Count number of subset with a given difference.cpp

Author: KrishnaRA03

Count the number of subset with a given difference.cpp

@@ -0,0 +1,52 @@
+/* Count the number of subset with a given difference - Top Down Approach*/
+
+/*Let sum of subset 1 be s1 and subset 2 with s2
+s1 - s2 = diff (given)
+s1 + s2=sum of array (logical)
+Therefore adding both eq we get :
+2s1= diff + sum of array
+s1= (diff + sum of array)/2;
+Problem reduces to find no of subsets with given sum*/
+
+#include <bits/stdc++.h>
+using namespace std;
+#define int long long int
+#define fast                    \
+    cin.sync_with_stdio(false); \
+    cin.tie(NULL);              \
+    cout.tie(NULL);
+
+int dp[1001][10001];
+int CountOfSubsetSumGivenDiff(int n, int arr[], int sum)
+{
+    for (int i = 0; i <= n; i++)
+        dp[i][0] = 1;
+    for (int i = 1; i <= sum; i++)
+        dp[0][i] = 0;
+    for (int i = 1; i <= n; i++)
+    {
+        for (int j = 1; j <= sum; j++)
+        {
+            if (arr[i - 1] <= j)
+                dp[i][j] = dp[i - 1][j - arr[i - 1]] + dp[i - 1][j];
+            else
+                dp[i][j] = dp[i - 1][j];
+        }
+    }
+    return dp[n][sum];
+}
+signed main()
+{
+    fast;
+    int n = 4;
+    int arr[4] = {1, 2, 3, 3};
+    int diff = 1;
+    int sum = accumulate(arr, arr + n, 0);
+    int s1 = (diff + sum) / 2;
+    //memonization
+    memset(dp, -1, sizeof(dp));
+
+    cout << CountOfSubsetSumGivenDiff(n, arr, s1);
+
+    return 0;
+}