Count ways to split array into two subsets having difference between their sum equal to K
Last Updated :
29 Apr, 2023
Given an array A[] of size N and an integer diff, the task is to count the number of ways to split the array into two subsets (non-empty subset is possible) such that the difference between their sums is equal to diff.
Examples:
Input: A[] = {1, 1, 2, 3}, diff = 1
Output: 3
Explanation: All possible combinations are as follows:
- {1, 1, 2} and {3}
- {1, 3} and {1, 2}
- {1, 2} and {1, 3}
All partitions have difference between their sums equal to 1. Therefore, the count of ways is 3.
Input: A[] = {1, 6, 11, 5}, diff=1
Output: 2
Naive Approach: The simplest approach to solve the problem is based on the following observations:
Let the sum of elements in the partition subsets S1 and S2 be sum1 and sum2 respectively.
Let sum of the array A[] be X.
Given, sum1 – sum2 = diff – (1)
Also, sum1 + sum2 = X – (2)
From equations (1) and (2),
sum1 = (X + diff)/2
Therefore, the task is reduced to finding the number of subsets with a given sum.
Therefore, the simplest approach is to solve this problem is by generating all the possible subsets and checking whether the subset has the required sum.
Time Complexity: O(2N)
Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach, the idea is to use Dynamic Programming. Initialize a dp[][] table of size N*X, where dp[i][C] stores the number of subsets of the sub-array A[i…N-1] such that their sum is equal to C. Thus, the recurrence is very trivial as there are only two choices i.e. either consider the ith element in the subset or don’t. So the recurrence relation will be:
dp[i][C] = dp[i – 1][C – A[i]] + dp[i-1][C]
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int countSubset( int arr[], int n, int diff)
{
int sum = 0;
for ( int i = 0; i < n; i++)
sum += arr[i];
sum += diff;
sum = sum / 2;
int t[n + 1][sum + 1];
for ( int j = 0; j <= sum; j++)
t[0][j] = 0;
for ( int i = 0; i <= n; i++)
t[i][0] = 1;
for ( int i = 1; i <= n; i++) {
for ( int j = 1; j <= sum; j++) {
if (arr[i - 1] > j)
t[i][j] = t[i - 1][j];
else {
t[i][j] = t[i - 1][j]
+ t[i - 1][j - arr[i - 1]];
}
}
}
return t[n][sum];
}
int main()
{
int diff = 1, n = 4;
int arr[] = { 1, 1, 2, 3 };
cout << countSubset(arr, n, diff);
}
|
Java
import java.io.*;
public class GFG
{
static int countSubset( int []arr, int n, int diff)
{
int sum = 0 ;
for ( int i = 0 ; i < n; i++)
sum += arr[i];
sum += diff;
sum = sum / 2 ;
int t[][] = new int [n + 1 ][sum + 1 ];
for ( int j = 0 ; j <= sum; j++)
t[ 0 ][j] = 0 ;
for ( int i = 0 ; i <= n; i++)
t[i][ 0 ] = 1 ;
for ( int i = 1 ; i <= n; i++) {
for ( int j = 1 ; j <= sum; j++) {
if (arr[i - 1 ] > j)
t[i][j] = t[i - 1 ][j];
else {
t[i][j] = t[i - 1 ][j]
+ t[i - 1 ][j - arr[i - 1 ]];
}
}
}
return t[n][sum];
}
public static void main(String[] args)
{
int diff = 1 , n = 4 ;
int arr[] = { 1 , 1 , 2 , 3 };
System.out.print(countSubset(arr, n, diff));
}
}
|
Python3
def countSubset(arr, n, diff):
sum = 0
for i in range (n):
sum + = arr[i]
sum + = diff
sum = sum / / 2
t = [[ 0 for i in range ( sum + 1 )]
for i in range (n + 1 )]
for j in range ( sum + 1 ):
t[ 0 ][j] = 0
for i in range (n + 1 ):
t[i][ 0 ] = 1
for i in range ( 1 , n + 1 ):
for j in range ( 1 , sum + 1 ):
if (arr[i - 1 ] > j):
t[i][j] = t[i - 1 ][j]
else :
t[i][j] = t[i - 1 ][j] + t[i - 1 ][j - arr[i - 1 ]]
return t[n][ sum ]
if __name__ = = '__main__' :
diff, n = 1 , 4
arr = [ 1 , 1 , 2 , 3 ]
print (countSubset(arr, n, diff))
|
C#
using System;
public class GFG
{
static int countSubset( int []arr, int n, int diff)
{
int sum = 0;
for ( int i = 0; i < n; i++)
sum += arr[i];
sum += diff;
sum = sum / 2;
int [,]t = new int [n + 1, sum + 1];
for ( int j = 0; j <= sum; j++)
t[0,j] = 0;
for ( int i = 0; i <= n; i++)
t[i,0] = 1;
for ( int i = 1; i <= n; i++) {
for ( int j = 1; j <= sum; j++) {
if (arr[i - 1] > j)
t[i,j] = t[i - 1,j];
else {
