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 diffint 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 Codeint main(){ // Given Input int diff = 1, n = 4; int arr[] = { 1, 1, 2, 3 }; // Function Call cout << countSubset(arr, n, diff);} |
Java
// Java program for the above approachimport 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
# 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 diffdef 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 Codeif __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#
// C# program for the above approachusing 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 |
Javascript
<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 difffunction 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 Inputvar diff = 1;var n = 4;var arr = [ 1, 1, 2, 3 ];// Function Calldocument.write(countSubset(arr, n, diff));</script> |
Output:
3
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 diffint 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 Codeint 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
// Java program for above approachimport 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)); }} |
Python3
# 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 diffdef 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 Inputdiff = 1n = 4arr = [1, 1, 2, 3]# Function Callprint(countSubset(arr, n, diff)) |
C#
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)); }} |
Javascript
// 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 difffunction 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 Codelet diff = 1, n = 4;let arr = [1, 1, 2, 3];// Function Callconsole.log(countSubset(arr, n, diff)); |
Output
3
Time Complexity: O(S*N), where S = sum of array elements + K/2
Auxiliary Space: O(S)
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