Count ways to split array into pair of subsets with difference between their sum equal to K
Last Updated :
12 Apr, 2023
Given an array arr[] consisting of N integers and an integer K, the task is to find the number of ways to split the array into a pair of subsets such that the difference between their sum is K.
Examples:
Input: arr[] = {1, 1, 2, 3}, K = 1
Output: 3
Explanation:
Following splits into a pair of subsets satisfies the given condition:
- {1, 1, 2}, {3}, difference = (1 + 1 + 2) – 3 = 4 – 3 = 1.
- {1, 3} {1, 2}, difference = (1 + 3) – (1 + 2) = 4 – 3 = 1.
- {1, 3} {1, 2}, difference = (1 + 3) – (1 + 2) = 4 – 3 = 1.
Therefore, the count of ways to split is 3.
Input: arr[] = {1, 2, 3}, K = 2
Output: 1
Explanation:
The only possible split into a pair of subsets satisfying the given condition is {1, 3}, {2}, where the difference = (1 + 3) – 2 = 4 – 2 =2.
Therefore, the count of ways to split is 1.
Naive Approach: The simple approach to solve the given problem is to generate all the possible subsets and store the sum of each subset in an array say subset[]. Then, check if there exist any pair exists in the array subset[] whose difference is K. After checking for all the pairs, print the total count of such pairs as the result.
Time Complexity: O(2N)
Auxiliary Space: O(2N)
Efficient Approach: The above approach can be optimized using the following observations.
Let the sum of the first and second subsets be S1 and S2 respectively, and the sum of array elements be Y.
Now, the sum of both the subsets must be equal to the sum of the array elements.
Therefore, S1 + S2 = Y — (1)
To satisfy the given condition, their difference must be equal to K.
Therefore, S1 – S2 = K — (2)
Adding (1) & (2), the equation obtained is
S1 = (K + Y)/2 — (3)
Therefore, for a pair of subsets having sums S1 and S2, equation (3) must hold true, i.e., the sum of elements of the subset must be equal to (K + Y)/2. Now, the problem reduces to counting the number of subsets with a given sum. This problem can be solved using Dynamic Programming, whose recurrence relation is as follows:
dp[i][C] = dp[i + 1][C – arr[i]] + dp[i + 1][C]
Here, dp[i][C] stores the number of subsets of the subarray arr[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 array element in the subset or don’t.
Below is the implementation of the above approach :
C++
#include <bits/stdc++.h>
using namespace std;
#define maxN 20
#define maxSum 50
#define minSum 50
#define base 50
int dp[maxN][maxSum + minSum];
bool v[maxN][maxSum + minSum];
int findCnt(int* arr, int i,
int required_sum,
int n)
{
if (i == n) {
if (required_sum == 0)
return 1;
else
return 0;
}
if (v[i][required_sum + base])
return dp[i][required_sum + base];
v[i][required_sum + base] = 1;
dp[i][required_sum + base]
= findCnt(arr, i + 1,
required_sum, n)
+ findCnt(arr, i + 1,
required_sum - arr[i], n);
return dp[i][required_sum + base];
}
void countSubsets(int* arr, int K,
int n)
{
int sum = 0;
for (int i = 0; i < n; i++) {
sum += arr[i];
}
int S1 = (sum + K) / 2;
cout << findCnt(arr, 0, S1, n);
}
int main()
{
int arr[] = { 1, 1, 2, 3 };
int N = sizeof(arr) / sizeof(int);
int K = 1;
countSubsets(arr, K, N);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG
{
static int maxN = 20;
static int maxSum = 50;
static int minSum = 50;
static int Base = 50;
static int[][] dp = new int[maxN][maxSum + minSum];
static boolean[][] v = new boolean[maxN][maxSum + minSum];
static int findCnt(int[] arr, int i,
int required_sum,
int n)
{
if (i == n) {
if (required_sum == 0)
return 1;
else
return 0;
}
if (v[i][required_sum + Base])
