Partition of a set into K subsets with equal sum

Given an integer array of N elements, the task is to divide this array into K non-empty subsets such that the sum of elements in every subset is same. All elements of this array should be part of exactly one partition. 
Examples:

Input : arr = [2, 1, 4, 5, 6], K = 3
Output : Yes
we can divide above array into 3 parts with equal
sum as [[2, 4], [1, 5], [6]]

Input  : arr = [2, 1, 5, 5, 6], K = 3
Output : No
It is not possible to divide above array into 3
parts with equal sum

We can solve this problem recursively, we keep an array for sum of each partition and a boolean array to check whether an element is already taken into some partition or not. 
First we need to check some base cases, 
If K is 1, then we already have our answer, complete array is only subset with same sum. 
If N < K, then it is not possible to divide array into subsets with equal sum, because we can’t divide the array into more than N parts. 
If sum of array is not divisible by K, then it is not possible to divide the array. We will proceed only if k divides sum. Our goal reduces to divide array into K parts where sum of each part should be array_sum/K 
In below code a recursive method is written which tries to add array element into some subset. If sum of this subset reaches required sum, we iterate for next part recursively, otherwise we backtrack for different set of elements. If number of subsets whose sum reaches the required sum is (K-1), we flag that it is possible to partition array into K parts with equal sum, because remaining elements already have a sum equal to required sum. 

C++

// C++ program to check whether an array can be
// partitioned into K subsets of equal sum
#include <bits/stdc++.h>
using namespace std;

// Recursive Utility method to check K equal sum
// subsetition of array
/**
    array           - given input array
    subsetSum array   - sum to store each subset of the array
    taken           - boolean array to check whether element
                      is taken into sum partition or not
    K               - number of partitions needed
    N               - total number of element in array
    curIdx          - current subsetSum index
    limitIdx        - lastIdx from where array element should
                      be taken */
bool isKPartitionPossibleRec(int arr[], int subsetSum[], bool taken[],
                   int subset, int K, int N, int curIdx, int limitIdx)
{
    if (subsetSum[curIdx] == subset)
    {
        /*  current index (K - 2) represents (K - 1) subsets of equal
            sum last partition will already remain with sum 'subset'*/
        if (curIdx == K - 2)
            return true;

        //  recursive call for next subsetition
        return isKPartitionPossibleRec(arr, subsetSum, taken, subset,
                                            K, N, curIdx + 1, N - 1);
    }

    //  start from limitIdx and include elements into current partition
    for (int i = limitIdx; i >= 0; i--)
    {
        //  if already taken, continue
        if (taken[i])
            continue;
        int tmp = subsetSum[curIdx] + arr[i];

        // if temp is less than subset then only include the element
        // and call recursively
        if (tmp <= subset)
        {
            //  mark the element and include into current partition sum
            taken[i] = true;
            subsetSum[curIdx] += arr[i];
            bool nxt = isKPartitionPossibleRec(arr, subsetSum, taken,
                                            subset, K, N, curIdx, i - 1);

            // after recursive call unmark the element and remove from
            // subsetition sum
            taken[i] = false;
            subsetSum[curIdx] -= arr[i];
            if (nxt)
                return true;
        }
    }
    return false;
}

//  Method returns true if arr can be partitioned into K subsets
// with equal sum
bool isKPartitionPossible(int arr[], int N, int K)
{
    //  If K is 1, then complete array will be our answer
    if (K == 1)
        return true;

    //  If total number of partitions are more than N, then
    // division is not possible
    if (N < K)
        return false;

    // if array sum is not divisible by K then we can't divide
    // array into K partitions
    int sum = 0;
    for (int i = 0; i < N; i++)
        sum += arr[i];
    if (sum % K != 0)
        return false;

    //  the sum of each subset should be subset (= sum / K)
    int subset = sum / K;
    int subsetSum[K];
    bool taken[N];

    //  Initialize sum of each subset from 0
    for (int i = 0; i < K; i++)
        subsetSum[i] = 0;

    //  mark all elements as not taken
    for (int i = 0; i < N; i++)
        taken[i] = false;

    // initialize first subset sum as last element of
    // array and mark that as taken
    subsetSum[0] = arr[N - 1];
    taken[N - 1] = true;

    //  call recursive method to check K-substitution condition
    return isKPartitionPossibleRec(arr, subsetSum, taken,
                                     subset, K, N, 0, N - 1);
}

