给定N个元素的整数数组，任务是将该数组划分为K个非空子集，以使每个子集中的元素总和相同。该数组的所有元素都应该是一个分区的一部分。
例子：
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
我们可以递归地解决此问题，我们为每个分区的和保留一个数组，并为一个布尔数组检查一个元素是否已被带入某个分区。
首先，我们需要检查一些基本情况，
如果K为1，那么我们已经有了答案，完整的数组只是具有相同总和的子集。
如果N
在下面的代码中，编写了一种递归方法，该方法试图将数组元素添加到某个子集中。如果此子集的总和达到要求的总和，我们将递归迭代下一部分，否则我们将回溯不同的元素集。如果总和达到所需总和的子集数为(K-1)，则我们标记为可以将数组划分为具有相等总和的K个部分，因为其余元素的总和已等于所需总和。
C++
// C++ program to check whether an array can be
// partitioned into K subsets of equal sum
#include
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";
} 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 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
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 SinghPython3
# 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 - ean 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 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
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 SHUBHAMSINGH8410C#
// 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 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
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.输出：
Partitions into equal sum is possible.