diff --git a/OBST_DSA b/OBST_DSA
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+/*
+
+Problem Statement: Given sequence k = k1 <k2 < … <kn of n sorted keys, with a search probability pi for each key ki . Build the Binary search tree that has the least search cost given the access probability for each key?
+
+*/
+
+// BEGINNING OF CODE
+#include<iostream>
+using namespace std;
+
+int find(int, int);
+void print(int, int);
+int c[20][20], w[20][20], r[20][20], p[20], q[20], k, m, i, j, n;
+char idtr[10][7];
+
+int main()
+{
+    cout<<"\nEnter number of identifiers: ";
+    cin>>n;
+    for(i = 1; i <= n; i++)
+    {
+        cout<<"Enter Identifier "<<i<<": ";
+        cin>>idtr[i];
+    }
+    for(i = 1; i <= n; i++)
+    {
+        cout<<"Enter successful probability of "<<i<<": ";
+        cin>>p[i];
+    }
+    for(i = 0; i <= n; i++)
+    {
+        cout<<"Enter unsuccessful probability of "<<i<<": ";
+        cin>>q[i];
+    }
+    for(i = 0; i <= n; i++)
+    {
+        w[i][i] = q[i];
+        c[i][i] = r[i][i] = 0;
+        cout<<"\nW: "<<w[i][i]<<" | c: "<<c[i][i]<<" | r: "<<r[i][i];
+    }
+    for(i = 0; i < n; i++)
+    {
+        j = i + 1;
+        w[i][j] = p[j] + q[j] + q[i];
+        c[i][j] = w[i][j];
+        r[i][j] = j;
+        cout<<"\nW: "<<w[i][j]<<" | c: "<<c[i][j]<<" | r: "<<r[i][j];
+    }
+    for(m = 2; m <= n; m++)
+    {
+        for(i = 0; i <= n-m; i++)
+        {
+            j = i + m;
+            w[i][j] = p[j] + q[j] + w[i][j-1];
+            c[i][j] = 1000;
+            for(k = i + 1; k <= j; k++)
+            {
+                int cost = c[i][k-1] + c[k][j];
+                if(cost < c[i][j])
+                {
+                    c[i][j] = cost;
+                    r[i][j] = k;
+                }
+            }
+            c[i][j] += w[i][j];
+            cout<<"\nW: "<<w[i][j]<<" | c: "<<c[i][j]<<" | r: "<<r[i][j];
+        }
+    }
+    cout<<"\nFinal OBST is: ";
+    print(0, n);
+    return 0;
+}
+
+void print(int i, int j)
+{
+    if(i < j)
+        cout<<"\n"<<idtr[r[i][j]];
+    else
+        return;
+    print(i, r[i][j] - 1);
+    print(r[i][j], j);
+}
+// END OF CODE