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Check if K can be obtained by performing arithmetic operations on any permutation of an Array
  • Last Updated : 24 Sep, 2020

Given an array arr[] of N integers and an integer K, the task is to check if the expression formed for any permutation of the given array after assigning arithmetic operators(+, -, /, *) gives the value K or not. If found to be true, then print the order of operations performed. Otherwise, print “-1”.
Note: After division operator ‘/’ is applied, the result can be a floating-point number. 

Examples: 

Input: arr[] = {2, 0, 0, 2}, K = 4
Output: 2 + 0 => 2 + 2 => 4 + 0 => 4

Input: arr[] = {1, 2, 4, 7}, K = 50
Output: -1

Approach: The idea is to use Backtracking and find the value of the resultant equation by generating all possible combinations of assigning the operators. Below are the steps: 



  1. Traverse the array And choose the first two elements and apply every possible operation. Let the result of this above operation be X.
  2. Now, perform all possible operations on X and the next element of the array and so on.
  3. Repeat all the above steps until all the possible combinations of operations are performed.
  4. Now, if any combination of operations generates the result K, then print that combination.
  5. Otherwise, print “-1”.

Below is the implementation of the above approach: 

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function that finds the string of
// possible combination of operations
bool find(vector<double>& v, int n,
          double dest, string s)
{
    // If only one number left then
    // check for result
    if (n == 1) {
 
        // If resultant value is K
        if (abs(v[0] - dest) <= 0.0000001) {
 
            cout << s + to_string(int(dest))
                 << " ";
            return 1;
        }
 
        // Else return 0
        return 0;
    }
 
    // Choose all combination of numbers
    // and operators and operate them
    for (int i = 0; i < n; i++) {
 
        for (int j = i + 1; j < n; j++) {
 
            double a = v[i], b = v[j];
 
            // Choose the first two and
            // operate it with '+'
            string p = s;
            v[i] = a + b;
 
            // Place it to 0th position
            v[j] = v[n - 1];
 
            // Place (n-1)th element on
            // 1st position
            s = (s + to_string(int(a)) + '+'
                 + to_string(int(b))
                 + " => ");
 
            // Evaluate the expression
            // with current combination
            if (find(v, n - 1, dest, s))
                return 1;
 
            // Now, we have N - 1 elements
            s = p;
 
            v[i] = a - b;
 
            // Try '-' operation
            v[j] = v[n - 1];
 
            s = (s + to_string(int(a)) + '-'
                 + to_string(int(b))
                 + " => ");
 
            // Evaluate the expression
            // with current combination
            if (find(v, n - 1, dest, s))
                return 1;
            s = p;
 
            v[i] = b - a;
 
            // Try reverse '-'
            v[j] = v[n - 1];
 
            s = (s + to_string(int(b)) + '-'
                 + to_string(int(a))
                 + " => ");
 
            // Evaluate the expression
            // with current combination
            if (find(v, n - 1, dest, s))
                return 1;
            s = p;
 
            v[i] = a * b;
 
            // Try '*' operation
            v[j] = v[n - 1];
            s = (s + to_string(int(a))
                 + '*'
                 + to_string(int(b))
                 + " => ");
 
            // Evaluate the expression
            // with current combination
            if (find(v, n - 1, dest, s))
                return 1;
 
            s = p;
 
            if (b != 0) {
 
                v[i] = a / b;
 
                // Try '/'
                v[j] = v[n - 1];
 
                s = (s + to_string(int(a))
                     + '/'
                     + to_string(int(b))
                     + " => ");
 
                // Evaluate the expression
                // with current combination
                if (find(v, n - 1, dest, s))
                    return 1;
            }
 
            s = p;
 
            if (a != 0) {
 
                v[i] = b / a;
 
                // Try reverse '/'
                v[j] = v[n - 1];
                s = (s + to_string(int(b))
                     + '/'
                     + to_string(int(a))
                     + " => ");
 
