d

```
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 final class SubsetCount {
private 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 static 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 static 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];
}
}
```