Posted on by Kalkicode
Code Backtracking

# K partition with equal sum

The "K Partition with Equal Sum" problem involves partitioning a given collection of positive integers into K subsets, such that the sum of elements in each subset is equal. The goal is to find such a partition if it exists and print the subsets with equal sums. This problem can be solved using a backtracking approach that explores all possible combinations of subsets.

## Problem Statement

Given an array of positive integers and an integer K, the task is to find if it's possible to partition the array into K subsets, each with an equal sum.

## Example

Let's take an example to illustrate the problem. Consider the array:

[6, 2, 7, 1, 8, 4, 5, 3, 9, 15]
• For K = 2, the subsets with equal sum are: [6, 2, 7, 1, 5, 9] and [8, 4, 3, 15].
• For K = 4, the subsets with equal sum are: [6, 2, 7], [1, 5, 9], [8, 4, 3], and [15].
• For K = 3, it's not possible to partition the array into subsets with equal sums.

## Idea to Solve

The problem can be solved using a backtracking approach. We'll explore all possible combinations of subsets by placing elements in each subset and recursively checking if the sum constraint is satisfied. We'll backtrack when the sum exceeds the target sum or all elements have been assigned.

## Pseudocode

function findSolution(collection, result, visit, n, k, sum, i, j):
if i >= n or j >= k:
return 0
if sum == result[j]:
if j + 1 == k:
return 1
else:
return findSolution(collection, result, visit, n, k, sum, 0, j + 1)
for x = i to n - 1:
if visit[x] == 0 and result[j] + collection[x] <= sum:
visit[x] = j + 1
result[j] = result[j] + collection[x]
if findSolution(collection, result, visit, n, k, sum, i + 1, j) == 1:
return 1
result[j] = result[j] - collection[x]
visit[x] = 0
return 0

function partition(collection, n, k):
if k <= 0 or k > n:
print "Given", k, "partition not possible"
else:
sum = 0
for i = 0 to n - 1:
sum = sum + collection[i]
if sum % k == 0:
result[k]
visit[n]
for i = 0 to n - 1:
if i < k:
result[i] = 0
visit[i] = 0
if findSolution(collection, result, visit, n, k, sum / k, 0, 0) == 1:
print "Total Sum:", sum
print "Partition size:", k
print "Equal sum is:", sum / k
print "Output"
print "------------------"
for j = 1 to k:
print "Set", j, ":"
for i = 0 to n - 1:
if visit[i] == j:
print collection[i]
else:
print "Partition of size", k, "cannot be divided into equal amounts"
else:
print "Partition with", k, "parts does not produce equal sum"

function main():
collection = [6, 2, 7, 1, 8, 4, 5, 3, 9, 15]
n = size of collection
k = 2
partition(collection, n, k)
k = 4
partition(collection, n, k)
k = 3
partition(collection, n, k)

main()

## Algorithm Explanation

1. findSolution(collection, result, visit, n, k, sum, i, j): This function recursively explores all possible combinations of subsets and checks if they form equal sums. It takes several parameters:

• collection: The array of positive integers.
• result: An array to store the sum of elements in each subset.
• visit: An array to track visited elements.
• n: The number of elements in the collection.
• k: The number of subsets to be formed.
• sum: The target sum for each subset.
• i: The index of the current element being considered.
• j: The index of the current subset being formed.
2. The base case is when i becomes greater than or equal to n or j becomes greater than or equal to k. This indicates that we've either exhausted the collection or formed all subsets.

3. If the sum of the current subset matches the target sum, we check if we've formed all subsets. If so, we return 1, indicating success. If not, we move on to the next subset by recursively calling the function with j + 1.

4. We iterate through the collection from index i to n - 1. If the element at index x has not been visited and adding it to the current subset's sum won't exceed the target sum, we update the visit and result arrays and recursively call the function with the next element.

5. If a solution is found in the recursive tree, we return 1. If not, we backtrack by resetting the values of the result and visit arrays.

6. partition(collection, n, k): This function calculates the total sum of the collection and checks if the sum can be divided evenly among k subsets. If so, it initializes arrays and calls the findSolution function.

