class Solution {
    
    /** Recursive code */
    static boolean equalPartition(int arr[]) {
        int n = arr.length;
        
        int sum = 0;
        for(int ele : arr)
            sum += ele;
        
        // if even sum -- then we can divide
        if(sum%2 != 0)
            return false;
        
        sum = sum/2;
        return findSubsetSum(arr, sum, n);
    }
    
    static boolean findSubsetSum(int[] arr, int sum, int n)
    {
        // when we reach end of the array, so couldn't found a subset
        if(n == 0)
            return false;
 
        // when sum becomes 0 that means found a subset
        if(sum == 0)
            return true;
 
        if(arr[n-1] <= sum)
        {
            // pick the element in subset
            boolean pick = findSubsetSum(arr, sum - arr[n-1], n-1);
            // do not pick the element in subset
            boolean notPick = findSubsetSum(arr, sum, n-1);
            return pick || notPick;
        }
        // when current element exceeds given sum, so skip the element
        return findSubsetSum(arr, sum, n-1);
    }

    /** Memoization code */
    static boolean equalPartition(int arr[]) {
        int n = arr.length;
        
        int sum = 0;
        for(int ele : arr)
            sum += ele;
        
        // if even sum -- then we can divide
        if(sum%2 != 0)
            return false;
        
        sum = sum/2;
        
        boolean[][] t = new boolean[n+1][sum+1];
        
        // when array of size will increase but sum needs to always 0, then we can have always empty subset
        for(int i=0; i<n+1; i++)
        {
            Arrays.fill(t[i], true);
        }
 
        // when sum increases but size of array is always 0, then we cannot have any subset
        for(int j=0; j<n+1; j++)
        {
            Arrays.fill(t[j], false);
        }

        
        return findSubsetSum(arr, sum, n, t);
    }
    
    static boolean findSubsetSum(int[] arr, int sum, int n, boolean[][] t)
    {
        // when we reach end of the array, so couldn't found a subset
        if(n == 0)
            return false;
 
        // when sum becomes 0 that means found a subset
        if(sum == 0)
            return true;
 
        if(t[n][sum] != false)
            return t[n][sum];
 
        if(arr[n-1] <= sum)
        {
            // pick the element in subset
            boolean pick = findSubsetSum(arr, sum - arr[n-1], n-1, t);
            // do not pick the element in subset
            boolean notPick = findSubsetSum(arr, sum, n-1, t);
            t[n][sum] = pick || notPick;
            return t[n][sum];
        }
        // when current element exceeds given sum, so skip the element
        t[n][sum] = findSubsetSum(arr, sum, n-1, t);
        return t[n][sum];
    }

    /** Top-down approach */
     static boolean equalPartition(int arr[]) {
        int n = arr.length;
        
        int sum = 0;
        for(int ele : arr)
            sum += ele;
        
        // if even sum -- then we can divide
        if(sum%2 != 0)
            return false;
        
        sum = sum/2;

        boolean[][] t = new boolean[n+1][sum+1];
 
        // intialise the matrix on basis of base condition
        for(int i=0; i<n+1; i++)
        {
            for(int j=0; j<sum+1; j++)
            {
                // when sum increases but size of array is always 0, then we cannot have any subset
                if(i==0)
                {
                    t[i][j] = false;
                }
                // when array of size will increase but sum needs to always 0, then we can have always empty subset
                if(j==0)
                {
                     t[i][j] = true;
                }
            }
        }
 
        for(int i=1; i<n+1; i++)
        {
            for(int j=1; j<sum+1; j++)
            {
                if(arr[i-1] <= j)
                {
                    boolean pick = t[i-1][j - arr[i-1]];
                    boolean notPick = t[i-1][j];
                    t[i][j] = pick || notPick;
                }
                else
                    t[i][j] = t[i-1][j];
            }
        }
        return t[n][sum];
    }