Check if sum of array can be made equal to X by removing either the first or last digits of every array element

Given an array arr[] consisting of N positive integers and a positive integer X, the task is to check if the sum of the given array can be made equal to X by removing either the first or last digit of every array element. If it is possible to make the sum of array elements equal to X, then print “Yes”. Otherwise, print “No”.

Examples:

Input: arr[] = {545, 433, 654, 23}, X = 134
Output: Yes
Explanation:
Following removal of digits makes the sum of array equal to X:

• Removing the first digit of arr[0] modifies arr[0] to 45.
• Removing the first digit of arr[1] modifies to 33.
• Removing the first digit of arr[2] modifies arr[2] to 54.
• Removing the last digit of arr[3] modifies arr[3] to 2.

The modified array is {45, 33, 54, 2}. Therefore, sum of the modified array = 45 + 33 + 54 + 2 = 134(= X).

Input: arr[] = {54, 22, 76, 432}, X = 48
Output: No

Approach: The given problem can be solved using recursion by generating all possible ways to remove the first or the last digit for every array element. Follow the steps below to solve the given problem.

• Define a recursive function, say makeSumX(arr, S, i, X) that takes the array arr[], the current sum S, the current index i, and the required sum X as the parameters and perform the following operations:
  • If the current index equals the length of the array and the current sum S is equal to the required sum X, then return true. Otherwise, return false.
  • Convert the integer arr[i] to a string.
  • Remove the last digit of the integer arr[i] and store it in a variable say L.
  • Remove the first digit of the integer arr[i] and store it in a variable say R.
  • Return the value from the function as the Bitwise OR of makeSumX(arr, S + L, i + 1, X) and makeSumX(arr, S + R, i + 1, X).
• If the value returned by the function makeSumX(arr, 0, 0, X) is true, then print “Yes”. Otherwise, print “No”.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Utility Function to check if the sum
// of the array elements can be made
// equal to X by removing either the first
// or last digits of every array element
bool makeSumX(int arr[], int X, int S,
              int i, int N)
{

    // Base Case
    if (i == N)
    {
        return S == X;
    }

    // Convert arr[i] to string
    string a = to_string(arr[i]);

    // Remove last digit
    int l = stoi(a.substr(0, a.length() - 1));

    // Remove first digit
    int r = stoi(a.substr(1));

    // Recursive function call
    bool x = makeSumX(arr, X, S + l, i + 1, N);
    bool y = makeSumX(arr, X, S + r, i + 1, N);

    return (x || y);
}

// Function to check if sum of given
// array can be made equal to X or not
void Check(int arr[], int X, int N)
{

    if (makeSumX(arr, X, 0, 0, N))
    {
        cout << "Yes";
    }
    else
    {
        cout << "No";
    }
}

// Driver code
int main()
{
    int arr[] = { 545, 433, 654, 23 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int X = 134;

    // Function Call
    Check(arr, X, N);

    return 0;
}

// This code is contributed by Kingash

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG{

// Utility Function to check if the sum
// of the array elements can be made
// equal to X by removing either the first
// or last digits of every array element
static boolean makeSumX(int arr[], int X,
                        int S, int i)
{

    // Base Case
    if (i == arr.length)
    {
        return S == X;
    }

    // Convert arr[i] to string
    String a = Integer.toString(arr[i]);

    // Remove last digit
    int l = Integer.parseInt(
        a.substring(0, a.length() - 1));

    // Remove first digit
    int r = Integer.parseInt(a.substring(1));

    // Recursive function call
    boolean x = makeSumX(arr, X, S + l, i + 1);
    boolean y = makeSumX(arr, X, S + r, i + 1);

    return (x || y);
}

// Function to check if sum of given
// array can be made equal to X or not
static void Check(int arr[], int X)
{

    if (makeSumX(arr, X, 0, 0))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 545, 433, 654, 23 };
    int X = 134;

    // Function Call
    Check(arr, X);
}
}

// This code is contributed by Kingash

Python3

# Python3 program for the above approach

# Utility Function to check if the sum
# of the array elements can be made
# equal to X by removing either the first
# or last digits of every array element
def makeSumX(arr, X, S, i):

      # Base Case
    if i == len(arr):
        return S == X

    # Convert arr[i] to string
    a = str(arr[i])

    # Remove last digit
    l = int(a[:-1])

    # Remove first digit
    r = int(a[1:])

    # Recursive function call
    x = makeSumX(arr, X, S + l, i + 1)
    y = makeSumX(arr, X, S + r, i + 1)

    return x | y

# Function to check if sum of given
# array can be made equal to X or not
def Check(arr, X):

    if (makeSumX(arr, X, 0, 0)):
      print("Yes")
    else:
      print("No")


# Driver Code

arr = [545, 433, 654, 23]
X = 134

Check(arr, X)

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Utility Function to check if the sum
// of the array elements can be made
// equal to X by removing either the first
// or last digits of every array element
static bool makeSumX(int[] arr, int X,
                     int S, int i)
{

    // Base Case
    if (i == arr.Length)
    {
        return S == X;
    }

    // Convert arr[i] to string
    string a = Convert.ToString(arr[i]);

    // Remove last digit
    int l = Int32.Parse(
        a.Substring(0, a.Length - 1));

    // Remove first digit
    int r = Int32.Parse(a.Substring(1));

    // Recursive function call
    bool x = makeSumX(arr, X, S + l, i + 1);
    bool y = makeSumX(arr, X, S + r, i + 1);

    return (x || y);
}

// Function to check if sum of given
// array can be made equal to X or not
static void Check(int[] arr, int X)
{
    if (makeSumX(arr, X, 0, 0))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}

// Driver Code
public static void Main()
{
    int[] arr = { 545, 433, 654, 23 };
    int X = 134;

    // Function Call
    Check(arr, X);
}
}

// This code is contributed by sanjoy_62

Javascript

<script>
       // Javascript program for the above approach

       // Utility Function to check if the sum
       // of the array elements can be made
       // equal to X by removing either the first
       // or last digits of every array element
       function makeSumX(arr, X, S,
           i, N) {

           // Base Case
           if (i === N) {
               return S === X
           }

           // Convert arr[i] to string
           let a = (arr[i]).toString()

           // Remove last digit
           let l = parseInt((a.substr(0, a.length - 1)))

           // Remove first digit
           let r = parseInt((a.substr(1)))

           // Recursive function call
           let x = makeSumX(arr, X, S + l, i + 1, N);
           let y = makeSumX(arr, X, S + r, i + 1, N);

           return (x || y);
       }

       // Function to check if sum of given
       // array can be made equal to X or not
       function Check(arr, X, N) {

           if (makeSumX(arr, X, 0, 0, N)) {
               document.write("Yes")
           }
           else {
               document.write("No")
           }
       }

       // Driver code

       let arr = [545, 433, 654, 23];
       let N = arr.length;
       let X = 134;

       // Function Call
       Check(arr, X, N);

       // This code is contributed by Hritik
   </script>

Output:

Yes

Time Complexity: O(2^N)
Auxiliary Space: O(N)