diff --git a/dynamic_programming/optimal_binary_search_tree.py b/dynamic_programming/optimal_binary_search_tree.py
index b4f1181ac11c..cc2eac3c53b5 100644
--- a/dynamic_programming/optimal_binary_search_tree.py
+++ b/dynamic_programming/optimal_binary_search_tree.py
@@ -102,7 +102,7 @@ def find_optimal_binary_search_tree(nodes):
     # This 2D array stores the overall tree cost (which's as minimized as possible);
     # for a single key, cost is equal to frequency of the key.
     dp = [[freqs[i] if i == j else 0 for j in range(n)] for i in range(n)]
-    # sum[i][j] stores the sum of key frequencies between i and j inclusive in nodes
+    # total[i][j] stores the sum of key frequencies between i and j inclusive in nodes
     # array
     total = [[freqs[i] if i == j else 0 for j in range(n)] for i in range(n)]
     # stores tree roots that will be used later for constructing binary search tree
@@ -115,11 +115,17 @@ def find_optimal_binary_search_tree(nodes):
             dp[i][j] = sys.maxsize  # set the value to "infinity"
             total[i][j] = total[i][j - 1] + freqs[j]
 
-            # Apply Knuth's optimization
-            # Loop without optimization: for r in range(i, j + 1):
-            for r in range(root[i][j - 1], root[i + 1][j] + 1):  # r is a temporal root
-                left = dp[i][r - 1] if r != i else 0  # optimal cost for left subtree
-                right = dp[r + 1][j] if r != j else 0  # optimal cost for right subtree
+            # Apply Knuth's optimization with safe boundary handling
+            r_start = root[i][j - 1] if j > i else i
+            r_end = root[i + 1][j] if i < j else j
+
+            # Ensure r_start and r_end are within valid bounds
+            r_start = max(i, min(r_start, j))
+            r_end = min(j, max(r_end, i))
+
+            for r in range(r_start, r_end + 1):
+                left = dp[i][r - 1] if r > i else 0  # optimal cost for left subtree
+                right = dp[r + 1][j] if r < j else 0  # optimal cost for right subtree
                 cost = left + total[i][j] + right
 
                 if dp[i][j] > cost: