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Add boundary checking to optimal bst (#11771)

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* Add boundary checking to optimal bst

* Add additional checks to prevent IndexError

---------

Co-authored-by: Christian Clauss <cclauss@me.com>
This commit is contained in:
Andrey Ivanov
2026-09-23 14:48:00 +02:00
committed by GitHub
co-authored by Christian Clauss
parent d182474c4f
commit 4e817a8ef3
1 changed files with 12 additions and 6 deletions
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dynamic_programming/optimal_binary_search_tree.py
+12 -6
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@@ -102,7 +102,7 @@ def find_optimal_binary_search_tree(nodes) -> None:
# 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) -> None:
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: