Check if it is possible to get back to 12’0 clock only by adding or subtracting given seconds

Given N seconds. The task is to check if it is possible to start from the 12’0 clock and get back to 12 only by adding or subtracting the given seconds. We need to use all given seconds exactly once, we can either add an element or subtract it.
Examples:

Input: a[] = {60, 60, 120} 
Output: YES 
Add the first two seconds and 
subtract the last one to get back to 0. 

Input : a[] = {10, 20, 60, 180} 
Output : NO 

Simple Approach: Generate all possible combinations to solve the above problem. Hence generate the power set of N numbers. Check if anyone’s sum%(24*60) is equal to zero or not, if it is then it is possible else not. 
Below is the implementation of the above approach: 

C++

// C++ program to check if we come back to
// zero or not in a clock
#include <bits/stdc++.h>
using namespace std;

// Function to check all combinations
bool checkCombinations(int a[], int n)
{

    // Generate all power sets
    int pow_set_size = pow(2, n);
    int counter, j;

    // Check for every combination
    for (counter = 0; counter < pow_set_size; counter++) {

        // Store sum for all combinations
        int sum = 0;
        for (j = 0; j < n; j++) {

            /* Check if jth bit in the counter is set
             If set then print jth element from set */
            if (counter & (1 << j))
                sum += a[j]; // if set then consider as '+'
            else
                sum -= a[j]; // else consider as '-'
        }

        // If we can get back to 0
        if (sum % (24 * 60) == 0)
            return true;
    }
    return false;
}
// Driver Code
int main()
{
    int a[] = { 60, 60, 120 };
    int n = sizeof(a) / sizeof(a[0]);

    if (checkCombinations(a, n))
        cout << "YES";
    else
        cout << "NO";
    return 0;
}

Java

// Java program to check if we come
// back to zero or not in a clock
import java.lang.Math;

class GfG
{

    // Function to check all combinations
    static boolean checkCombinations(int a[], int n)
    {
        // Generate all power sets
        int pow_set_size = (int)Math.pow(2, n);
        int counter, j;

        // Check for every combination
        for (counter = 0; counter < pow_set_size; counter++)
        {

            // Store sum for all combinations
            int sum = 0;
            for (j = 0; j < n; j++)
            {

                /* Check if jth bit in the counter is set
                If set then print jth element from set */
                if ((counter & (1 << j)) != 0)
                    sum += a[j]; // if set then consider as '+'
                else
                    sum -= a[j]; // else consider as '-'
            }

            // If we can get back to 0
            if (sum % (24 * 60) == 0)
                return true;
        }
        return false;
    }

    // Driver code
    public static void main(String []args)
    {
        int a[] = { 60, 60, 120 };
        int n = a.length;

        if (checkCombinations(a, n))
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}

// This code is contributed by Rituraj Jain

Python 3

# Python 3 program to check if we come
# back to zero or not in a clock

# Function to check all combinations
def checkCombinations(a, n):

    # Generate all power sets
    pow_set_size = pow(2, n)

    # Check for every combination
    for counter in range(pow_set_size):

        # Store sum for all combinations
        sum = 0
        for j in range(n) :

            # Check if jth bit in the counter is set
            # If set then print jth element from set
            if (counter & (1 << j)):
                sum += a[j] # if set then consider as '+'
            else:
                sum -= a[j] # else consider as '-'

        # If we can get back to 0
        if (sum % (24 * 60) == 0):
            return True
    return False

# Driver Code
if __name__ == "__main__":

    a = [ 60, 60, 120 ]
    n = len(a)

    if (checkCombinations(a, n)):
        print("YES")
    else:
        print("NO")

# This code is contributed by ita_c

C#

// C# program to check if we come
// back to zero or not in a clock
using System;

class GfG
{

    // Function to check all combinations
    static bool checkCombinations(int [] a, int n)
    {
        // Generate all power sets
        int pow_set_size = (int)Math.Pow(2, n);
        int counter, j;

        // Check for every combination
        for (counter = 0; counter < pow_set_size; counter++)
        {

            // Store sum for all combinations
            int sum = 0;
            for (j = 0; j < n; j++)
            {

                /* Check if jth bit in the counter is set
                If set then print jth element from set */
                if ((counter & (1 << j)) != 0)
                    sum += a[j]; // if set then consider as '+'
                else
                    sum -= a[j]; // else consider as '-'
            }

            // If we can get back to 0
            if (sum % (24 * 60) == 0)
                return true;
        }
        return false;
    }

    // Driver code
    public static void Main()
    {
        int [] a = { 60, 60, 120 };
        int n = a.Length;

        if (checkCombinations(a, n))
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
    }
}

// This code is contributed by ihritik

PHP

<?php
// PHP program to check if we come back to
// zero or not in a clock

// Function to check all combinations
function checkCombinations($a, $n)
{

    // Generate all power sets
    $pow_set_size = pow(2, $n);

    // Check for every combination
    for ($counter = 0;
         $counter < $pow_set_size; $counter++)
    {

        // Store sum for all combinations
        $sum = 0;
        for ($j = 0; $j < $n; $j++)
        {

            /* Check if jth bit in the counter is set
            If set then print jth element from set */
            if ($counter & (1 << $j))
                $sum += $a[$j]; // if set then consider as '+'
            else
                $sum -= $a[$j]; // else consider as '-'
        }

        // If we can get back to 0
        if ($sum % (24 * 60) == 0)
            return true;
    }
    return false;
}

// Driver Code
$a = array( 60, 60, 120 );
$n = sizeof($a);

if (checkCombinations($a, $n))
    echo "YES";
else
    echo "NO";

// This code is contributed by Ryuga
?>

Javascript

<script>

// JavaScript program to check if we come
// back to zero or not in a clock

    // Function to check all combinations
    function checkCombinations(a , n) {
        // Generate all power sets
        var pow_set_size = parseInt( Math.pow(2, n));
        var counter, j;

        // Check for every combination
        for (counter = 0; counter < pow_set_size; counter++)
        {

            // Store sum for all combinations
            var sum = 0;
            for (j = 0; j < n; j++) {

                /*
                 * Check if jth bit in the counter is
                 set If set then print jth element from set
                 */
                if ((counter & (1 << j)) != 0)
                    sum += a[j]; // if set then consider as '+'
                else
                    sum -= a[j]; // else consider as '-'
            }

            // If we can get back to 0
            if (sum % (24 * 60) == 0)
                return true;
        }
        return false;
    }

    // Driver code

        var a = [ 60, 60, 120 ];
        var n = a.length;

        if (checkCombinations(a, n))
            document.write("YES");
        else
            document.write("NO");

// This code contributed by Rajput-Ji

</script>

Output:

YES

Time Complexity: O(N*2^N), as we are using a nested loop to traverse 2^N*N times. Where N is the number of elements in the array.

Auxiliary Space: O(1), as we are not using any extra space.

If we take a closer look, we can notice that this problem is basically a variation of the Partition Problem. So we can optimize it using Dynamic Programming (Please refer to method 2 of Partition Problem).