## Code Solution

// C Program
// K partition with equal sum
#include <stdio.h>

// Find subset of given sum
int findSolution(int collection[], int result[], int visit[], int n, int k, int sum, int i, int j)
{
if (i >= n || j >= k)
{
return 0;
}
if (sum == result[j])
{
if (j + 1 == k)
{
// When all subsets exist
return 1;
}
else
{
// Backtrack next subset sum
return findSolution(collection, result, visit, n, k, sum, 0, j + 1);
}
}
// Execute loop through by size n
for (int x = i; x < n; ++x)
{
if (visit[x] == 0 && result[j] + collection[x] <= sum)
{
// Active visit status
visit[x] = j + 1;
result[j] = result[j] + collection[x];
if (findSolution(collection, result, visit, n, k, sum, i + 1, j) == 1)
{
// When solution is found
return 1;
}
// Back to previous values
result[j] = result[j] - collection[x];
visit[x] = 0;
}
}
return 0;
}
// Handles the request of partition of k with equal sum
void partition(int collection[], int n, int k)
{
if (k <= 0 || k > n)
{
printf("\n Given %d partition not possible\n", k);
}
else
{
int i = 0;
int j = 0;
int sum = 0;
// Sum of elements
for (i = 0; i < n; ++i)
{
sum += collection[i];
}
if (sum % k == 0)
{
// When partition possible of equal sum
// This is used to handle partition sum info
int result[k];
// This are used to track visited element
int visit[n];
// Through the loop set initial values
for (i = 0; i < n; ++i)
{
if (i < k)
{
result[i] = 0;
}
visit[i] = 0;
}
if (findSolution(collection, result, visit, n, k, sum / k, 0, 0))
{
// Display calculated result
printf("\n Total Sum : %d", sum);
printf("\n Partition size : %d", k);
printf("\n Equal sum is : %d", sum / k);
printf("\n Output");
printf("\n ------------------");
// Display resultant set
for (j = 1; j <= k; ++j)
{
printf("\n Set %d : ", j);
for (i = 0; i < n; ++i)
{
if (visit[i] == j)
{
printf("  %d", collection[i]);
}
}
}
printf("\n\n");
}
else
{
printf("\n Partition of size %d cannot be divided into equal amount\n", k);
}
}
else
{
printf("\n Partition with %d parts is not produce equal sum\n", k);
}
}
}
int main(int argc, char
const *argv[])
{
// Define collection of positive elements
int collection[] = {
6 , 2 , 7 , 1 , 8 , 4 , 5 , 3 , 9 , 15
};
// Get the number of elements
int n = sizeof(collection) / sizeof(collection[0]);
// Test cases
int k = 2;
partition(collection, n, k);
k = 4;
partition(collection, n, k);
k = 3;
partition(collection, n, k);
return 0;
}

#### Output

Total Sum : 60
Partition size : 2
Equal sum is : 30
Output
------------------
Set 1 :   6  2  7  1  5  9
Set 2 :   8  4  3  15

Total Sum : 60
Partition size : 4
Equal sum is : 15
Output
------------------
Set 1 :   6  2  7
Set 2 :   1  5  9
Set 3 :   8  4  3
Set 4 :   15

Total Sum : 60
Partition size : 3
Equal sum is : 20
Output
------------------
Set 1 :   6  2  7  1  4
Set 2 :   8  3  9
Set 3 :   5  15
/*
Java Program for
K partition with equal sum
*/
class Subset
{
// Find subset of given sum
public boolean findSolution(int[] collection, int[] result, int[] visit, int n, int k, int sum, int i, int j)
{
if (i >= n || j >= k)
{
return false;
}
if (sum == result[j])
{
if (j + 1 == k)
{
// When all subsets exist
return true;
}
else
{
// Backtrack next subset sum
return findSolution(collection, result, visit, n, k, sum, 0, j + 1);
}
}
// Execute loop through by size n
for (int x = i; x < n; ++x)
{
if (visit[x] == 0 && result[j] + collection[x] <= sum)
{
// Active visit status
visit[x] = j + 1;
result[j] = result[j] + collection[x];
if (findSolution(collection, result, visit, n, k, sum, i + 1, j) == true)
{
// When solution is found
return true;
}
// Back to previous values
result[j] = result[j] - collection[x];
visit[x] = 0;
}
}
return false;
}
// Handles the request of partition of k with equal sum
public void partition(int[] collection, int n, int k)
{
if (k <= 0 || k > n)
{
System.out.print("\n Given " + k + " partition not possible\n");
}
else
{
// Loop controlling variable i and j
int i = 0;
int j = 0;
int sum = 0;
// Sum of elements
for (i = 0; i < n; ++i)
{
sum += collection[i];
}
if (sum % k == 0)
{
// When partition possible of equal sum
// This is used to handle partition sum info
int[] result = new int[k];
// This are used to track visited element
int[] visit = new int[n];
// Through the loop set initial values
for (i = 0; i < n; ++i)
{
if (i < k)
{
result[i] = 0;
}
visit[i] = 0;
}
if (findSolution(collection, result, visit, n, k, sum / k, 0, 0) == true)
{
// Display calculated result
System.out.print("\n Total Sum : " + sum + "");
System.out.print("\n Partition size : " + k + "");
System.out.print("\n Equal sum is : " + sum / k + "");
System.out.print("\n Output");
System.out.print("\n ------------------");
// Display resultant set
for (j = 1; j <= k; ++j)
{
System.out.print("\n Set " + j  + " : ");
for (i = 0; i < n; ++i)
{
if (visit[i] == j)
{
System.out.print(" " + collection[i]);
}
}
}
System.out.print("\n\n");
}
else
{
System.out.print("\n Partition of size " + k + " cannot be divided into equal amount\n");
}
}
else
{
System.out.print("\n Partition with " + k + " parts is not produce equal sum\n");
}
}
}
public static void main(String[] args)
{
// Define collection of positive elements
int[] collection = {
6 , 2 , 7 , 1 , 8 , 4 , 5 , 3 , 9 , 15
};
// Get the number of elements
int n = collection.length;
// Test cases
int k = 2;
k = 4;
k = 3;
}
}