//  Driver code to test above methods
int main()
{
    int arr[] = {2, 1, 4, 5, 3, 3};
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 3;

    if (isKPartitionPossible(arr, N, K))
        cout << "Partitions into equal sum is possible.\n";
    else
        cout << "Partitions into equal sum is not possible.\n";
}

Java

// Java program to check whether an array can be
// partitioned into K subsets of equal sum
class GFG
{

// Recursive Utility method to check K equal sum
// subsetition of array
/**
    array         - given input array
    subsetSum array - sum to store each subset of the array
    taken         - boolean array to check whether element
                    is taken into sum partition or not
    K             - number of partitions needed
    N             - total number of element in array
    curIdx         - current subsetSum index
    limitIdx     - lastIdx from where array element should
                    be taken */
static boolean isKPartitionPossibleRec(int arr[], int subsetSum[], boolean taken[],
                int subset, int K, int N, int curIdx, int limitIdx)
{
    if (subsetSum[curIdx] == subset)
    {
        /* current index (K - 2) represents (K - 1) subsets of equal
            sum last partition will already remain with sum 'subset'*/
        if (curIdx == K - 2)
            return true;

        // recursive call for next subsetition
        return isKPartitionPossibleRec(arr, subsetSum, taken, subset,
                                            K, N, curIdx + 1, N - 1);
    }

    // start from limitIdx and include elements into current partition
    for (int i = limitIdx; i >= 0; i--)
    {
        // if already taken, continue
        if (taken[i])
            continue;
        int tmp = subsetSum[curIdx] + arr[i];

        // if temp is less than subset then only include the element
        // and call recursively
        if (tmp <= subset)
        {
            // mark the element and include into current partition sum
            taken[i] = true;
            subsetSum[curIdx] += arr[i];
            boolean nxt = isKPartitionPossibleRec(arr, subsetSum, taken,
                                            subset, K, N, curIdx, i - 1);

            // after recursive call unmark the element and remove from
            // subsetition sum
            taken[i] = false;
            subsetSum[curIdx] -= arr[i];
            if (nxt)
                return true;
        }
    }
    return false;
}

// Method returns true if arr can be partitioned into K subsets
// with equal sum
static boolean isKPartitionPossible(int arr[], int N, int K)
{
    // If K is 1, then complete array will be our answer
    if (K == 1)
        return true;

    // If total number of partitions are more than N, then
    // division is not possible
    if (N < K)
        return false;

    // if array sum is not divisible by K then we can't divide
    // array into K partitions
    int sum = 0;
    for (int i = 0; i < N; i++)
        sum += arr[i];
    if (sum % K != 0)
        return false;

    // the sum of each subset should be subset (= sum / K)
    int subset = sum / K;
    int []subsetSum = new int[K];
    boolean []taken = new boolean[N];

    // Initialize sum of each subset from 0
    for (int i = 0; i < K; i++)
        subsetSum[i] = 0;

    // mark all elements as not taken
    for (int i = 0; i < N; i++)
        taken[i] = false;

    // initialize first subset sum as last element of
    // array and mark that as taken
    subsetSum[0] = arr[N - 1];
    taken[N - 1] = true;

    // call recursive method to check K-substitution condition
    return isKPartitionPossibleRec(arr, subsetSum, taken,
                                    subset, K, N, 0, N - 1);
}

// Driver code
public static void main(String[] args)
{
    int arr[] = {2, 1, 4, 5, 3, 3};
    int N = arr.length;
    int K = 3;

    if (isKPartitionPossible(arr, N, K))
        System.out.println("Partitions into equal sum is possible.");
    else
        System.out.println("Partitions into equal sum is not possible.");
}
}

// This code is contributed by Princi Singh

Python3

# Python3 program to check whether an array can be
# partitioned into K subsets of equal sum

# Recursive Utility method to check K equal sum
# subsetition of array

"""*
array     - given input array
subsetSum array - sum to store each subset of the array
taken     -boolean array to check whether element
is taken into sum partition or not
K         - number of partitions needed
N         - total number of element in array
curIdx     - current subsetSum index
limitIdx     - lastIdx from where array element should
be taken """

def isKPartitionPossibleRec(arr, subsetSum, taken,
                            subset, K, N, curIdx, limitIdx):
    if subsetSum[curIdx] == subset:

        """ current index (K - 2) represents (K - 1)
        subsets of equal sum last partition will
        already remain with sum 'subset'"""
        if (curIdx == K - 2):
            return True

        # recursive call for next subsetition
        return isKPartitionPossibleRec(arr, subsetSum, taken,
                                       subset, K, N, curIdx + 1 , N - 1)