                // Evaluate the expression
                // with current combination
                if (find(v, n - 1, dest, s))
                    return 1;
            }
 
            s = p;
 
            // Backtracking Step
            v[i] = a;
            v[j] = b;
        }
    }
 
    // Return 0 if there doesnt exist
    // any combination that gives K
    return 0;
}
 
// Function that finds the possible
// combination of operation to get the
// resultant value K
void checkPossibleOperation(
    vector<double>& arr, double K)
{
    // Store the resultant operation
    string s = "";
 
    // Function Call
    if (!find(arr, arr.size(), K, s)) {
        cout << "-1";
    }
}
 
// Driver Code
int main()
{
    // Given array arr[]
    vector<double> arr = { 2, 0, 0, 2 };
 
    // Resultant value K
    double K = 4;
 
    // Function Call
    checkPossibleOperation(arr, K);
 
    return 0;
}

Java




// Java program for above approach
class GFG{
     
// Function that finds the string of
// possible combination of operations
static boolean find(double[] v, int n,
                    double dest, String s)
{
     
    // If only one number left then
    // check for result
    if (n == 1)
    {
         
        // If resultant value is K
        if (Math.abs(v[0] - dest) <= 0.0000001)
        {
            System.out.println(s +
            String.valueOf((int)dest) + " ");
                     
            return true;
        }
 
    // Else return 0
    return false;
    }
     
    // Choose all combination of numbers
    // and operators and operate them
    for(int i = 0; i < n; i++)
    {
        for(int j = i + 1; j < n; j++)
        {
            double a = v[i], b = v[j];
             
            // Choose the first two and
            // operate it with '+'
            String p = s;
            v[i] = a + b;
 
            // Place it to 0th position
            v[j] = v[n - 1];
 
            // Place (n-1)th element on
            // 1st position
            s = (s + String.valueOf((int)a) +
               '+' + String.valueOf((int)b) +
                " => ");
             
            // Evaluate the expression
            // with current combination
            if (find(v, n - 1, dest, s))
                return true;
 
            // Now, we have N - 1 elements
            s = p;
 
            v[i] = a - b;
 
            // Try '-' operation
            v[j] = v[n - 1];
 
            s = (s + String.valueOf((int)a) +
               '-' + String.valueOf((int)b) +
                " => ");
 
            // Evaluate the expression
            // with current combination
            if (find(v, n - 1, dest, s))
                return true;
                 
            s = p;
            v[i] = b - a;
 
            // Try reverse '-'
            v[j] = v[n - 1];
 
            s = (s + String.valueOf((int)b) +
               '-' + String.valueOf((int)a) +
                " => ");
 
            // Evaluate the expression
            // with current combination
            if (find(v, n - 1, dest, s))
                return true;
                 
            s = p;
            v[i] = a * b;
 
            // Try '*' operation
            v[j] = v[n - 1];
            s = (s + String.valueOf((int)a) +
               '*' + String.valueOf((int)b) +
                " => ");
 
            // Evaluate the expression
            // with current combination
            if (find(v, n - 1, dest, s))
                return true;
 
            s = p;
 
            if (b != 0)
            {
                v[i] = a / b;
 
                // Try '/'
                v[j] = v[n - 1];
 
                s = (s + String.valueOf((int)a) +
                   '/' + String.valueOf((int)b) +
                    " => ");
 
                // Evaluate the expression
                // with current combination
                if (find(v, n - 1, dest, s))
                    return true;
            }
            s = p;
 
            if (a != 0)
            {
                v[i] = b / a;
 
                // Try reverse '/'
                v[j] = v[n - 1];
                s = (s + String.valueOf((int)b) +
                   '/' + String.valueOf((int)a) +
                    " => ");
 
                // Evaluate the expression
                // with current combination
                if (find(v, n - 1, dest, s))
                    return true;
            }
            s = p;
 
