package com.thealgorithms.dynamicprogramming;

/**
 * Find the number of subsets present in the given array with a sum equal to target.
 * Based on Solution discussed on
 * StackOverflow(https://stackoverflow.com/questions/22891076/count-number-of-subsets-with-sum-equal-to-k)
 * @author Samrat Podder(https://github.com/samratpodder)
 */
public class SubsetCount {

    /**
     * Dynamic Programming Implementation.
     * Method to find out the number of subsets present in the given array with a sum equal to
     * target. Time Complexity is O(n*target) and Space Complexity is O(n*target)
     * @param arr is the input array on which subsets are  to searched
     * @param target is the sum of each element of the subset taken together
     *
     */
    public int getCount(int[] arr, int target) {
        /**
         * Base Cases - If target becomes zero, we have reached the required sum for the subset
         * If we reach the end of the array arr then, either if target==arr[end], then we add one to
         * the final count Otherwise we add 0 to the final count
         */
        int n = arr.length;
        int[][] dp = new int[n][target + 1];
        for (int i = 0; i < n; i++) {
            dp[i][0] = 1;
        }
        if (arr[0] <= target) dp[0][arr[0]] = 1;
        for (int t = 1; t <= target; t++) {
            for (int idx = 1; idx < n; idx++) {
                int notpick = dp[idx - 1][t];
                int pick = 0;
                if (arr[idx] <= t) pick += dp[idx - 1][target - t];
                dp[idx][target] = pick + notpick;
            }
        }
        return dp[n - 1][target];
    }

    /**
     * This Method is a Space Optimized version of the getCount(int[], int) method and solves the
     * same problem This approach is a bit better in terms of Space Used Time Complexity is
     * O(n*target) and Space Complexity is O(target)
     * @param arr is the input array on which subsets are  to searched
     * @param target is the sum of each element of the subset taken together
     */
    public int getCountSO(int[] arr, int target) {
        int n = arr.length;
        int[] prev = new int[target + 1];
        prev[0] = 1;
        if (arr[0] <= target) prev[arr[0]] = 1;
        for (int ind = 1; ind < n; ind++) {
            int[] cur = new int[target + 1];
            cur[0] = 1;
            for (int t = 1; t <= target; t++) {
                int notTaken = prev[t];
                int taken = 0;
                if (arr[ind] <= t) taken = prev[t - arr[ind]];
                cur[t] = notTaken + taken;
            }
            prev = cur;
        }
        return prev[target];
    }
}
