Add DP Solution to Subset Count (#3580)
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package com.thealgorithms.dynamicprogramming;
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/**
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* Find the number of subsets present in the given array with a sum equal to target.
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* Based on Solution discussed on StackOverflow(https://stackoverflow.com/questions/22891076/count-number-of-subsets-with-sum-equal-to-k)
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* @author Samrat Podder(https://github.com/samratpodder)
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*/
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public class SubsetCount {
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/**
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* Dynamic Programming Implementation.
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* Method to find out the number of subsets present in the given array with a sum equal to target.
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* Time Complexity is O(n*target) and Space Complexity is O(n*target)
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* @param arr is the input array on which subsets are to searched
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* @param target is the sum of each element of the subset taken together
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*
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*/
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public int getCount(int[] arr, int target){
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/**
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* Base Cases - If target becomes zero, we have reached the required sum for the subset
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* If we reach the end of the array arr then, either if target==arr[end], then we add one to the final count
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* Otherwise we add 0 to the final count
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*/
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int n = arr.length;
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int[][] dp = new int[n][target+1];
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for (int i = 0; i < n; i++) {
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dp[i][0] = 1;
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}
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if(arr[0]<=target) dp[0][arr[0]] = 1;
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for(int t=1;t<=target;t++){
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for (int idx = 1; idx < n; idx++) {
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int notpick = dp[idx-1][t];
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int pick =0;
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if(arr[idx]<=t) pick+=dp[idx-1][target-t];
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dp[idx][target] = pick+notpick;
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}
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}
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return dp[n-1][target];
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}
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/**
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* This Method is a Space Optimized version of the getCount(int[], int) method and solves the same problem
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* This approach is a bit better in terms of Space Used
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* Time Complexity is O(n*target) and Space Complexity is O(target)
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* @param arr is the input array on which subsets are to searched
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* @param target is the sum of each element of the subset taken together
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*/
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public int getCountSO(int[] arr, int target){
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int n = arr.length;
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int prev[]=new int[target+1];
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prev[0] =1;
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if(arr[0]<=target) prev[arr[0]] = 1;
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for(int ind = 1; ind<n; ind++){
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int cur[]=new int[target+1];
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cur[0]=1;
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for(int t= 1; t<=target; t++){
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int notTaken = prev[t];
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int taken = 0;
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if(arr[ind]<=t) taken = prev[t-arr[ind]];
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cur[t]= notTaken + taken;
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}
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prev = cur;
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}
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return prev[target];
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}
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}
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package com.thealgorithms.dynamicprogramming;
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import static org.junit.jupiter.api.Assertions.assertEquals;
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import org.junit.jupiter.api.Test;
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public class SubsetCountTest {
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public static SubsetCount obj = new SubsetCount();
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@Test
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void hasMultipleSubset(){
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int[] arr = new int[]{1,2,3,3};
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assertEquals(3, obj.getCount(arr, 6));
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}
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@Test
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void singleElementSubset(){
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int[] arr = new int[]{1,1,1,1};
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assertEquals(4, obj.getCount(arr, 1));
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}
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@Test
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void hasMultipleSubsetSO(){
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int[] arr = new int[]{1,2,3,3};
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assertEquals(3, obj.getCountSO(arr, 6));
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}
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@Test
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void singleSubsetSO(){
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int[] arr = new int[]{1,1,1,1};
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assertEquals(1,obj.getCountSO(arr, 4));
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}
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}
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