            // Backtracking Step
            v[i] = a;
            v[j] = b;
        }
    }
     
    // Return 0 if there doesnt exist
    // any combination that gives K
    return false;
}
 
// Function that finds the possible
// combination of operation to get the
// resultant value K
static void checkPossibleOperation(double[] arr,
                                   double K)
{
     
    // Store the resultant operation
    String s = "";
 
    // Function call
    if (!find(arr, arr.length, K, s))
    {
        System.out.println("-1");
    }
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array arr[]
    double[] arr = { 2, 0, 0, 2 };
     
    // Resultant value K
    double K = 4;
 
    // Function call
    checkPossibleOperation(arr, K);
}
}
 
// This code is contributed by Dadi Madhav

Python3




# Python3 program for the above approach
# Function that finds the string of
# possible combination of operations
def find(v, n, dest, s):
   
    # If only one number left then
    # check for result
    if (n == 1):
 
        # If resultant value is K
        if (abs(v[0] - dest) <= 0.0000001):
            print (s + str(int(dest)) ,end = " ")
            return 1
 
        #;Else return 0
        return 0
 
    # Choose all combination of numbers
    # and operators and operate them
    for i in range (n):
        for j in range (i + 1,  n):
            a = v[i]
            b = v[j]
 
            # Choose the first two and
            # operate it with '+'
            p = s
            v[i] = a + b
 
            # Place it to 0th position
            v[j] = v[n - 1]
 
            # Place (n-1)th element on
            # 1st position
            s = (s + str(int(a)) + '+' +
                 str(int(b)) + " => ")
 
            # Evaluate the expression
            # with current combination
            if (find(v, n - 1, dest, s)):
                return 1
 
            # Now, we have N - 1 elements
            s = p
 
            v[i] = a - b
 
            # Try '-' operation
            v[j] = v[n - 1]
 
            s = (s + str(int(a)) + '-' +
                 str(int(b)) +" => ")
 
            # Evaluate the expression
            # with current combination
            if (find(v, n - 1, dest, s)):
                return 1
            s = p
            v[i] = b - a
 
            # Try reverse '-'
            v[j] = v[n - 1]
 
            s = (s + str(int(b)) + '-' +
                 str(int(a)) + " => ")
 
            # Evaluate the expression
            # with current combination
            if (find(v, n - 1, dest, s)):
                return 1
            s = p
            v[i] = a * b
 
            # Try '*' operation
            v[j] = v[n - 1]
            s = (s + str(int(a)) + '*' +
                 str(int(b)) + " => ");
 
            # Evaluate the expression
            # with current combination
            if (find(v, n - 1, dest, s)):
                return 1
            s = p
             
            if (b != 0):
                v[i] = a // b
 
                # Try '/'
                v[j] = v[n - 1]
 
                s = (s + str(int(a)) + '/' +
                     str(int(b)) + " => ")
 
                # Evaluate the expression
                # with current combination
                if (find(v, n - 1, dest, s)):
                    return 1
            s = p
 
            if (a != 0):
                v[i] = b // a
 
                # Try reverse '/'
                v[j] = v[n - 1]
                s = (s + str(int(b)) + '/' +
                     str(int(a)) + " => ")
 
                # Evaluate the expression
                # with current combination
                if (find(v, n - 1, dest, s)):
                    return 1         
            s = p
 
            # Backtracking Step
            v[i] = a
            v[j] = b
 
    # Return 0 if there doesnt exist
    # any combination that gives K
    return 0
 
# Function that finds the possible
# combination of operation to get the
# resultant value K
def checkPossibleOperation(arr, K):
 
