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++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the number of ways to divide
// the array into two subsets and such that the
// difference between their sums is equal to diff
int countSubset(int arr[], int n, int diff)
{
// Store the sum of the set S1
int sum = 0;
for (int i = 0; i < n; i++)
sum += arr[i];
sum += diff;
sum = sum / 2;
// Initializing the matrix
int t[n + 1][sum + 1];
// Number of ways to get sum
// using 0 elements is 0
for (int j = 0; j <= sum; j++)
t[0][j] = 0;
// Number of ways to get sum 0
// using i elements is 1
for (int i = 0; i <= n; i++)
t[i][0] = 1;
// Traverse the 2D array
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= sum; j++) {
// If the value is greater
// than the sum store the
// value of previous state
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 the result
return t[n][sum];
}
// Driver Code
int main()
{
// Given Input
int diff = 1, n = 4;
int arr[] = { 1, 1, 2, 3 };
// Function Call
cout << countSubset(arr, n, diff);
}
// Java program for the above approach
import java.io.*;
public class GFG
{
// Function to count the number of ways to divide
// the array into two subsets and such that the
// difference between their sums is equal to diff
static int countSubset(int []arr, int n, int diff)
{
// Store the sum of the set S1
int sum = 0;
for (int i = 0; i < n; i++)
sum += arr[i];
sum += diff;
sum = sum / 2;
// Initializing the matrix
int t[][] = new int[n + 1][sum + 1];
// Number of ways to get sum
// using 0 elements is 0
for (int j = 0; j <= sum; j++)
t[0][j] = 0;
// Number of ways to get sum 0
// using i elements is 1
for (int i = 0; i <= n; i++)
t[i][0] = 1;
// Traverse the 2D array
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= sum; j++) {
// If the value is greater
// than the sum store the
// value of previous state
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 the result
return t[n][sum];
}
// Driver Code
public static void main(String[] args)
{
// Given Input
int diff = 1, n = 4;
int arr[] = { 1, 1, 2, 3 };
// Function Call
System.out.print(countSubset(arr, n, diff));
}
}
// This code is contributed by AnkThon
# Python3 program for the above approach
# Function to count the number of ways to divide
# the array into two subsets and such that the
# difference between their sums is equal to diff
def countSubset(arr, n, diff):
# Store the sum of the set S1
sum = 0
for i in range(n):
sum += arr[i]
sum += diff
sum = sum // 2
# Initializing the matrix
t = [[0 for i in range(sum + 1)]
for i in range(n + 1)]
# Number of ways to get sum
# using 0 elements is 0
for j in range(sum + 1):
t[0][j] = 0
# Number of ways to get sum 0
# using i elements is 1
for i in range(n + 1):
t[i][0] = 1
# Traverse the 2D array
for i in range(1, n + 1):
for j in range(1, sum + 1):
# If the value is greater
# than the sum store the
# value of previous state
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 the result
return t[n][sum]
# Driver Code
if __name__ == '__main__':
# Given Input
diff, n = 1, 4
arr = [ 1, 1, 2, 3 ]
# Function Call
print (countSubset(arr, n, diff))
# This code is contributed by mohit kumar 29
// C# program for the above approach
using System;
public class GFG
{
// Function to count the number of ways to divide
// the array into two subsets and such that the
// difference between their sums is equal to diff
static int countSubset(int []arr, int n, int diff)
{
// Store the sum of the set S1
int sum = 0;
for (int i = 0; i < n; i++)
sum += arr[i];
sum += diff;
sum = sum / 2;
// Initializing the matrix
int [,]t = new int[n + 1, sum + 1];
// Number of ways to get sum
// using 0 elements is 0
for (int j = 0; j <= sum; j++)
t[0,j] = 0;
// Number of ways to get sum 0
// using i elements is 1
for (int i = 0; i <= n; i++)
t[i,0] = 1;
// Traverse the 2D array
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= sum; j++) {
// If the value is greater
// than the sum store the
// value of previous state
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 the result
return t[n,sum];
}
// Driver Code
public static void Main(string[] args)
{
// Given Input
int diff = 1, n = 4;
int []arr = { 1, 1, 2, 3 };
// Function Call
Console.Write(countSubset(arr, n, diff));
}
}
// This code is contributed by AnkThon
<script>
// JavaScript program for the above approach
// Function to count the number of ways to divide
// the array into two subsets and such that the
// difference between their sums is equal to diff
function countSubset(arr, n, diff)
{
// Store the sum of the set S1
var sum = 0;
for (var i = 0; i < n; i++){
sum += arr[i];
}
sum += diff;
sum = sum / 2;
// Initializing the matrix
//int t[n + 1][sum + 1];
var t = new Array(n + 1);
// Loop to create 2D array using 1D array
for (var i = 0; i < t.length; i++) {
t[i] = new Array(sum + 1);
}
// Loop to initialize 2D array elements.
