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Analysis

T(n) $\leq$ $T(\lceil \frac{n}{5} \rceil)$ + T(7n/10 + 6) + O(n)

To get T(7n/10 + 6), note that the smallest partition size would be

$3(\lceil \frac{1}{2} \lceil \frac{n}{5} \rceil \rceil \;-\; 2) \;>=\; (\frac{3n}{10} \;-\; 6)$

7n/10 + 6 denotes number of elements in larger partition

Assume T(n) $\leq$ cn for $\lceil n/5 \rceil$ and 7n/10+6

T(n) = O(n)

T(n) $\leq$ c(n/5) + c(7n/10 + 6) + O(n)

= cn/5 + c7n/10 + c6 + O(n)

= 1/10(2cn + 7cn + 60c + O(n)) $\leq$ cn

2cn + 7cn + 60c + O(n) $\leq$ 10cn

60c + O(n) $\leq$ cn

c(n - 60) $\geq$ O(n)

c $\geq$ O(n) / (n - 60)

c is a valid constant for large enough n.

Stopping condition: T(n) = $\Theta(1)$ , if n $\leq$ 80

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