1. **Problem Statement:** We are given an array of integers, including both positive and negative numbers. We need to find the contiguous subarray within this array that has the maximum possible sum. 2. **Formula and Approach:** This is a classic problem often solved using Kadane's Algorithm. The key idea is to iterate through the array, keeping track of the current subarray sum and the maximum sum found so far. 3. **Kadane's Algorithm Steps:** - Initialize two variables: $current\_sum = 0$ and $max\_sum = -\infty$ (or the smallest possible number). - Iterate through each element $x$ in the array: - Update $current\_sum = \max(x, current\_sum + x)$. - Update $max\_sum = \max(max\_sum, current\_sum)$. 4. **Explanation:** - At each step, we decide whether to add the current element to the existing subarray or start a new subarray from the current element. - $max\_sum$ keeps track of the best sum found so far. 5. **Apply to the example array:** $A = [-2, 1, -3, 4, -1, 2, 1, -5, 4]$ - Start: $current\_sum = 0$, $max\_sum = -\infty$ - $x = -2$: $current\_sum = \max(-2, 0 + (-2)) = -2$, $max\_sum = -2$ - $x = 1$: $current\_sum = \max(1, -2 + 1) = 1$, $max\_sum = 1$ - $x = -3$: $current\_sum = \max(-3, 1 + (-3)) = -2$, $max\_sum = 1$ - $x = 4$: $current\_sum = \max(4, -2 + 4) = 4$, $max\_sum = 4$ - $x = -1$: $current\_sum = \max(-1, 4 + (-1)) = 3$, $max\_sum = 4$ - $x = 2$: $current\_sum = \max(2, 3 + 2) = 5$, $max\_sum = 5$ - $x = 1$: $current\_sum = \max(1, 5 + 1) = 6$, $max\_sum = 6$ - $x = -5$: $current\_sum = \max(-5, 6 + (-5)) = 1$, $max\_sum = 6$ - $x = 4$: $current\_sum = \max(4, 1 + 4) = 5$, $max\_sum = 6$ 6. **Result:** The maximum sum is $6$, and the subarray that produces this sum is $[4, -1, 2, 1]$. **Final answer:** Maximum sum = $6$ from subarray $[4, -1, 2, 1]$.