#### Output

Total Sum : 60
Partition size : 2
Equal sum is : 30
Output
------------------
Set 1 :  6 2 7 1 5 9
Set 2 :  8 4 3 15

Total Sum : 60
Partition size : 4
Equal sum is : 15
Output
------------------
Set 1 :  6 2 7
Set 2 :  1 5 9
Set 3 :  8 4 3
Set 4 :  15

Total Sum : 60
Partition size : 3
Equal sum is : 20
Output
------------------
Set 1 :  6 2 7 1 4
Set 2 :  8 3 9
Set 3 :  5 15
#include <iostream>

using namespace std;
/*
C++ Program for
K partition with equal sum
*/
class Subset
{
public:
// Find subset of given sum
bool findSolution(int collection[], int result[], int visit[], int n, int k, int sum, int i, int j)
{
if (i >= n || j >= k)
{
return false;
}
if (sum == result[j])
{
if (j + 1 == k)
{
// When all subsets exist
return true;
}
else
{
// Backtrack next subset sum
return this->findSolution(collection, result, visit, n, k, sum, 0, j + 1);
}
}
// Execute loop through by size n
for (int x = i; x < n; ++x)
{
if (visit[x] == 0 && result[j] + collection[x] <= sum)
{
// Active visit status
visit[x] = j + 1;
result[j] = result[j] + collection[x];
if (this->findSolution(collection, result, visit, n, k, sum, i + 1, j) == true)
{
// When solution is found
return true;
}
// Back to previous values
result[j] = result[j] - collection[x];
visit[x] = 0;
}
}
return false;
}
// Handles the request of partition of k with equal sum
void partition(int collection[], int n, int k)
{
if (k <= 0 || k > n)
{
cout << "\n Given " << k << " partition not possible\n";
}
else
{
// Loop controlling variable i and j
int i = 0;
int j = 0;
int sum = 0;
// Sum of elements
for (i = 0; i < n; ++i)
{
sum += collection[i];
}
if (sum % k == 0)
{
// When partition possible of equal sum
// This is used to handle partition sum info
int result[k];
// This are used to track visited element
int visit[n];
// Through the loop set initial values
for (i = 0; i < n; ++i)
{
if (i < k)
{
result[i] = 0;
}
visit[i] = 0;
}
if (this->findSolution(collection, result, visit, n, k, sum / k, 0, 0) == true)
{
// Display calculated result
cout << "\n Total Sum : " << sum << "";
cout << "\n Partition size : " << k << "";
cout << "\n Equal sum is : " << sum / k << "";
cout << "\n Output";
cout << "\n ------------------";
// Display resultant set
for (j = 1; j <= k; ++j)
{
cout << "\n Set " << j << " : ";
for (i = 0; i < n; ++i)
{
if (visit[i] == j)
{
cout << " " << collection[i];
}
}
}
cout << "\n\n";
}
else
{
cout << "\n Partition of size " << k << " cannot be divided into equal amount\n";
}
}
else
{
cout << "\n Partition with " << k << " parts is not produce equal sum\n";
}
}
}
};
int main()
{
// Define collection of positive elements
int collection[] = {
6 , 2 , 7 , 1 , 8 , 4 , 5 , 3 , 9 , 15
};
// Get the number of elements
int n = sizeof(collection) / sizeof(collection[0]);
// Test cases
int k = 2;
k = 4;
k = 3;
return 0;
}

#### Output

Total Sum : 60
Partition size : 2
Equal sum is : 30
Output
------------------
Set 1 :  6 2 7 1 5 9
Set 2 :  8 4 3 15

Total Sum : 60
Partition size : 4
Equal sum is : 15
Output
------------------
Set 1 :  6 2 7
Set 2 :  1 5 9
Set 3 :  8 4 3
Set 4 :  15

