Given a set of non negative integers and a target value, find if there exists a subset in the given set whose sum is target.
Solution 1. Enumerate all possible subsets and check if their sum is the target
The runtime of this solution is O(2^n). This enumeration algorithm is similar with the problem Subsets. The difference is
that Subsets has to get all possible subsets. But this problem can terminate the check earlier if for one element arr[startIdx]
including it in the subset and not including it both returns a false check. This is correct because for any subsets, it either has
arr[startIdx] or does not have it.
1 import java.util.Arrays;
2 public class Subsets {
3 public boolean findIfExistSubset(int[] arr, int target){
4 if(target == 0){
5 return true;
6 }
7 if(arr == null || arr.length == 0 || target < 0){
8 return false;
9 }
10 Arrays.sort(arr);
11 if(target < arr[0] || target >= arr[arr.length - 1] * arr.length){
12 return false;
13 }
14 return helper(arr, 0, 0, target);
15 }
16 private boolean helper(int[] arr, int startIdx, int currSum, int target){
17 if(currSum == target){
18 return true;
19 }
20 if(currSum > target){
21 return false;
22 }
23 if(startIdx >= arr.length){
24 return false;
25 }
26 currSum += arr[startIdx];
27 if(helper(arr, startIdx + 1, currSum, target)){
28 return true;
29 }
30 currSum -= arr[startIdx];
31 if(helper(arr, startIdx + 1, currSum, target)){
32 return true;
33 }
34 return false;
35 }
36 }
Solution 2. Dynamic Programming, runtime is O(arr.length * target), space complexity is O(arr.length * target)
State: T[i][j]: if total sum j can be found from a subset from arr[0......i - 1];
Function: T[i][j] = T[i - 1][j] || T[i - 1][j - arr[i - 1]], if j >= arr[i - 1]; if the current target j is >= arr[i - 1], it means that we
can possibly select arr[i - 1] as part of the subset. T[i][j] will be either not selecting arr[i - 1] (T[i - 1][j]) or selecting arr[i - 1] (T[i - 1][j - arr[i - 1]]);
T[i][j] = T[i - 1][j], if j < arr[i - 1]; if j < arr[i - 1], it means we can‘t select arr[i - 1].
i - 1 here indicates that for each element in arr[], it can only be selected at most once.
Init: T[i][0] = true for i in [0, arr.length]; this means when the target is 0, then for set arr[0.... i - 1] we always have the empty set that sums up to 0.
Answer: T[arr.length][target]
1 public boolean findIfExistSubsetDp(int[] arr, int target){
2 boolean[][] T = new boolean[arr.length + 1][target + 1];
3 for(int i = 0; i <= arr.length; i++){
4 T[i][0] = true;
5 }
6 for(int i = 1; i <= arr.length; i++){
7 for(int j = 1; j <= target; j++){
8 if(j >= arr[i - 1]){
9 T[i][j] = T[i - 1][j] || T[i - 1][j - arr[i - 1]];
10 }
11 else{
12 T[i][j] = T[i - 1][j];
13 }
14 }
15 }
16 return T[arr.length][target];
17 }
Optimized Dp solution with O(target) space
1 public boolean findIfExistSubsetDp(int[] arr, int target){
2 boolean[][] T = new boolean[2][target + 1];
3 T[0][0] = true;
4 for(int i = 1; i <= arr.length; i++){
5 T[i % 2][0] = true;
6 for(int j = 1; j <= target; j++){
7 if(j >= arr[i - 1]){
8 T[i % 2][j] = T[(i - 1) % 2][j] || T[(i - 1) % 2][j - arr[i - 1]];
9 }
10 else{
11 T[i % 2][j] = T[(i - 1) % 2][j];
12 }
13 }
14 }
15 return T[arr.length % 2][target];
16 }
Follow up question:
What about if we change the condition so that each element in the input set can be selected more than once?
The change needs to make is that when the current element arr[i - 1] can be selected, we don‘t exclude it from
being selected again.
T[i][j] = T[i - 1][j] || T[i][j - arr[i - 1]];
1 public boolean findIfExistSubsetDp(int[] arr, int target){
2 boolean[][] T = new boolean[arr.length + 1][target + 1];
3 for(int i = 0; i <= arr.length; i++){
4 T[i][0] = true;
5 }
6 for(int i = 1; i <= arr.length; i++){
7 for(int j = 1; j <= target; j++){
8 if(j >= arr[i - 1]){
9 T[i][j] = T[i - 1][j] || T[i][j - arr[i - 1]];
10 }
11 else{
12 T[i][j] = T[i - 1][j];
13 }
14 }
15 }
16 return T[arr.length][target];
17 }
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