Partition Problem (Dynamic Programming) added

dynamic_programming/partition.cc

@@ -0,0 +1,52 @@
+// Partition Problem (Dynamic Programming)
+
+// Description:
+//  Given an integer array a, determine whether a can be partitioned into two subsets
+//  s1 and s2, such that sum(s1) = sum(s2)
+
+// Example:
+//  input: arr[] = {4,1,2,3}
+//  output: True
+//  The array can be partitioned to {4,1} and {2, 3}
+
+#include <iostream>
+
+using namespace std;
+
+bool partition(int arr[], int n){
+  int sum = 0;
+  for (int i=0; i<n ;i++){
+    sum += arr[i];
+  }
+  if (sum % 2 != 0){
+    return false;
+  }
+  sum /= 2;
+  int matrix[n+1][sum+1];
+  for (int i=0; i<=n; i++){
+    for (int j=0; j<=sum; j++){
+      if (j==0){
+        matrix[i][j] = true;
+      }
+      else if (i==0){
+        matrix[i][j] = false;
+      }
+      else {
+        matrix[i][j] = matrix[i-1][j];
+        if (j >= arr[i-1]){
+          matrix[i][j] = matrix[i][j] || matrix[i-1][j-arr[i-1]];
+        }
+      }
+    }
+  }
+  return matrix[n][sum];
+}
+int main(){
+  int arr1[] = {3,1,1,2,2,1};
+  int arr2[] = {4,2,1,3,4};
+  int arr3[] = {2,4,5,5};
+  cout<<partition(arr1, sizeof(arr1)/sizeof(arr1[0]))<<endl;
+  cout<<partition(arr2, sizeof(arr2)/sizeof(arr2[0]))<<endl;
+  cout<<partition(arr3, sizeof(arr3)/sizeof(arr3[0]))<<endl;
+  return 0;
+}