    # start from limitIdx and include
    # elements into current partition
    for i in range(limitIdx, -1, -1):

        # if already taken, continue
        if (taken[i]):
            continue
        tmp = subsetSum[curIdx] + arr[i]

        # if temp is less than subset, then only
        # include the element and call recursively
        if (tmp <= subset):

            # mark the element and include into
            # current partition sum
            taken[i] = True
            subsetSum[curIdx] += arr[i]
            nxt = isKPartitionPossibleRec(arr, subsetSum, taken,
                                          subset, K, N, curIdx, i - 1)

            # after recursive call unmark the element and
            # remove from subsetition sum
            taken[i] = False
            subsetSum[curIdx] -= arr[i]
            if (nxt):
                return True
    return False

# Method returns True if arr can be
# partitioned into K subsets with equal sum
def isKPartitionPossible(arr, N, K):

    # If K is 1,
    # then complete array will be our answer
    if (K == 1):
        return True

    # If total number of partitions are more than N,
    # then division is not possible
    if (N < K):
        return False

    # if array sum is not divisible by K then
    # we can't divide array into K partitions
    sum = 0
    for i in range(N):
        sum += arr[i]
    if (sum % K != 0):
        return False

    # the sum of each subset should be subset (= sum / K)
    subset = sum // K
    subsetSum = [0] * K
    taken = [0] * N

    # Initialize sum of each subset from 0
    for i in range(K):
        subsetSum[i] = 0

    # mark all elements as not taken
    for i in range(N):
        taken[i] = False

    # initialize first subset sum as
    # last element of array and mark that as taken
    subsetSum[0] = arr[N - 1]
    taken[N - 1] = True

    # call recursive method to check
    # K-substitution condition
    return isKPartitionPossibleRec(arr, subsetSum, taken,
                                   subset, K, N, 0, N - 1)

# Driver Code
arr = [2, 1, 4, 5, 3, 3 ]
N = len(arr)
K = 3
if (isKPartitionPossible(arr, N, K)):
    print("Partitions into equal sum is possible.\n")
else:
    print("Partitions into equal sum is not possible.\n")

# This code is contributed by SHUBHAMSINGH8410

C#

// C# program to check whether an array can be
// partitioned into K subsets of equal sum
using System;

class GFG
{

// Recursive Utility method to check K equal sum
// subsetition of array
/**
    array     - given input array
    subsetSum array - sum to store each subset of the array
    taken     - boolean array to check whether element
                    is taken into sum partition or not
    K         - number of partitions needed
    N         - total number of element in array
    curIdx     - current subsetSum index
    limitIdx     - lastIdx from where array element should
                    be taken */
static bool isKPartitionPossibleRec(int []arr, int []subsetSum, bool []taken,
                int subset, int K, int N, int curIdx, int limitIdx)
{
    if (subsetSum[curIdx] == subset)
    {
        /* current index (K - 2) represents (K - 1) subsets of equal
            sum last partition will already remain with sum 'subset'*/
        if (curIdx == K - 2)
            return true;

        // recursive call for next subsetition
        return isKPartitionPossibleRec(arr, subsetSum, taken, subset,
                                            K, N, curIdx + 1, N - 1);
    }

    // start from limitIdx and include elements into current partition
    for (int i = limitIdx; i >= 0; i--)
    {
        // if already taken, continue
        if (taken[i])
            continue;
        int tmp = subsetSum[curIdx] + arr[i];

        // if temp is less than subset then only include the element
        // and call recursively
        if (tmp <= subset)
        {
            // mark the element and include into current partition sum
            taken[i] = true;
            subsetSum[curIdx] += arr[i];
            bool nxt = isKPartitionPossibleRec(arr, subsetSum, taken,
                                            subset, K, N, curIdx, i - 1);

            // after recursive call unmark the element and remove from
            // subsetition sum
            taken[i] = false;
            subsetSum[curIdx] -= arr[i];
            if (nxt)
                return true;
        }
    }
    return false;
}

// Method returns true if arr can be partitioned into K subsets
// with equal sum
static bool isKPartitionPossible(int []arr, int N, int K)
{
    // If K is 1, then complete array will be our answer
    if (K == 1)
        return true;

    // If total number of partitions are more than N, then
    // division is not possible
    if (N < K)
        return false;

    // if array sum is not divisible by K then we can't divide
    // array into K partitions
    int sum = 0;
    for (int i = 0; i < N; i++)
        sum += arr[i];
    if (sum % K != 0)
        return false;

    // the sum of each subset should be subset (= sum / K)
    int subset = sum / K;
    int []subsetSum = new int[K];
    bool []taken = new bool[N];

