Consider a sequence of 14 elements: $A = [-5, -10, 6, 3, -1, -2, 13, 4, -9, -1, 4, 12, -3, 0]$. The subsequence sum $S(i,j) = \sum_{k=i}^j A[k]$ . Determine the maximum of $S(i,j)$, where $0 \le i \le j < 14$. (Divide and conquer approach may be used.) Answer: _______.

We need to find the maximum subarray sum of the array:

Using Kadane's algorithm (or the Divide-and-Conquer maximum subarray sum approach):

1. At index $0$ ($A[0] = -5$): $\text{current\_max} = -5$, $\text{global\_max} = -5$

2. At index $1$ ($A[1] = -10$): $\text{current\_max} = -10$, $\text{global\_max} = -5$

3. At index $2$ ($A[2] = 6$): $\text{current\_max} = 6$, $\text{global\_max} = 6$

4. At index $3$ ($A[3] = 3$): $\text{current\_max} = 9$, $\text{global\_max} = 9$

5. At index $4$ ($A[4] = -1$): $\text{current\_max} = 8$, $\text{global\_max} = 9$

6. At index $5$ ($A[5] = -2$): $\text{current\_max} = 6$, $\text{global\_max} = 9$

7. At index $6$ ($A[6] = 13$): $\text{current\_max} = 19$, $\text{global\_max} = 19$

8. At index $7$ ($A[7] = 4$): $\text{current\_max} = 23$, $\text{global\_max} = 23$

9. At index $8$ ($A[8] = -9$): $\text{current\_max} = 14$, $\text{global\_max} = 23$

10. At index $9$ ($A[9] = -1$): $\text{current\_max} = 13$, $\text{global\_max} = 23$

11. At index $10$ ($A[10] = 4$): $\text{current\_max} = 17$, $\text{global\_max} = 23$

12. At index $11$ ($A[11] = 12$): $\text{current\_max} = 29$, $\text{global\_max} = 29$

13. At index $12$ ($A[12] = -3$): $\text{current\_max} = 26$, $\text{global\_max} = 29$

14. At index $13$ ($A[13] = 0$): $\text{current\_max} = 26$, $\text{global\_max} = 29$

The contiguous subarray giving this maximum is from index $i = 2$ to $j = 11$: