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 => 4Input: 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:
- Traverse the array And choose the first two elements and apply every possible operation. Let the result of this above operation be X.
- Now, perform all possible operations on X and the next element of the array and so on.
- Repeat all the above steps until all the possible combinations of operations are performed.
- Now, if any combination of operations generates the result K, then print that combination.
- 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 operationsbool 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 Kvoid checkPossibleOperation( vector<double>& arr, double K){ // Store the resultant operation string s = ""; // Function Call if (!find(arr, arr.size(), K, s)) { cout << "-1"; }}// Driver Codeint 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 approachclass GFG{ // Function that finds the string of// possible combination of operationsstatic 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 Kstatic 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 Codepublic 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 operationsdef 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 Kdef checkPossibleOperation(arr, K): # Store the resultant operation s = "" # Function Call if (not find(arr, len(arr), K, s)): print ("-1")# Driver Codeif __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 approachusing System;class GFG{ // Function that finds the string of// possible combination of operationsstatic 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 Kstatic 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 Codepublic 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 |
Javascript
<script>// Javascript program for above approach// Function that finds the string of// possible combination of operationsfunction find(v,n,dest,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) { document.write(s + (dest).toString() + " <br>"); return true; } // Else return 0 return false; } // Choose all combination of numbers // and operators and operate them for(let i = 0; i < n; i++) { for(let j = i + 1; j < n; j++) { let a = v[i], b = v[j]; // Choose the first two and // operate it with '+' let 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 + a.toString() + '+' + b.toString() + " => "); // 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 + a.toString() + '-' + b.toString() + " => "); // 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 + b.toString() + '-' + a.toString() + " => "); // 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 + a.toString() + '*' + b.toString() + " => "); // 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 + a.toString() + '/' + b.toString() + " => "); // 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 + b.toString() + '/' + a.toString() + " => "); // 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 Kfunction checkPossibleOperation(arr,k){ // Store the resultant operation let s = ""; // Function call if (!find(arr, arr.length, K, s)) { document.write("-1"); }}// Driver Code// Given array arr[]let arr=[2, 0, 0, 2 ];// Resultant value Klet K = 4;// Function callcheckPossibleOperation(arr, K);// This code is contributed by avanitrachhadiya2155</script> |
Output:
2+0 => 2+2 => 4+0 => 4
Time Complexity: O(N!*4N)
Auxiliary Space: O(N)
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