    # Store the resultant operation
    s = ""
   
    # Function Call
    if (not find(arr, len(arr), K, s)):
        print ("-1")
 
# Driver Code
if __name__ == "__main__":
 
    # Given array arr[]
    arr = [2, 0, 0, 2]
 
    # Resultant value K
    K = 4
 
    # Function Call
    checkPossibleOperation(arr, K)
 
#This code is contributed by Chitranayal

C#




// C# program for
// the above approach
using System;
class GFG{
     
// Function that finds the string of
// possible combination of operations
static bool find(double[] v, int n,
                 double dest, String s)
{
  // If only one number left then
  // check for result
  if (n == 1)
  {
    // If resultant value is K
    if (Math.Abs(v[0] - dest) <= 0.0000001)
    {
      Console.WriteLine(s + String.Join("",
                                        (int)dest) + " ");
      return true;
    }
 
    // Else return 0
    return false;
  }
 
  // Choose all combination of numbers
  // and operators and operate them
  for(int i = 0; i < n; i++)
  {
    for(int j = i + 1; j < n; j++)
    {
      double a = v[i], b = v[j];
 
      // Choose the first two and
      // operate it with '+'
      String p = s;
      v[i] = a + b;
 
      // Place it to 0th position
      v[j] = v[n - 1];
 
      // Place (n-1)th element on
      // 1st position
      s = (s + String.Join("", (int)a) +
           '+' + String.Join("", (int)b) +
           " => ");
 
      // Evaluate the expression
      // with current combination
      if (find(v, n - 1, dest, s))
        return true;
 
      // Now, we have N - 1 elements
      s = p;
 
      v[i] = a - b;
 
      // Try '-' operation
      v[j] = v[n - 1];
 
      s = (s + String.Join("", (int)a) +
           '-' + String.Join("", (int)b) +
           " => ");
 
      // Evaluate the expression
      // with current combination
      if (find(v, n - 1, dest, s))
        return true;
 
      s = p;
      v[i] = b - a;
 
      // Try reverse '-'
      v[j] = v[n - 1];
 
      s = (s + String.Join("", (int)b) +
           '-' + String.Join("", (int)a) +
           " => ");
 
      // Evaluate the expression
      // with current combination
      if (find(v, n - 1, dest, s))
        return true;
 
      s = p;
      v[i] = a * b;
 
      // Try '*' operation
      v[j] = v[n - 1];
      s = (s + String.Join("", (int)a) +
           '*' + String.Join("", (int)b) +
           " => ");
 
      // Evaluate the expression
      // with current combination
      if (find(v, n - 1, dest, s))
        return true;
 
      s = p;
 
      if (b != 0)
      {
        v[i] = a / b;
 
        // Try '/'
        v[j] = v[n - 1];
 
        s = (s + String.Join("", (int)a) +
             '/' + String.Join("", (int)b) +
             " => ");
 
        // Evaluate the expression
        // with current combination
        if (find(v, n - 1, dest, s))
          return true;
      }
      s = p;
 
      if (a != 0)
      {
        v[i] = b / a;
 
        // Try reverse '/'
        v[j] = v[n - 1];
        s = (s + String.Join("", (int)b) +
             '/' + String.Join("", (int)a) +
             " => ");
 
        // Evaluate the expression
        // with current combination
        if (find(v, n - 1, dest, s))
          return true;
      }
      s = p;
 
      // Backtracking Step
      v[i] = a;
      v[j] = b;
    }
  }
 
  // Return 0 if there doesnt exist
  // any combination that gives K
  return false;
}
 
// Function that finds the possible
// combination of operation to get the
// resultant value K
static void checkPossibleOperation(double[] arr,
                                   double K)
{
  // Store the resultant operation
  String s = "";
 
  // Function call
  if (!find(arr, arr.Length, K, s))
  {
    Console.WriteLine("-1");
  }
}
 
// Driver Code
public static void Main(String[] args)
{   
  // Given array []arr
  double[] arr = {2, 0, 0, 2};
 
  // Resultant value K
  double K = 4;
 
  // Function call
  checkPossibleOperation(arr, K);
}
}
 
// This code is contributed by Princi Singh
Output: 
2+0 => 2+2 => 4+0 => 4


Time Complexity: O(N!*4N)
Auxiliary Space: O(N)

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