Complete the above code.
max_price, prices[i] + rod_cut(prices,len - i - 1)
max_price, prices[i - 1].
max_price, rod_cut(prices, len - i - 1)
max_price, prices[i - 1] + rod_cut(prices,len - i - 1)
Correct Answer: A — max_price, prices[i] + rod_cut(prices,len - i - 1)
Explanation:
max_price, prices[i] + rod_cut(prices,len - i - 1) completes the above code.
