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Copy path04 Subset sum(Knapsack variation).cpp
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04 Subset sum(Knapsack variation).cpp
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// https://www.geeksforgeeks.org/subset-sum-problem-dp-25/
#include <iostream>
using namespace std;
bool isSubsetPossible(int arr[], int n, int sum) {
bool t[n + 1][sum + 1]; // DP - matrix
// initialization
// here we are setting 1st row and 1st column
// i denotes the size of the array
// j denotes the target sum (subset sum)
for (int i = 0; i <= n; i++) { // itereate as long it is less then length of the array
for (int j = 0; j <= sum; j++) {
if (i == 0)// when array(i) is empty than there is no meaning of sum of elements so return false
t[i][j] = false;
if (j == 0) // when sum(j) is zero and there is always a chance of empty subset so return it as true;
t[i][j] = true;
}
}
// start from 1 since 1st row and column is already considerd
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= sum; j++) {
if (arr[i - 1] <= j)
// after taking and element substract from the (sum) i.e -> in {3,8} 3 is taken then we want 11-3=8in the array
t[i][j] = t[i - 1][j - arr[i - 1]] || t[i - 1][j]; // either take or(||) do not take
else // if sum is less than array size just leave and increment
t[i][j] = t[i - 1][j];
}
}
return t[n][sum]; // at last return T/F
}
int main() {
int n; cin >> n;
int arr[n];
for (int i = 0; i < n; i++)
cin >> arr[i];
int sum; cin >> sum;
isSubsetPossible(arr, n, sum) ? cout << "Yes\n" : cout << "No\n";
return 0;
}
/*
Complexity Analysis:
Time Complexity: O(sum*n), where sum is the ‘target sum’ and ‘n’ is the size of array.
Auxiliary Space: O(sum*n), as the size of 2-D array is sum*n.
*/