diff --git a/Recursion DP/EqualSubsetSumPartition.cpp b/Recursion DP/EqualSubsetSumPartition.cpp
new file mode 100644
index 0000000..7d100f6
--- /dev/null
+++ b/Recursion DP/EqualSubsetSumPartition.cpp	
@@ -0,0 +1,61 @@
+#include <bits/stdc++.h>
+using namespace std;
+class PartitionSet {
+public:
+  bool canPartition(const vector<int> &num) {
+    int n = num.size();
+    // find the total sum
+    int sum = 0;
+    for (int i = 0; i < n; i++) {
+      sum += num[i];
+    }
+
+    // if 'sum' is a an odd number, we can't have two subsets with same total
+    if (sum % 2 != 0) {
+      return false;
+    }
+
+    // we are trying to find a subset of given numbers that has a total sum of ‘sum/2’.
+    sum /= 2;
+
+    vector<vector<bool>> dp(n, vector<bool>(sum + 1));
+
+    // populate the sum=0 column, as we can always have '0' sum without including any element
+    for (int i = 0; i < n; i++) {
+      dp[i][0] = true;
+    }
+
+    // with only one number, we can form a subset only when the required sum is
+    // equal to its value
+    for (int s = 1; s <= sum; s++) {
+      dp[0][s] = (num[0] == s ? true : false);
+    }
+
+    // process all subsets for all sums
+    for (int i = 1; i < n; i++) {
+      for (int s = 1; s <= sum; s++) {
+        // if we can get the sum 's' without the number at index 'i'
+        if (dp[i - 1][s]) {
+          dp[i][s] = dp[i - 1][s];
+        } else if (s >= num[i]) { // else if we can find a subset to get the remaining sum
+          dp[i][s] = dp[i - 1][s - num[i]];
+        }
+      }
+    }
+
+    // the bottom-right corner will have our answer.
+    return dp[n - 1][sum];
+  }
+};
+
+int main(int argc, char *argv[]) {
+  PartitionSet ps;
+  vector<int> num = {1, 2, 3, 4};
+  cout << ps.canPartition(num) << endl;
+  num = vector<int>{1, 1, 3, 4, 7};
+  cout << ps.canPartition(num) << endl;
+  num = vector<int>{2, 3, 4, 6};
+  cout << ps.canPartition(num) << endl;
+}
+
+// The Code is contributed by Ritika Rawat