return dp[i][required_sum + Base];
v[i][required_sum + Base] = true;
dp[i][required_sum + Base]
= findCnt(arr, i + 1,
required_sum, n)
+ findCnt(arr, i + 1,
required_sum - arr[i], n);
return dp[i][required_sum + Base];
}
static void countSubsets(int[] arr, int K,
int n)
{
int sum = 0;
for (int i = 0; i < n; i++)
{
sum += arr[i];
}
int S1 = (sum + K) / 2;
System.out.print(findCnt(arr, 0, S1, n));
}
public static void main(String[] args)
{
int[] arr = { 1, 1, 2, 3 };
int N = arr.length;
int K = 1;
countSubsets(arr, K, N);
}
}
|
Python3
maxN = 20;
maxSum = 50;
minSum = 50;
Base = 50;
dp = [[0 for i in range(maxSum + minSum)] for j in range(maxN)];
v = [[False for i in range(maxSum + minSum)] for j in range(maxN)];
def findCnt(arr, i, required_sum, n):
if (i == n):
if (required_sum == 0):
return 1;
else:
return 0;
if (v[i][required_sum + Base]):
return dp[i][required_sum + Base];
v[i][required_sum + Base] = True;
dp[i][required_sum + Base] = findCnt(arr, i + 1, required_sum, n)\
+ findCnt(arr, i + 1, required_sum - arr[i], n);
return dp[i][required_sum + Base];
def countSubsets(arr, K, n):
sum = 0;
for i in range(n):
sum += arr[i];
S1 = (sum + K) // 2;
print(findCnt(arr, 0, S1, n));
if __name__ == '__main__':
arr = [1, 1, 2, 3];
N = len(arr);
K = 1;
countSubsets(arr, K, N);
|
C#
using System;
class GFG {
static int maxN = 20;
static int maxSum = 50;
static int minSum = 50;
static int Base = 50;
static int[,] dp = new int[maxN, maxSum + minSum];
static bool[,] v = new bool[maxN, maxSum + minSum];
static int findCnt(int[] arr, int i,
int required_sum,
int n)
{
if (i == n) {
if (required_sum == 0)
return 1;
else
return 0;
}
if (v[i, required_sum + Base])
return dp[i, required_sum + Base];
v[i,required_sum + Base] = true;
dp[i,required_sum + Base]
= findCnt(arr, i + 1,
required_sum, n)
+ findCnt(arr, i + 1,
required_sum - arr[i], n);
return dp[i,required_sum + Base];
}
static void countSubsets(int[] arr, int K,
int n)
{
int sum = 0;
for (int i = 0; i < n; i++)
{
sum += arr[i];
}
int S1 = (sum + K) / 2;
Console.Write(findCnt(arr, 0, S1, n));
}
static void Main()
{
int[] arr = { 1, 1, 2, 3 };
int N = arr.Length;
int K = 1;
countSubsets(arr, K, N);
}
}
|
Javascript
<script>
var maxN = 20;
var maxSum = 50;
var minSum = 50;
var base = 50;
var dp = Array.from(Array(maxN),()=> Array(maxSum+minSum));
var v = Array.from(Array(maxN), ()=> Array(maxSum+minSum));
function findCnt(arr, i, required_sum, n)
{
if (i == n) {
if (required_sum == 0)
return 1;
else
return 0;
}
if (v[i][required_sum + base])
return dp[i][required_sum + base];
v[i][required_sum + base] = 1;
dp[i][required_sum + base]
= findCnt(arr, i + 1,
required_sum, n)
+ findCnt(arr, i + 1,
required_sum - arr[i], n);
return dp[i][required_sum + base];
}
function countSubsets(arr, K, n)
{
var sum = 0;
for (var i = 0; i < n; i++) {
sum += arr[i];
}
var S1 = (sum + K) / 2;
document.write( findCnt(arr, 0, S1, n));
}
var arr = [ 1, 1, 2, 3 ];
var N = arr.length;
var K = 1;
countSubsets(arr, K, N);
</script>
|
Time Complexity: O(N*K)
Auxiliary Space: O(N*K)
Efficient approach : Using DP Tabulation method ( Iterative approach )
The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memorization(top-down) because memorization method needs extra stack space of recursion calls.
Steps to solve this problem :
- Calculate the total sum of elements in the given array.
- Calculate the value of S1 using the formula (sum + K) / 2, where K is the given difference.
- Create a 2D dp array of size (n+1)x(S1+1), where n is the size of the given array.
- Initialize the base case of dp[i][0] as 1 for all i from 0 to n.