Total Sum : 60
Partition size : 3
Equal sum is : 20
Output
------------------
Set 1 :  6 2 7 1 4
Set 2 :  8 3 9
Set 3 :  5 15
// Include namespace system
using System;
/*
C# Program for
K partition with equal sum
*/
public class Subset
{
// Find subset of given sum
public Boolean findSolution(int[] collection, int[] result, int[] visit, int n, int k, int sum, int i, int j)
{
if (i >= n || j >= k)
{
return false;
}
if (sum == result[j])
{
if (j + 1 == k)
{
// When all subsets exist
return true;
}
else
{
// Backtrack next subset sum
return findSolution(collection, result, visit, n, k, sum, 0, j + 1);
}
}
// Execute loop through by size n
for (int x = i; x < n; ++x)
{
if (visit[x] == 0 && result[j] + collection[x] <= sum)
{
// Active visit status
visit[x] = j + 1;
result[j] = result[j] + collection[x];
if (findSolution(collection, result, visit, n, k, sum, i + 1, j) == true)
{
// When solution is found
return true;
}
// Back to previous values
result[j] = result[j] - collection[x];
visit[x] = 0;
}
}
return false;
}
// Handles the request of partition of k with equal sum
public void partition(int[] collection, int n, int k)
{
if (k <= 0 || k > n)
{
Console.Write("\n Given " + k + " partition not possible\n");
}
else
{
// Loop controlling variable i and j
int i = 0;
int j = 0;
int sum = 0;
// Sum of elements
for (i = 0; i < n; ++i)
{
sum += collection[i];
}
if (sum % k == 0)
{
// When partition possible of equal sum
// This is used to handle partition sum info
int[] result = new int[k];
// This are used to track visited element
int[] visit = new int[n];
// Through the loop set initial values
for (i = 0; i < n; ++i)
{
if (i < k)
{
result[i] = 0;
}
visit[i] = 0;
}
if (findSolution(collection, result, visit, n, k, sum / k, 0, 0) == true)
{
// Display calculated result
Console.Write("\n Total Sum : " + sum + "");
Console.Write("\n Partition size : " + k + "");
Console.Write("\n Equal sum is : " + sum / k + "");
Console.Write("\n Output");
Console.Write("\n ------------------");
// Display resultant set
for (j = 1; j <= k; ++j)
{
Console.Write("\n Set " + j + " : ");
for (i = 0; i < n; ++i)
{
if (visit[i] == j)
{
Console.Write(" " + collection[i]);
}
}
}
Console.Write("\n\n");
}
else
{
Console.Write("\n Partition of size " + k + " cannot be divided into equal amount\n");
}
}
else
{
Console.Write("\n Partition with " + k + " parts is not produce equal sum\n");
}
}
}
public static void Main(String[] args)
{
// Define collection of positive elements
int[] collection = {
6 , 2 , 7 , 1 , 8 , 4 , 5 , 3 , 9 , 15
};
// Get the number of elements
int n = collection.Length;
// Test cases
int k = 2;
k = 4;
k = 3;
}
}

#### Output

Total Sum : 60
Partition size : 2
Equal sum is : 30
Output
------------------
Set 1 :  6 2 7 1 5 9
Set 2 :  8 4 3 15

Total Sum : 60
Partition size : 4
Equal sum is : 15
Output
------------------
Set 1 :  6 2 7
Set 2 :  1 5 9
Set 3 :  8 4 3
Set 4 :  15

Total Sum : 60
Partition size : 3
Equal sum is : 20
Output
------------------
Set 1 :  6 2 7 1 4
Set 2 :  8 3 9
Set 3 :  5 15
<?php
/*
Php Program for
K partition with equal sum
*/
class Subset
{
// Find subset of given sum
public  function findSolution( & \$collection, & \$result, & \$visit, \$n, \$k, \$sum, \$i, \$j)
{
if (\$i >= \$n || \$j >= \$k)
{
return false;
}
if (\$sum == \$result[\$j])
{
if (\$j + 1 == \$k)
{
// When all subsets exist
return true;
}
else
{
// Backtrack next subset sum
return \$this->findSolution(\$collection, \$result, \$visit, \$n, \$k, \$sum, 0, \$j + 1);
}
}
// Execute loop through by size n
for (\$x = \$i; \$x < \$n; ++\$x)
{
if (\$visit[\$x] == 0 && \$result[\$j] + \$collection[\$x] <= \$sum)
{
// Active visit status
\$visit[\$x] = \$j + 1;
\$result[\$j] = \$result[\$j] + \$collection[\$x];
if (\$this->findSolution(\$collection, \$result, \$visit, \$n, \$k, \$sum, \$i + 1, \$j) == true)
{
// When solution is found
return true;
}
// Back to previous values
\$result[\$j] = \$result[\$j] - \$collection[\$x];
\$visit[\$x] = 0;
}
}
return false;
}
// Handles the request of partition of k with equal sum
public  function partition( & \$collection, \$n, \$k)
{
if (\$k <= 0 || \$k > \$n)
{
echo "\n Given ". \$k ." partition not possible\n";
}
else
{
// Loop controlling variable i and j
\$i = 0;
\$j = 0;
\$sum = 0;
// Sum of elements
for (\$i = 0; \$i < \$n; ++\$i)
{
\$sum += \$collection[\$i];
}
if (\$sum % \$k == 0)
{
// When partition possible of equal sum
// This is used to handle partition sum info
\$result = array_fill(0, \$k, 0);
// This are used to track visited element
\$visit = array_fill(0, \$n, 0);
if (\$this->findSolution(\$collection, \$result, \$visit, \$n, \$k, intval(\$sum / \$k), 0, 0) == true)
{
// Display calculated result
echo "\n Total Sum : ". \$sum ."";
echo "\n Partition size : ". \$k ."";
echo "\n Equal sum is : ". intval(\$sum / \$k) ."";
echo "\n Output";
echo "\n ------------------";
// Display resultant set
for (\$j = 1; \$j <= \$k; ++\$j)
{
echo "\n Set ". \$j ." : ";
for (\$i = 0; \$i < \$n; ++\$i)
{
if (\$visit[\$i] == \$j)
{
echo " ". \$collection[\$i];
}
}
}
echo "\n\n";
}
else
{
echo "\n Partition of size ". \$k ." cannot be divided into equal amount\n";
}
}
else
{
echo "\n Partition with ". \$k ." parts is not produce equal sum\n";
}
}
}
}

function main()
{
// Define collection of positive elements
\$collection = array(6, 2, 7, 1, 8, 4, 5, 3, 9, 15);
// Get the number of elements
\$n = count(\$collection);
// Test cases
\$k = 2;
\$k = 4;
\$k = 3;
}
main();

