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Count ways to place ‘+’ and ‘-‘ in front of array elements to obtain sum K
  • Last Updated : 24 Jun, 2021

Given an array A[] consisting of N non-negative integers, and an integer K, the task is to find the number of ways ‘+’ and ‘-‘ operators can be placed in front of elements of the array A[] such that the sum of the array becomes K.

Examples:

Input: A[] = {1, 1, 2, 3}, N = 4, K = 1
Output: 3
Explanation: Three possible ways are:

  1. + 1 + 1 + 2 – 3 = 1
  2. + 1 – 1 – 2 + 3 = 1
  3. – 1 + 1 – 1 + 3 = 1

Input: A[] = {1, 1, 1, 1, 1}, N = 5, K = 3
Output: 5

 

Approach: The problem can be solved based on the following observations:



  1. Store the sum of elements having ‘+’ in front of that element and ‘-‘ in front of that element in variables, say P1 and P2, such that the sum of the array becomes K.
  2. Store the total sum of the array A[] in a variable, say K.
  3. Therefore, following equations arises:
    1. P1 + P2 = sum
    2. P1 – P2 = K
       
  4. Solving the above equations obtains P1 = (sum + K) / 2.
  5. Therefore, the problem has transformed into finding the number of subsets with sum P1.
  6. If an element of A is equal to 0, both ‘+’ and ‘-‘ operators work in valid arrangements, thus, the 0s can be safely ignored and separately calculated.

Hence, the problem can be solved using Dynamic Programming. Follow the steps below to solve the problem:

  1. Calculate and store the sum of elements of the array A[] and the number of 0s in A[] in variables sum and c respectively.
  2. If K is greater than sum or (sum + K) is odd, return 0.
  3. Add K to sum and divide it by 2, i.e. sum = (sum + K) / 2, which is the required sum. Find the number of subsets equal to that sum.
  4. Create a 2D dp array of dimensions N*sum. where dp[i][j] represents the number of subsets up to i-1 that have sum j.
  5. The base cases required to be considered are as follows:
    1. dp[0][i] = 0, for 0 <= i <= sum, as no elements from the array A[] has been considered
    2. dp[i][0] = 1, for 0 <= i <= N, as obtaining a sum 0 is always possible.
  6. Iterate from 1 to N, and for each current index i, perform the following operations:
    1. Iterate from 1 to sum and for each current index j, perform the following transitions:
      1. If A[i – 1] is less than j and A[i – 1] is not equal to 0, set dp[i][j] = dp[i – 1][j] + dp[i – 1][j – A[i – 1]].
      2. Otherwise, copy the previous state, i.e. dp[i][j] = dp[i – 1][j].
  7. Finally, return the product of dp[N][sum] and 2c (to account for the 0s) i.e dp[N][sum]*2c.

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 number of ways
// '+' and '-' operators can be placed
// in front of array elements to make
// the sum of array elements equal to K
int solve(int A[], int N, int K)
{
    // Stores sum of the array
    int sum = 0;
 
    // Stores count of 0s in A[]
    int c = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Update sum
        sum += A[i];
 
        // Update count of 0s
        if (A[i] == 0)
            c++;
    }
    // Conditions where no arrangements
    // are possible which adds up to K
    if (K > sum || (sum + K) % 2)
        return 0;
 
    // Required sum
    sum = (sum + K) / 2;
 
    // Dp array
    int dp[N + 1][sum + 1];
    // Base cases
    for (int i = 0; i <= sum; i++)
        dp[0][i] = 0;
 
    for (int i = 0; i <= N; i++)
        dp[i][0] = 1;
 
    // Fill the dp array
    for (int i = 1; i <= N; i++) {
        for (int j = 1; j <= sum; j++) {
 
            if (A[i - 1] <= j && A[i - 1])
                dp[i][j] = dp[i - 1][j]
                           + dp[i - 1][j - A[i - 1]];
            else
                dp[i][j] = dp[i - 1][j];
        }
    }
 
    // Return answer
    return dp[N][sum] + pow(2, c);
}
 
// Driver Code
int main()
{
    // Input
    int A[] = { 1, 1, 2, 3 };
    int N = sizeof(A) / sizeof(A[0]);
    int K = 3;
 
    // Function call
    cout << solve(A, N, K) << endl;
 
    return 0;
}


Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to count number of ways
// '+' and '-' operators can be placed
// in front of array elements to make
// the sum of array elements equal to K
static int solve(int A[], int N, int K)
{
   
    // Stores sum of the array
    int sum = 0;
 
    // Stores count of 0s in A[]
    int c = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Update sum
        sum += A[i];
 
