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Update divide-and-conquer-dp.md
typo, quadrangle was spelled "qudrangle"
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src/dynamic_programming/divide-and-conquer-dp.md

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@@ -21,7 +21,7 @@ time. Then the straightforward evaluation of the above recurrence is $O(m n^2)$.
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are $m \times n$ states, and $n$ transitions for each state.
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Let $opt(i, j)$ be the value of $k$ that minimizes the above expression. Assuming that the
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cost function satisfies the qudrangle inequality, we can show that
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cost function satisfies the quadrangle inequality, we can show that
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$opt(i, j) \leq opt(i, j + 1)$ for all $i, j$. This is known as the _monotonicity condition_.
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Then, we can apply divide and conquer DP. The optimal
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"splitting point" for a fixed $i$ increases as $j$ increases.

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