#### Output

Total Sum : 60
Partition size : 2
Equal sum is : 30
Output
------------------
Set 1 :  6 2 7 1 5 9
Set 2 :  8 4 3 15

Total Sum : 60
Partition size : 4
Equal sum is : 15
Output
------------------
Set 1 :  6 2 7
Set 2 :  1 5 9
Set 3 :  8 4 3
Set 4 :  15

Total Sum : 60
Partition size : 3
Equal sum is : 20
Output
------------------
Set 1 :  6 2 7 1 4
Set 2 :  8 3 9
Set 3 :  5 15
/*
Node Js Program for
K partition with equal sum
*/
class Subset
{
// Find subset of given sum
findSolution(collection, result, visit, n, k, sum, i, j)
{
if (i >= n || j >= k)
{
return false;
}
if (sum == result[j])
{
if (j + 1 == k)
{
// When all subsets exist
return true;
}
else
{
// Backtrack next subset sum
return this.findSolution(collection, result, visit, n, k, sum, 0, j + 1);
}
}
// Execute loop through by size n
for (var x = i; x < n; ++x)
{
if (visit[x] == 0 && result[j] + collection[x] <= sum)
{
// Active visit status
visit[x] = j + 1;
result[j] = result[j] + collection[x];
if (this.findSolution(collection, result, visit, n, k, sum, i + 1, j) == true)
{
// When solution is found
return true;
}
// Back to previous values
result[j] = result[j] - collection[x];
visit[x] = 0;
}
}
return false;
}
// Handles the request of partition of k with equal sum
partition(collection, n, k)
{
if (k <= 0 || k > n)
{
process.stdout.write("\n Given " + k + " partition not possible\n");
}
else
{
// Loop controlling variable i and j
var i = 0;
var j = 0;
var sum = 0;
// Sum of elements
for (i = 0; i < n; ++i)
{
sum += collection[i];
}
if (sum % k == 0)
{
// When partition possible of equal sum
// This is used to handle partition sum info
var result = Array(k).fill(0);
// This are used to track visited element
var visit = Array(n).fill(0);
if (this.findSolution(collection, result, visit, n, k, parseInt(sum / k), 0, 0) == true)
{
// Display calculated result
process.stdout.write("\n Total Sum : " + sum + "");
process.stdout.write("\n Partition size : " + k + "");
process.stdout.write("\n Equal sum is : " + parseInt(sum / k) + "");
process.stdout.write("\n Output");
process.stdout.write("\n ------------------");
// Display resultant set
for (j = 1; j <= k; ++j)
{
process.stdout.write("\n Set " + j + " : ");
for (i = 0; i < n; ++i)
{
if (visit[i] == j)
{
process.stdout.write(" " + collection[i]);
}
}
}
process.stdout.write("\n\n");
}
else
{
process.stdout.write("\n Partition of size " + k + " cannot be divided into equal amount\n");
}
}
else
{
process.stdout.write("\n Partition with " + k + " parts is not produce equal sum\n");
}
}
}
}

function main()
{
// Define collection of positive elements
var collection = [6, 2, 7, 1, 8, 4, 5, 3, 9, 15];
// Get the number of elements
var n = collection.length;
// Test cases
var k = 2;
k = 4;
k = 3;
}
main();

#### Output

Total Sum : 60
Partition size : 2
Equal sum is : 30
Output
------------------
Set 1 :  6 2 7 1 5 9
Set 2 :  8 4 3 15

Total Sum : 60
Partition size : 4
Equal sum is : 15
Output
------------------
Set 1 :  6 2 7
Set 2 :  1 5 9
Set 3 :  8 4 3
Set 4 :  15

Total Sum : 60
Partition size : 3
Equal sum is : 20
Output
------------------
Set 1 :  6 2 7 1 4
Set 2 :  8 3 9
Set 3 :  5 15
#   Python 3 Program for
#   K partition with equal sum

class Subset :
#  Find subset of given sum
def findSolution(self, collection, result, visit, n, k, sum, i, j) :
if (i >= n or j >= k) :
return False

if (sum == result[j]) :
if (j + 1 == k) :
#  When all subsets exist
return True
else :
#  Backtrack next subset sum
return self.findSolution(collection, result, visit, n, k, sum, 0, j + 1)

#  Execute loop through by size n
x = i
while (x < n) :
if (visit[x] == 0 and result[j] + collection[x] <= sum) :
#  Active visit status
visit[x] = j + 1
result[j] = result[j] + collection[x]
if (self.findSolution(collection, result, visit, n, k, sum, i + 1, j) == True) :
#  When solution is found
return True