- Fill the dp array using the choice diagram approach. For each element in the array, we have two choices: include it or exclude it. If we include it, we reduce the sum by the value of that element, and if we exclude it, we leave the sum as it is. So, we update the dp array accordingly.
- Return the value of dp[n][S1], which represents the count of subsets with the required sum S1.
Implementation :
C++
#include <bits/stdc++.h>
using namespace std;
int countSubsets(int* arr, int K, int n) {
int sum = 0;
for (int i = 0; i < n; i++)
sum += arr[i];
int S1 = (sum + K) / 2;
vector<vector<int>> dp(n+1, vector<int>(S1+1, 0));
for (int i = 0; i <= n; i++)
dp[i][0] = 1;
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= S1; j++) {
if (arr[i-1] <= j) {
dp[i][j] = dp[i-1][j] + dp[i-1][j-arr[i-1]];
}
else {
dp[i][j] = dp[i-1][j];
}
}
}
return dp[n][S1];
}
int main() {
int arr[] = {1, 1, 2, 3};
int N = sizeof(arr) / sizeof(int);
int K = 1;
cout << countSubsets(arr, K, N) << endl;
return 0;
}
|
Java
import java.util.*;
class Main
{
static int countSubsets(int[] arr, int K, int n)
{
int sum = 0;
for (int i = 0; i < n; i++)
sum += arr[i];
int S1 = (sum + K) / 2;
int[][] dp = new int[n+1][S1+1];
for (int i = 0; i <= n; i++)
dp[i][0] = 1;
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= S1; j++) {
if (arr[i-1] <= j) {
dp[i][j] = dp[i-1][j] + dp[i-1][j-arr[i-1]];
}
else {
dp[i][j] = dp[i-1][j];
}
}
}
return dp[n][S1];
}
public static void main(String[] args) {
int[] arr = {1, 1, 2, 3};
int N = arr.length;
int K = 1;
System.out.println(countSubsets(arr, K, N));
}
}
|
Python3
def countSubsets(arr, K, n):
sum = 0
for i in range(n):
sum += arr[i]
S1 = (sum + K) // 2
dp = [[0 for j in range(S1+1)] for i in range(n+1)]
for i in range(n+1):
dp[i][0] = 1
for i in range(1, n+1):
for j in range(S1+1):
if arr[i-1] <= j:
dp[i][j] = dp[i-1][j] + dp[i-1][j-arr[i-1]]
else:
dp[i][j] = dp[i-1][j]
return dp[n][S1]
if __name__ == '__main__':
arr = [1, 1, 2, 3]
N = len(arr)
K = 1
print(countSubsets(arr, K, N))
|
C#
using System;
class GFG {
static int countSubsets(int[] arr, int K, int n)
{
int sum = 0;
for (int i = 0; i < n; i++)
sum += arr[i];
int S1 = (sum + K) / 2;
int[, ] dp = new int[n + 1, S1 + 1];
for (int i = 0; i <= n; i++)
dp[i, 0] = 1;
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= S1; j++) {
if (arr[i - 1] <= j) {
dp[i, j] = dp[i - 1, j]
+ dp[i - 1, j - arr[i - 1]];
}
else {
dp[i, j] = dp[i - 1, j];
}
}
}
return dp[n, S1];
}
static void Main()
{
int[] arr = { 1, 1, 2, 3 };
int N = arr.Length;
int K = 1;
Console.WriteLine(countSubsets(arr, K, N));
}
}
|
Javascript
function countSubsets(arr, K, n)
{
let sum = 0;
for (let i = 0; i < n; i++) {
sum += arr[i];
}
let S1 = Math.floor((sum + K) / 2);
let dp = new Array(n + 1).fill().map(() => new Array(S1 + 1).fill(0));
for (let i = 0; i <= n; i++) {
dp[i][0] = 1;
}
for (let i = 1; i <= n; i++) {
for (let j = 0; j <= S1; j++) {
if (arr[i - 1] <= j) {
dp[i][j] = dp[i - 1][j] + dp[i - 1][j - arr[i - 1]];
} else {
dp[i][j] = dp[i - 1][j];
}
}
}
return dp[n][S1];
}
const arr = [1, 1, 2, 3];
const N = arr.length;
const K = 1;
console.log(countSubsets(arr, K, N));
|
Time Complexity: O(N*k)
Auxiliary Space: O(N*K)
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