for (var i = 0; i < t.length; i++) {
for (var j = 0; j < t[i].length; j++) {
t[i][j] = 0;
}
}
// Number of ways to get sum
// using 0 elements is 0
for (var j = 0; j <= sum; j++)
t[0][j] = 0;
// Number of ways to get sum 0
// using i elements is 1
for (var i = 0; i <= n; i++)
t[i][0] = 1;
// Traverse the 2D array
for (var i = 1; i <= n; i++) {
for (var j = 1; j <= sum; j++) {
// If the value is greater
// than the sum store the
// value of previous state
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 the result
return t[n][sum];
}
// Driver Code
// Given Input
var diff = 1;
var n = 4;
var arr = [ 1, 1, 2, 3 ];
// Function Call
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++
// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the number of ways to divide
// the array into two subsets and such that the
// difference between their sums is equal to diff
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;
// Initializing the vector Dp
int dp[sum + 1] = {0};
// Base Case
dp[0] = 1;
// iterate over subproblems to get the current computation
for (int i = 0; i < n; i++) {
for (int j = sum; j >= arr[i]; j--) {
// update DP from previous values
dp[j] += dp[j - arr[i]];
}
}
// return answer
return dp[sum];
}
// Driver Code
int main()
{
// Given Input
int diff = 1, n = 4;
int arr[] = { 1, 1, 2, 3 };
// Function Call
cout << countSubset(arr, n, diff);
}
// this code is contributed by bhardwajji
// Java program for above approach
import java.util.*;
public class Main
{
// Function to count the number of ways to divide
// the array into two subsets and such that the
// difference between their sums is equal to diff
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;
// Initializing the vector Dp
int dp[] = new int[sum + 1];
// Base Case
dp[0] = 1;
// iterate over subproblems to get the current computation
for (int i = 0; i < n; i++) {
for (int j = sum; j >= arr[i]; j--) {
// update DP from previous values
dp[j] += dp[j - arr[i]];
}
}
// return answer
return dp[sum];
}
// Driver Code
public static void main(String[] args)
{
// Given Input
int diff = 1, n = 4;
int arr[] = { 1, 1, 2, 3 };
// Function Call
System.out.println(countSubset(arr, n, diff));
}
}
# Function to count the number of ways to divide
# the array into two subsets and such that the
# difference between their sums is equal to diff
def countSubset(arr, n, diff):
# Calculating the sum of all elements
sum = 0
for i in range(n):
sum += arr[i]
sum += diff
# If sum is odd, then no such subset can exist
if sum % 2 != 0:
return 0
# Initializing the vector Dp
dp = [0] * (sum // 2 + 1)
# Base Case
dp[0] = 1
# iterate over subproblems to get the current computation
for i in range(n):
for j in range(sum // 2, arr[i] - 1, -1):
# update DP from previous values
dp[j] += dp[j - arr[i]]
# return answer
return dp[sum // 2]
# Given Input
diff = 1
n = 4
arr = [1, 1, 2, 3]
# Function Call
print(countSubset(arr, n, diff))
using System;
class MainClass
{
// Function to count the number of ways to divide
// the array into two subsets and such that the
// difference between their sums is equal to diff
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;
// Initializing the vector Dp
int[] dp = new int[sum + 1];
// Base Case
dp[0] = 1;
// Iterate over subproblems to get the current
// computation
for (int i = 0; i < n; i++) {
for (int j = sum; j >= arr[i]; j--) {
// Update DP from previous values
dp[j] += dp[j - arr[i]];
}
}
// Return answer
return dp[sum];
}
// Driver Code
public static void Main()
{
// Given Input
int diff = 1, n = 4;
int[] arr = { 1, 1, 2, 3 };
// Function Call
Console.WriteLine(countSubset(arr, n, diff));
}
}
// Function to count the number of ways to divide
// the array into two subsets and such that the
// difference between their sums is equal to diff
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);
// Initializing the vector Dp
let dp = new Array(sum + 1).fill(0);
// Base Case
dp[0] = 1;
// Iterate over subproblems to get the current
// computation
for (let i = 0; i < n; i++) {
for (let j = sum; j >= arr[i]; j--) {
// Update DP from previous values
dp[j] += dp[j - arr[i]];
}
}
// Return answer
return dp[sum];
}
// Driver Code
let diff = 1,
n = 4;
let arr = [1, 1, 2, 3];
// Function Call
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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