#  Back to previous values
result[j] = result[j] - collection[x]
visit[x] = 0

x += 1

return False

#  Handles the request of partition of k with equal sum
def partition(self, collection, n, k) :
if (k <= 0 or k > n) :
print("\n Given ", k ," partition not possible")
else :
#  Loop controlling variable i and j
i = 0
j = 0
sum = 0
#  Sum of elements
while (i < n) :
sum += collection[i]
i += 1

if (sum % k == 0) :
#  When partition possible of equal sum
#  This is used to handle partition sum info
result = [0] * (k)
#  This are used to track visited element
visit = [0] * (n)
if (self.findSolution(collection, result, visit, n, k, int(sum / k), 0, 0) == True) :
#  Display calculated result
print("\n Total Sum : ", sum ,"", end = "")
print("\n Partition size : ", k ,"", end = "")
print("\n Equal sum is : ", int(sum / k) ,"", end = "")
print("\n Output", end = "")
print("\n ------------------", end = "")
#  Display resultant set
j = 1
while (j <= k) :
print("\n Set ", j ," : ", end = "")
i = 0
while (i < n) :
if (visit[i] == j) :
print(" ", collection[i], end = "")

i += 1

j += 1

print("\n")
else :
print("\n Partition of size ", k ," cannot be divided into equal amount")

else :
print("\n Partition with ", k ," parts is not produce equal sum")

def main() :
#  Define collection of positive elements
collection = [6, 2, 7, 1, 8, 4, 5, 3, 9, 15]
#  Get the number of elements
n = len(collection)
#  Test cases
k = 2
k = 4
k = 3

if __name__ == "__main__": main()

#### Output

Total Sum :  60
Partition size :  2
Equal sum is :  30
Output
------------------
Set  1  :   6  2  7  1  5  9
Set  2  :   8  4  3  15

Total Sum :  60
Partition size :  4
Equal sum is :  15
Output
------------------
Set  1  :   6  2  7
Set  2  :   1  5  9
Set  3  :   8  4  3
Set  4  :   15

Total Sum :  60
Partition size :  3
Equal sum is :  20
Output
------------------
Set  1  :   6  2  7  1  4
Set  2  :   8  3  9
Set  3  :   5  15
#   Ruby Program for
#   K partition with equal sum

class Subset
#  Find subset of given sum
def findSolution(collection, result, visit, n, k, sum, i, j)
if (i >= n || j >= k)
return false
end

if (sum == result[j])
if (j + 1 == k)
#  When all subsets exist
return true
else
#  Backtrack next subset sum
return self.findSolution(collection, result, visit, n, k, sum, 0, j + 1)
end

end

#  Execute loop through by size n
x = i
while (x < n)
if (visit[x] == 0 && result[j] + collection[x] <= sum)
#  Active visit status
visit[x] = j + 1
result[j] = result[j] + collection[x]
if (self.findSolution(collection, result, visit, n, k, sum, i + 1, j) == true)
#  When solution is found
return true
end

#  Back to previous values
result[j] = result[j] - collection[x]
visit[x] = 0
end

x += 1
end

return false
end

#  Handles the request of partition of k with equal sum
def partition(collection, n, k)
if (k <= 0 || k > n)
print("\n Given ", k ," partition not possible\n")
else
#  Loop controlling variable i and j
i = 0
j = 0
sum = 0
#  Sum of elements
while (i < n)
sum += collection[i]
i += 1
end

if (sum % k == 0)
#  When partition possible of equal sum
#  This is used to handle partition sum info
result = Array.new(k) {0}
#  This are used to track visited element
visit = Array.new(n) {0}
if (self.findSolution(collection, result, visit, n, k, sum / k, 0, 0) == true)
#  Display calculated result
print("\n Total Sum : ", sum ,"")
print("\n Partition size : ", k ,"")
print("\n Equal sum is : ", sum / k ,"")
print("\n Output")
print("\n ------------------")
#  Display resultant set
j = 1
while (j <= k)
print("\n Set ", j ," : ")
i = 0
while (i < n)
if (visit[i] == j)
print(" ", collection[i])
end

i += 1
end

j += 1
end

print("\n\n")
else
print("\n Partition of size ", k ," cannot be divided into equal amount\n")
end

else
print("\n Partition with ", k ," parts is not produce equal sum\n")
end

end

end

end

def main()
#  Define collection of positive elements
collection = [6, 2, 7, 1, 8, 4, 5, 3, 9, 15]
#  Get the number of elements
n = collection.length
#  Test cases
k = 2
k = 4
k = 3
end

main()

#### Output

Total Sum : 60
Partition size : 2
Equal sum is : 30
Output
------------------
Set 1 :  6 2 7 1 5 9
Set 2 :  8 4 3 15

Total Sum : 60
Partition size : 4
Equal sum is : 15
Output
------------------
Set 1 :  6 2 7
Set 2 :  1 5 9
Set 3 :  8 4 3
Set 4 :  15

Total Sum : 60
Partition size : 3
Equal sum is : 20
Output
------------------
Set 1 :  6 2 7 1 4
Set 2 :  8 3 9
Set 3 :  5 15

