# 4.1 The maximum-subarray problem

## 4.1-1

What does $\text{FIND-MAXIMUM-SUBARRAY}$ return when all elements of $A$ are negative?

It will return the greatest element of $A$.

## 4.1-2

Write pseudocode for the brute-force method of solving the maximum-subarray problem. Your procedure should run in $\Theta(n^2)$ time.

  1 2 3 4 5 6 7 8 9 10 11 12 BRUTE-FORCE-FIND-MAXIMUM-SUBARRAY(A) n = A.length max-sum = -∞ for l = 1 to n sum = 0 for h = l to n sum = sum + A[h] if sum > max-sum max-sum = sum low = l high = h return (low, high, max-sum) 

## 4.1-3

Implement both the brute-force and recursive algorithms for the maximum-subarray problem on your own computer. What problem size $n_0$ gives the crossover point at which the recursive algorithm beats the brute-force algorithm? Then, change the base case of the recursive algorithm to use the brute-force algorithm whenever the problem size is less than $n_0$. Does that change the crossover point?

On my computer, $n_0$ is $37$.

If the algorithm is modified to used divide and conquer when $n \ge 37$ and the brute-force approach when $n$ is less, the performance at the crossover point almost doubles. The performance at $n_0 - 1$ stays the same, though (or even gets worse, because of the added overhead).

What I find interesting is that if we set $n_0 = 20$ and used the mixed approach to sort $40$ elements, it is still faster than both.

## 4.1-4

Suppose we change the definition of the maximum-subarray problem to allow the result to be an empty subarray, where the sum of the values of an empty subarray is $0$. How would you change any of the algorithms that do not allow empty subarrays to permit an empty subarray to be the result?

If the original algorithm returns a negative sum, returning an empty subarray instead.

## 4.1-5

Use the following ideas to develop a nonrecursive, linear-time algorithm for the maximum-subarray problem. Start at the left end of the array, and progress toward the right, keeping track of the maximum subarray seen so far. Knowing a maximum subarray $A[1..j]$, extend the answer to find a maximum subarray ending at index $j + 1$ by using the following observation: a maximum subarray $A[i..j + 1]$, for some $1 \le i \le j + 1$. Determine a maximum subarray of the form $A[i..j + 1]$ in constant time based on knowing a maximum subarray ending at index $j$.

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ITERATIVE-FIND-MAXIMUM-SUBARRAY(A) n = A.length max-sum = -∞ sum = -∞ for j = 1 to n currentHigh = j if sum > 0 sum = sum + A[j] else currentLow = j sum = A[j] if sum > max-sum max-sum = sum low = currentLow high = currentHigh return (low, high, max-sum)