    // Initialize sum of each subset from 0
    for (int i = 0; i < K; i++)
        subsetSum[i] = 0;

    // mark all elements as not taken
    for (int i = 0; i < N; i++)
        taken[i] = false;

    // initialize first subset sum as last element of
    // array and mark that as taken
    subsetSum[0] = arr[N - 1];
    taken[N - 1] = true;

    // call recursive method to check K-substitution condition
    return isKPartitionPossibleRec(arr, subsetSum, taken,
                                    subset, K, N, 0, N - 1);
}

// Driver code
static public void Main ()
{

    int []arr = {2, 1, 4, 5, 3, 3};
    int N = arr.Length;
    int K = 3;

    if (isKPartitionPossible(arr, N, K))
        Console.WriteLine("Partitions into equal sum is possible.");
    else
        Console.WriteLine("Partitions into equal sum is not possible.");
}
}

// This code is contributed by ajit.

Javascript

<script>
// Javascript program to check whether an array can be
// partitioned into K subsets of equal sum


// Recursive Utility method to check K equal sum
// subsetition of array
/**
    array           - given input array
    subsetSum array   - sum to store each subset of the array
    taken           - boolean array to check whether element
                      is taken into sum partition or not
    K               - number of partitions needed
    N               - total number of element in array
    curIdx          - current subsetSum index
    limitIdx        - lastIdx from where array element should
                      be taken */
function isKPartitionPossibleRec(arr, subsetSum, taken, subset, K, N, curIdx, limitIdx) {
    if (subsetSum[curIdx] == subset) {
        /*  current index (K - 2) represents (K - 1) subsets of equal
            sum last partition will already remain with sum 'subset'*/
        if (curIdx == K - 2)
            return true;

        //  recursive call for next subsetition
        return isKPartitionPossibleRec(arr, subsetSum, taken, subset,
            K, N, curIdx + 1, N - 1);
    }

    //  start from limitIdx and include elements into current partition
    for (let i = limitIdx; i >= 0; i--) {
        //  if already taken, continue
        if (taken[i])
            continue;
        let tmp = subsetSum[curIdx] + arr[i];

        // if temp is less than subset then only include the element
        // and call recursively
        if (tmp <= subset) {
            //  mark the element and include into current partition sum
            taken[i] = true;
            subsetSum[curIdx] += arr[i];
            let nxt = isKPartitionPossibleRec(arr, subsetSum, taken,
                subset, K, N, curIdx, i - 1);

            // after recursive call unmark the element and remove from
            // subsetition sum
            taken[i] = false;
            subsetSum[curIdx] -= arr[i];
            if (nxt)
                return true;
        }
    }
    return false;
}

//  Method returns true if arr can be partitioned into K subsets
// with equal sum
function isKPartitionPossible(arr, N, K) {
    //  If K is 1, then complete array will be our answer
    if (K == 1)
        return true;

    //  If total number of partitions are more than N, then
    // division is not possible
    if (N < K)
        return false;

    // if array sum is not divisible by K then we can't divide
    // array into K partitions
    let sum = 0;
    for (let i = 0; i < N; i++)
        sum += arr[i];
    if (sum % K != 0)
        return false;

    //  the sum of each subset should be subset (= sum / K)
    let subset = sum / K;
    let subsetSum = new Array(K);
    let taken = new Array(N);

    //  Initialize sum of each subset from 0
    for (let i = 0; i < K; i++)
        subsetSum[i] = 0;

    //  mark all elements as not taken
    for (let i = 0; i < N; i++)
        taken[i] = false;

    // initialize first subset sum as last element of
    // array and mark that as taken
    subsetSum[0] = arr[N - 1];
    taken[N - 1] = true;

    //  call recursive method to check K-substitution condition
    return isKPartitionPossibleRec(arr, subsetSum, taken,
        subset, K, N, 0, N - 1);
}

//  Driver code to test above methods
let arr = [2, 1, 4, 5, 3, 3];
let N = arr.length;
let K = 3;

if (isKPartitionPossible(arr, N, K))
    document.write("Partitions into equal sum is possible");
else
    document.write("Partitions into equal sum is not possible")

    // This code is contributed by saurabh_jaiswal.
</script>

Output:

Partitions into equal sum is possible.

Complexity Analysis: 

Time Complexity: O(2^(N * K)). 
Because if we have K trees stacked on top of each other, the new height of the tree is K * n. i.e one subset is not independent from other.
Space Complexity: O(N). 
Extra space is required for visited array.
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