/*
Scala Program for
K partition with equal sum
*/
class Subset
{
// Find subset of given sum
def findSolution(collection: Array[Int], result: Array[Int], visit: Array[Int], n: Int, k: Int, sum: Int, i: Int, j: Int): Boolean = {
if (i >= n || j >= k)
{
return false;
}
if (sum == result(j))
{
if (j + 1 == k)
{
// When all subsets exist
return true;
}
else
{
// Backtrack next subset sum
return this.findSolution(collection, result, visit, n, k, sum, 0, j + 1);
}
}
// Execute loop through by size n
var x: Int = i;
while (x < n)
{
if (visit(x) == 0 && result(j) + collection(x) <= sum)
{
// Active visit status
visit(x) = j + 1;
result(j) = result(j) + collection(x);
if (this.findSolution(collection, result, visit, n, k, sum, i + 1, j) == true)
{
// When solution is found
return true;
}
// Back to previous values
result(j) = result(j) - collection(x);
visit(x) = 0;
}
x += 1;
}
return false;
}
// Handles the request of partition of k with equal sum
def partition(collection: Array[Int], n: Int, k: Int): Unit = {
if (k <= 0 || k > n)
{
print("\n Given " + k + " partition not possible\n");
}
else
{
// Loop controlling variable i and j
var i: Int = 0;
var j: Int = 0;
var sum: Int = 0;
// Sum of elements
while (i < n)
{
sum += collection(i);
i += 1;
}
if (sum % k == 0)
{
// When partition possible of equal sum
// This is used to handle partition sum info
var result: Array[Int] = Array.fill[Int](k)(0);
// This are used to track visited element
var visit: Array[Int] = Array.fill[Int](n)(0);
if (this.findSolution(collection, result, visit, n, k, (sum / k).toInt, 0, 0) == true)
{
// Display calculated result
print("\n Total Sum : " + sum + "");
print("\n Partition size : " + k + "");
print("\n Equal sum is : " + (sum / k).toInt + "");
print("\n Output");
print("\n ------------------");
// Display resultant set
j = 1;
while (j <= k)
{
print("\n Set " + j + " : ");
i = 0;
while (i < n)
{
if (visit(i) == j)
{
print(" " + collection(i));
}
i += 1;
}
j += 1;
}
print("\n\n");
}
else
{
print("\n Partition of size " + k + " cannot be divided into equal amount\n");
}
}
else
{
print("\n Partition with " + k + " parts is not produce equal sum\n");
}
}
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: Subset = new Subset();
// Define collection of positive elements
var collection: Array[Int] = Array(6, 2, 7, 1, 8, 4, 5, 3, 9, 15);
// Get the number of elements
var n: Int = collection.length;
// Test cases
var k: Int = 2;
k = 4;
k = 3;
}
}

#### Output

Total Sum : 60
Partition size : 2
Equal sum is : 30
Output
------------------
Set 1 :  6 2 7 1 5 9
Set 2 :  8 4 3 15

Total Sum : 60
Partition size : 4
Equal sum is : 15
Output
------------------
Set 1 :  6 2 7
Set 2 :  1 5 9
Set 3 :  8 4 3
Set 4 :  15

Total Sum : 60
Partition size : 3
Equal sum is : 20
Output
------------------
Set 1 :  6 2 7 1 4
Set 2 :  8 3 9
Set 3 :  5 15
/*
Swift 4 Program for
K partition with equal sum
*/
class Subset
{
// Find subset of given sum
func findSolution(_ collection: [Int], _ result: inout[Int], _ visit: inout[Int], _ n: Int, _ k: Int, _ sum: Int, _ i: Int, _ j: Int)->Bool
{
if (i >= n || j >= k)
{
return false;
}
if (sum == result[j])
{
if (j + 1 == k)
{
// When all subsets exist
return true;
}
else
{
// Backtrack next subset sum
return self.findSolution(collection, &result, &visit, n, k, sum, 0, j + 1);
}
}
// Execute loop through by size n
var x: Int = i;
while (x < n)
{
if (visit[x] == 0 && result[j] + collection[x] <= sum)
{
// Active visit status
visit[x] = j + 1;
result[j] = result[j] + collection[x];
if (self.findSolution(collection, &result, &visit, n, k, sum, i + 1, j) == true)
{
// When solution is found
return true;
}
// Back to previous values
result[j] = result[j] - collection[x];
visit[x] = 0;
}
x += 1;
}
return false;
}
// Handles the request of partition of k with equal sum
func partition(_ collection: [Int], _ n: Int, _ k: Int)
{
if (k <= 0 || k > n)
{
print("\n Given ", k ," partition not possible");
}
else
{
// Loop controlling variable i and j
var i: Int = 0;
var j: Int = 0;
var sum: Int = 0;
// Sum of elements
while (i < n)
{
sum += collection[i];
i += 1;
}
if (sum % k == 0)
{
// When partition possible of equal sum
// This is used to handle partition sum info
var result: [Int] = Array(repeating: 0, count: k);
// This are used to track visited element
var visit: [Int] = Array(repeating: 0, count: n);
if (self.findSolution(collection, &result, &visit, n, k, sum / k, 0, 0) == true)
{
// Display calculated result
print("\n Total Sum : ", sum ,"", terminator: "");
print("\n Partition size : ", k ,"", terminator: "");
print("\n Equal sum is : ", sum / k ,"", terminator: "");
print("\n Output", terminator: "");
print("\n ------------------", terminator: "");
// Display resultant set
j = 1;
while (j <= k)
{
print("\n Set ", j ," : ", terminator: "");
i = 0;
while (i < n)
{
if (visit[i] == j)
{
print(" ", collection[i], terminator: "");
}
i += 1;
}
j += 1;
}
print("\n");
}
else
{
print("\n Partition of size ", k ," cannot be divided into equal amount");
}
}
else
{
print("\n Partition with ", k ," parts is not produce equal sum");
}
}
}
}
func main()
{
// Define collection of positive elements
let collection: [Int] = [6, 2, 7, 1, 8, 4, 5, 3, 9, 15];
// Get the number of elements
let n: Int = collection.count;
// Test cases
var k: Int = 2;
k = 4;
k = 3;
}
main();

