` `                ``cout << ans[i] << ``" "``;` `        ``}` `    ``}` `}`   `// Driver Codea` `int` `main()` `{` `    ``int` `arr[] = { 1, 1, 2, 3, 4, 4 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``arrayPartition(arr, N);`   `    ``return` `0;` `}`

Java

 `// Java program for the above approach`   `import` `java.util.HashMap;`       `class` `GFG{`   `// Function to partition the array into` `// two subsets such that count of unique` `// elements in both subsets is the same` `public` `static` `void` `arrayPartition(``int` `a[], ``int` `n)` `{` `    ``// Stores the subset number for` `    ``// each array elements` `    ``int``[] ans = ``new` `int``[n];`   `    ``// Stores the count of unique` `    ``// array elements` `    ``int` `cnt = ``0``;` `    ``int` `ind = ``0``, flag = ``0``;`   `    ``// Stores the frequency of elements` `    ``HashMap mp = ``new` `HashMap();`   `    ``// Traverse the array` `    ``for` `(``int` `i = ``0``; i < n; i++) {` `        ``if``(mp.containsKey(a[i])){` `            ``mp.put(a[i], mp.get(a[i]) + ``1``);` `        ``}``else``{` `            ``mp.put(a[i], ``1``);` `        ``}` `    ``}`   `    ``// Count of elements having a` `    ``// frequency of 1` `    ``for` `(``int` `i = ``0``; i < n; i++) {`   `        ``if` `(mp.get(a[i]) == ``1``)` `            ``cnt++;`   `        ``// Check if there exists any` `        ``// element with frequency > 2` `        ``if` `(mp.get(a[i]) > ``2` `&& flag == ``0``) {` `            ``flag = ``1``;` `            ``ind = i;` `        ``}` `    ``}`   `    ``// Count of elements needed to` `    ``// have frequency exactly 1 in` `    ``// each subset` `    ``int` `p = (cnt + ``1``) / ``2``;` `    ``int` `ans1 = ``0``;`   `    ``// Initialize all values in the` `    ``/// array ans[] as 1` `    ``for` `(``int` `i = ``0``; i < n; i++)` `        ``ans[i] = ``1``;`   `    ``// Traverse the array ans[]` `    ``for` `(``int` `i = ``0``; i < n; i++) {`   `        ``// This array element is a` `        ``// part of first subset` `        ``if` `(mp.get(a[i]) == ``1` `&& ans1 < p) {` `            ``ans[i] = ``1``;` `            ``ans1++;` `        ``}`   `        ``// Half array elements with` `        ``// frequency 1 are part of` `        ``// the second subset` `        ``else` `if` `(mp.get(a[i]) == ``1``) {` `            ``ans[i] = ``2``;` `        ``}` `    ``}`   `    ``// If count of elements is exactly` `    ``// 1 are odd and has no element` `    ``// with frequency > 2` `    ``if` `(cnt % ``2` `== ``1` `&& flag == ``0``) {` `        ``System.out.println(-``1` `+ ``"\n"``);` `        ``return``;` `    ``}`   `    ``// If count of elements that occurs` `    ``// exactly once are even` `    ``if` `(cnt % ``2` `== ``0``) {`   `        ``// Print the result` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``System.out.print(ans[i] + ``" "``);` `        ``}` `    ``}`   `    ``// If the count of elements has` `    ``// exactly 1 frequency are odd` `    ``// and there is an element with` `    ``// frequency greater than 2` `    ``else` `{`   `        ``// Print the result` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``if` `(ind == i)` `            ``System.out.print(``2` `+ ``" "``);` `            ``else` `            ``System.out.print(ans[i] + ``" "``);` `        ``}` `    ``}` `}`   `// Driver Codea` `public` `static` `void` `main(String args[])` `{` `    ``int` `arr[] = { ``1``, ``1``, ``2``, ``3``, ``4``, ``4` `};` `    ``int` `N = arr.length;` `    ``arrayPartition(arr, N);`   `}` `}`   `// This code is contributed by gfgking.`