        // Update count of 0s
        if (A[i] == 0)
            c++;
    }
    // Conditions where no arrangements
    // are possible which adds up to K
    if ((K > sum) || (((sum + K) % 2) != 0))
        return 0;
 
    // Required sum
    sum = (sum + K) / 2;
 
    // Dp array
    int dp[][] = new int[N + 1][sum + 1];
    // Base cases
    for (int i = 0; i <= sum; i++)
        dp[0][i] = 0;
 
    for (int i = 0; i <= N; i++)
        dp[i][0] = 1;
 
    // Fill the dp array
    for (int i = 1; i <= N; i++) {
        for (int j = 1; j <= sum; j++) {
 
            if ((A[i - 1] <= j)  && (A[i - 1] != 0))
                dp[i][j] = dp[i - 1][j]
                           + dp[i - 1][j - A[i - 1]];
            else
                dp[i][j] = dp[i - 1][j];
        }
    }
 
    // Return answer
    return dp[N][sum] + (int)Math.pow(2, c);
}
 
// Driver Code
public static void main(String[] args)
{
    // Input
    int A[] = { 1, 1, 2, 3 };
    int N = A.length;
    int K = 3;
 
    // Function call
    System.out.print(solve(A, N, K));
 
}
}
 
// This code is contributed by sanjoy_62.


Python3




# Python3 program for the above approach
 
# Function to count number of ways
# '+' and '-' operators can be placed
# in front of array elements to make
# the sum of array elements equal to K
def solve(A, N, K):
     
    # Stores sum of the array
    sum = 0
 
    # Stores count of 0s in A[]
    c = 0
 
    # Traverse the array
    for i in range(N):
         
        # Update sum
        sum += A[i]
 
        # Update count of 0s
        if (A[i] == 0):
            c += 1
 
    # Conditions where no arrangements
    # are possible which adds up to K
    if (K > sum or (sum + K) % 2):
        return 0
 
    # Required sum
    sum = (sum + K) // 2
 
    # Dp array
    dp = [[0 for i in range(sum + 1)]
             for j in range(N + 1)]
              
    # Base cases
    for i in range(sum + 1):
        dp[0][i] = 0
 
    for i in range(N + 1):
        dp[i][0] = 1
 
    # Fill the dp array
    for i in range(1, N + 1, 1):
        for j in range(1, sum + 1, 1):
            if (A[i - 1] <= j and A[i - 1]):
                dp[i][j] = (dp[i - 1][j] +
                            dp[i - 1][j - A[i - 1]])
            else:
                dp[i][j] = dp[i - 1][j]
 
    # Return answer
    return dp[N][sum] + pow(2, c)
 
# Driver Code
if __name__ == '__main__':
     
    # Input
    A = [ 1, 1, 2, 3 ]
    N = len(A)
    K = 3
 
    # Function call
    print(solve(A, N, K))
     
# This code is contributed by SURENDRA_GANGWAR


C#




// C# program for the above approach
using System;
 
public class GFG
{
 
  // Function to count number of ways
  // '+' and '-' operators can be placed
  // in front of array elements to make
  // the sum of array elements equal to K
  static int solve(int[] A, int N, int K)
  {
 
    // Stores sum of the array
    int sum = 0;
 
    // Stores count of 0s in A[]
    int c = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
      // Update sum
      sum += A[i];
 
      // Update count of 0s
      if (A[i] == 0)
        c++;
    }
    // Conditions where no arrangements
    // are possible which adds up to K
    if ((K > sum) || (((sum + K) % 2) != 0))
      return 0;
 
    // Required sum
    sum = (sum + K) / 2;
 
    // Dp array
    int[, ] dp= new int[N + 1, sum + 1];
 
    // Base cases
    for (int i = 0; i <= sum; i++)
      dp[0, i] = 0;
 
    for (int i = 0; i <= N; i++)
      dp[i,0] = 1;
 
    // Fill the dp array
    for (int i = 1; i <= N; i++) {
      for (int j = 1; j <= sum; j++) {
 
        if ((A[i - 1] <= j)  && (A[i - 1] != 0))
          dp[i,j] = dp[i - 1,j]
          + dp[i - 1,j - A[i - 1]];
        else
          dp[i,j] = dp[i - 1,j];
      }
    }
 
    // Return answer
    return dp[N, sum] + (int)Math.Pow(2, c);
  }
 
  // Driver code
  static public void Main ()
  {
 
    // Input
    int[] A = { 1, 1, 2, 3 };
    int N = A.Length;
    int K = 3;
 
    // Function call
    Console.Write(solve(A, N, K));
  }
}
 
// This code is contributed by offbeat


Output: 

3

 

Time Complexity: O(N * sum)
Auxiliary Space: O(N * sum)

 

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