#### Output

Total Sum :  60
Partition size :  2
Equal sum is :  30
Output
------------------
Set  1  :   6  2  7  1  5  9
Set  2  :   8  4  3  15

Total Sum :  60
Partition size :  4
Equal sum is :  15
Output
------------------
Set  1  :   6  2  7
Set  2  :   1  5  9
Set  3  :   8  4  3
Set  4  :   15

Total Sum :  60
Partition size :  3
Equal sum is :  20
Output
------------------
Set  1  :   6  2  7  1  4
Set  2  :   8  3  9
Set  3  :   5  15
/*
Kotlin Program for
K partition with equal sum
*/
class Subset
{
// Find subset of given sum
fun findSolution(collection: Array <Int> , result: Array <Int> , visit: Array <Int> , n: Int, k: Int, sum: Int, i: Int, j: Int): Boolean
{
if (i >= n || j >= k)
{
return false;
}
if (sum == result[j])
{
if (j + 1 == k)
{
// When all subsets exist
return true;
}
else
{
// Backtrack next subset sum
return this.findSolution(collection, result, visit, n, k, sum, 0, j + 1);
}
}
// Execute loop through by size n
var x: Int = i;
while (x < n)
{
if (visit[x] == 0 && result[j] + collection[x] <= sum)
{
// Active visit status
visit[x] = j + 1;
result[j] = result[j] + collection[x];
if (this.findSolution(collection, result, visit, n, k, sum, i + 1, j) == true)
{
// When solution is found
return true;
}
// Back to previous values
result[j] = result[j] - collection[x];
visit[x] = 0;
}
x += 1;
}
return false;
}
// Handles the request of partition of k with equal sum
fun partition(collection: Array<Int> , n: Int, k: Int): Unit
{
if (k <= 0 || k > n)
{
print("\n Given " + k + " partition not possible\n");
}
else
{
// Loop controlling variable i and j
var i: Int = 0;
var j: Int ;
var sum: Int = 0;
// Sum of elements
while (i < n)
{
sum += collection[i];
i += 1;
}
if (sum % k == 0)
{
// When partition possible of equal sum
// This is used to handle partition sum info
var result: Array <Int> = Array(k)
{
0
};
// This are used to track visited element
var visit: Array<Int> = Array(n)
{
0
};
if (this.findSolution(collection, result, visit, n, k, sum / k, 0, 0) == true)
{
// Display calculated result
print("\n Total Sum : " + sum + "");
print("\n Partition size : " + k + "");
print("\n Equal sum is : " + sum / k + "");
print("\n Output");
print("\n ------------------");
// Display resultant set
j = 1;
while (j <= k)
{
print("\n Set " + j + " : ");
i = 0;
while (i < n)
{
if (visit[i] == j)
{
print(" " + collection[i]);
}
i += 1;
}
j += 1;
}
print("\n\n");
}
else
{
print("\n Partition of size " + k + " cannot be divided into equal amount\n");
}
}
else
{
print("\n Partition with " + k + " parts is not produce equal sum\n");
}
}
}
}
fun main(args: Array <String> ): Unit
{
// Define collection of positive elements
var collection: Array<Int> = arrayOf(6, 2, 7, 1, 8, 4, 5, 3, 9, 15);
// Get the number of elements
var n: Int = collection.count();
// Test cases
var k: Int = 2;
k = 4;
k = 3;
}

#### Output

Total Sum : 60
Partition size : 2
Equal sum is : 30
Output
------------------
Set 1 :  6 2 7 1 5 9
Set 2 :  8 4 3 15

Total Sum : 60
Partition size : 4
Equal sum is : 15
Output
------------------
Set 1 :  6 2 7
Set 2 :  1 5 9
Set 3 :  8 4 3
Set 4 :  15

Total Sum : 60
Partition size : 3
Equal sum is : 20
Output
------------------
Set 1 :  6 2 7 1 4
Set 2 :  8 3 9
Set 3 :  5 15

## Time Complexity

The time complexity of the algorithm mainly depends on the recursive search space. The worst-case scenario is when all possible combinations are explored. This results in an exponential time complexity of O(n^k), where n is the size of the collection and k is the number of subsets.

## Comment

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