Partition a set into K subsets with equal sum - algorithm

Question

I'm going through an exercise to partition a set into K subsets with equal sum.
Let's say
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]]
I found a solution here,
http://www.geeksforgeeks.org/partition-set-k-subsets-equal-sum/
// 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 subsubset 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";
}
This works in all scenarios.
Question
Let's say if I pass arr = [4, 4, 1, 3, 2, 3, 2, 1] and k = 4,the algorithm tried to solve it by adding 1+2+2 and then, 3+3 or 3+1 and so on. It doesn't gets the partition and finally solves it to [[4,1], [4,1], [3,2], [3,2]]. I am not sure how does this algorithm finds the alternative? I'm not able to follow up with the recursion.
What are the ways to solve it? Is the backtracking the only way?
Thanks!