Create 08 Count the number of subset with given difference.cpp

Author: skjha1

src/Algorithms/Dynamic Programming/08 Count the number of subset with given difference.cpp

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+#include <bits/stdc++.h>
+using namespace std;
+
+int CountSubsetsWithSum(int arr[], int n, int sum) {
+	int 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++) {
+		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 count of subset as 0;
+				t[i][j] = 0;
+			if (j == 0) // when sum(j) is zero and there is always a chance of empty subset so return count as 1;
+				t[i][j] = 1;
+		}
+	}
+
+	for (int i = 1; i <= n; i++) {
+		for (int j = 1; j <= sum; j++) {
+			if (arr[i - 1] <= j) // when element in the list is less then target sum 
+				t[i][j] = t[i - 1][j - arr[i - 1]] + t[i - 1][j]; // either exclude or inxlude and add both of them to get final count 
+			else
+				t[i][j] = t[i - 1][j]; // exclude when element in the list is greater then the sum 
+		}
+	}
+
+	return t[n][sum]; // finally return the last row and last column element 
+}
+
+int CountSubsetsWithDiff(int arr[], int n, int diff) {
+	int sumOfArray = 0;
+	for (int i = 0; i < n; i++)
+		sumOfArray += arr[i]; // taking sum of the array 
+
+	if ((sumOfArray + diff) % 2 != 0)
+		return 0;
+	else
+		return CountSubsetsWithSum(arr, n, (sumOfArray + diff) / 2);// we will get the number of array(subset) with particular sum 
+}
+
+signed main() {
+	int n; cin >> n;
+	int arr[n];
+	for (int i = 0; i < n; i++)
+		cin >> arr[i];
+	int diff; cin >> diff;
+
+	cout << CountSubsetsWithDiff(arr, n, diff) << endl;
+	return 0;
+}