Python3

 `# Python3 program for the above approach`   `# Function to partition the array into` `# two subsets such that count of unique` `# elements in both subsets is the same` `def` `arrayPartition(a, n):` `    `  `    ``# Stores the subset number for` `    ``# each array elements` `    ``ans ``=` `[``0``] ``*` `n`   `    ``# Stores the count of unique` `    ``# array elements` `    ``cnt ``=` `0` `    ``ind, flag ``=` `0``, ``0`   `    ``# Stores the frequency of elements` `    ``mp ``=` `{}`   `    ``# Traverse the array` `    ``for` `i ``in` `a:` `        ``mp[i] ``=` `mp.get(i, ``0``) ``+` `1`   `    ``# Count of elements having a` `    ``# frequency of 1` `    ``for` `i ``in` `range``(n):` `        ``if` `((a[i] ``in` `mp) ``and` `mp[a[i]] ``=``=` `1``):` `            ``cnt ``+``=` `1`   `        ``# Check if there exists any` `        ``# element with frequency > 2` `        ``if` `(mp[a[i]] > ``2` `and` `flag ``=``=` `0``):` `            ``flag ``=` `1` `            ``ind ``=` `i`   `    ``# Count of elements needed to` `    ``# have frequency exactly 1 in` `    ``# each subset` `    ``p ``=` `(cnt ``+` `1``) ``/``/` `2` `    ``ans1 ``=` `0`   `    ``# Initialize all values in the` `    ``# array ans[] as 1` `    ``for` `i ``in` `range``(n):` `        ``ans[i] ``=` `1`   `    ``# Traverse the array ans[]` `    ``for` `i ``in` `range``(n):` `        `  `        ``# This array element is a` `        ``# part of first subset` `        ``if` `((a[i] ``in` `mp) ``and` `mp[a[i]] ``=``=` `1` `and` `            ``ans1 < p):` `            ``ans[i] ``=` `1` `            ``ans1 ``+``=` `1`   `        ``# Half array elements with` `        ``# frequency 1 are part of` `        ``# the second subset` `        ``elif` `((a[i] ``in` `mp) ``and` `mp[a[i]] ``=``=` `1``):` `            ``ans[i] ``=` `2`   `    ``# If count of elements is exactly` `    ``# 1 are odd and has no element` `    ``# with frequency > 2` `    ``if` `(cnt ``%` `2` `=``=` `1` `and` `flag ``=``=` `0``):` `        ``print` `(``-``1``)` `        ``return`   `    ``# If count of elements that occurs` `    ``# exactly once are even` `    ``if` `(cnt ``%` `2` `=``=` `0``):`   `        ``# Print the result` `        ``print``(``*``ans)`   `    ``# If the count of elements has` `    ``# exactly 1 frequency are odd` `    ``# and there is an element with` `    ``# frequency greater than 2` `    ``else``:` `        `  `        ``# Print the result` `        ``for` `i ``in` `range``(n):` `            ``if` `(ind ``=``=` `i):` `                ``print``(``2``, end ``=` `" "``)` `            ``else``:` `                ``print``(ans[i], end ``=` `" "``)`   `# Driver Codea` `if` `__name__ ``=``=` `'__main__'``:` `    `  `    ``arr ``=` `[ ``1``, ``1``, ``2``, ``3``, ``4``, ``4` `]` `    ``N ``=` `len``(arr)` `    `  `    ``arrayPartition(arr, N)`   `# This code is contributed by mohit kumar 29`

C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG{`   `// Function to partition the array into` `// two subsets such that count of unique` `// elements in both subsets is the same` `static` `void` `arrayPartition(``int` `[]a, ``int` `n)` `{` `    ``// Stores the subset number for` `    ``// each array elements` `    ``int` `[]ans = ``new` `int``[n];`   `    ``// Stores the count of unique` `    ``// array elements` `    ``int` `cnt = 0;` `    ``int` `ind=0, flag = 0;`   `    ``// Stores the frequency of elements` `    ``Dictionary<``int``,``int``> mp = ``new` `Dictionary<``int``,``int``>();`   `    ``// Traverse the array` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``if``(mp.ContainsKey(a[i]))` `         ``mp[a[i]]++;` `        ``else` `          ``mp.Add(a[i],1);` `    ``}`   `    ``// Count of elements having a` `    ``// frequency of 1` `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``if` `(mp.ContainsKey(a[i]) && mp[a[i]] == 1)` `            ``cnt++;`   `        ``// Check if there exists any` `        ``// element with frequency > 2` `        ``if` `(mp.ContainsKey(a[i]) && mp[a[i]] > 2 && flag == 0) {` `            ``flag = 1;` `            ``ind = i;` `        ``}` `    ``}`   `    ``// Count of elements needed to` `    ``// have frequency exactly 1 in` `    ``// each subset` `    ``int` `p = (cnt + 1) / 2;` `    ``int` `ans1 = 0;`   `    ``// Initialize all values in the` `    ``/// array ans[] as 1` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``ans[i] = 1;`   `    ``// Traverse the array ans[]` `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``// This array element is a` `        ``// part of first subset` `        ``if` `(mp.ContainsKey(a[i]) && mp[a[i]] == 1 && ans1 < p) {` `            ``ans[i] = 1;` `            ``ans1++;` `        ``}`   `        ``// Half array elements with` `        ``// frequency 1 are part of` `        ``// the second subset` `        ``else` `if` `(mp.ContainsKey(a[i]) && mp[a[i]] == 1) {` `            ``ans[i] = 2;` `        ``}` `    ``}`   `    ``// If count of elements is exactly` `    ``// 1 are odd and has no element` `    ``// with frequency > 2` `    ``if` `(cnt % 2 == 1 && flag == 0) {` `        ``Console.Write(-1);` `        ``return``;` `    ``}`   `    ``// If count of elements that occurs` `    ``// exactly once are even` `    ``if` `(cnt % 2 == 0) {`   `        ``// Print the result` `        ``for` `(``int` `i = 0; i < n; i++) {` `            ``Console.Write(ans[i] + ``" "``);` `        ``}` `    ``}`   `    ``// If the count of elements has` `    ``// exactly 1 frequency are odd` `    ``// and there is an element with` `    ``// frequency greater than 2` `    ``else` `{`   `        ``// Print the result` `        ``for` `(``int` `i = 0; i < n; i++) {` `            ``if` `(ind == i)` `                ``Console.Write(2 + ``" "``);` `            ``else` `                ``Console.Write(ans[i] + ``" "``);` `        ``}` `    ``}` `}`   `// Driver Codea` `public` `static` `void` `Main()` `{` `    ``int` `[]arr = { 1, 1, 2, 3, 4, 4 };` `    ``int` `N = arr.Length;` `    ``arrayPartition(arr, N);` `}` `}`   `// This code is contributed by SURENDRA_GANGWAR.`

Javascript

 ``

Output:

`1 1 1 2 1 1`

Time Complexity: O(NlogN)
Auxiliary Space: O(N) as using extra space for Map

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