t[i,j] = t[i - 1,j]
+ t[i - 1,j - arr[i - 1]];
}
}
}
return t[n,sum];
}
public static void Main( string [] args)
{
int diff = 1, n = 4;
int []arr = { 1, 1, 2, 3 };
Console.Write(countSubset(arr, n, diff));
}
}
|
Javascript
<script>
function countSubset(arr, n, diff)
{
var sum = 0;
for ( var i = 0; i < n; i++){
sum += arr[i];
}
sum += diff;
sum = sum / 2;
var t = new Array(n + 1);
for ( var i = 0; i < t.length; i++) {
t[i] = new Array(sum + 1);
}
for ( var i = 0; i < t.length; i++) {
for ( var j = 0; j < t[i].length; j++) {
t[i][j] = 0;
}
}
for ( var j = 0; j <= sum; j++)
t[0][j] = 0;
for ( var i = 0; i <= n; i++)
t[i][0] = 1;
for ( var i = 1; i <= n; i++) {
for ( var j = 1; j <= sum; j++) {
if (arr[i - 1] > j)
t[i][j] = t[i - 1][j];
else {
t[i][j] = t[i - 1][j]
+ t[i - 1][j - arr[i - 1]];
}
}
}
return t[n][sum];
}
var diff = 1;
var n = 4;
var arr = [ 1, 1, 2, 3 ];
document.write(countSubset(arr, n, diff));
</script>
|
Time Complexity: O(S*N), where S = sum of array elements + K/2
Auxiliary Space: O(S*N)
Efficient approach: space optimization
To optimize space complexity we only need to keep track of the values of the previous row in the 2D array to compute the values of the current row. Hence, we can replace the 2D array t[n + 1][sum + 1] with a 1D array dp[sum + 1].
Implementation Steps :
- Create vector Dp of size sum + 1 and initialize it with 0.
- Now initialize DP with Base Case dp[0] =1.
- To calculate answer iterate over subproblems with the help of nested loops and get the current value from previous computation.
- At last return the answer stored in dp[sum].
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
int countSubset( int arr[], int n, int diff)
{
int sum = 0;
for ( int i = 0; i < n; i++)
sum += arr[i];
sum += diff;
sum = sum / 2;
int dp[sum + 1] = {0};
dp[0] = 1;
for ( int i = 0; i < n; i++) {
for ( int j = sum; j >= arr[i]; j--) {
dp[j] += dp[j - arr[i]];
}
}
return dp[sum];
}
int main()
{
int diff = 1, n = 4;
int arr[] = { 1, 1, 2, 3 };
cout << countSubset(arr, n, diff);
}
|
Java
import java.util.*;
public class Main
{
public static int countSubset( int arr[], int n, int diff)
{
int sum = 0 ;
for ( int i = 0 ; i < n; i++)
sum += arr[i];
sum += diff;
sum = sum / 2 ;
int dp[] = new int [sum + 1 ];
dp[ 0 ] = 1 ;
for ( int i = 0 ; i < n; i++) {
for ( int j = sum; j >= arr[i]; j--) {
dp[j] += dp[j - arr[i]];
}
}
return dp[sum];
}
public static void main(String[] args)
{
int diff = 1 , n = 4 ;
int arr[] = { 1 , 1 , 2 , 3 };
System.out.println(countSubset(arr, n, diff));
}
}
|
Python3
def countSubset(arr, n, diff):
sum = 0
for i in range (n):
sum + = arr[i]
sum + = diff
if sum % 2 ! = 0 :
return 0
dp = [ 0 ] * ( sum / / 2 + 1 )
dp[ 0 ] = 1
for i in range (n):
for j in range ( sum / / 2 , arr[i] - 1 , - 1 ):
dp[j] + = dp[j - arr[i]]
return dp[ sum / / 2 ]
diff = 1
n = 4
arr = [ 1 , 1 , 2 , 3 ]
print (countSubset(arr, n, diff))
|
C#
using System;
class MainClass
{
static int countSubset( int [] arr, int n, int diff)
{
int sum = 0;
for ( int i = 0; i < n; i++) {
sum += arr[i];
}
sum += diff;
sum = sum / 2;
int [] dp = new int [sum + 1];
dp[0] = 1;
for ( int i = 0; i < n; i++) {
for ( int j = sum; j >= arr[i]; j--) {
dp[j] += dp[j - arr[i]];
}
}
return dp[sum];
}
public static void Main()
{
int diff = 1, n = 4;
int [] arr = { 1, 1, 2, 3 };
Console.WriteLine(countSubset(arr, n, diff));
}
}
|
Javascript
function countSubset(arr, n, diff) {
let sum = 0;
for (let i = 0; i < n; i++) {
sum += arr[i];
}
sum += diff;
sum = Math.floor(sum / 2);
let dp = new Array(sum + 1).fill(0);
dp[0] = 1;
for (let i = 0; i < n; i++) {
for (let j = sum; j >= arr[i]; j--) {
dp[j] += dp[j - arr[i]];
}
}
return dp[sum];
}
let diff = 1,
n = 4;
let arr = [1, 1, 2, 3];
console.log(countSubset(arr, n, diff));
|
Time Complexity: O(S*N), where S = sum of array elements + K/2
Auxiliary Space: O(S)
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