Question: Given an array of integers, I can find the maximal subarray sum using Kadane's algorithm. In code this looks like: def kadane(arr, n): # initialize subarray_sum, max_subarray_sum and subarray_sum = 0 max_subarray_sum = np.int32(-2**31) # Just some initial value to check # for all negative values case finish = -1 # local variable local_start = 0 for i in range(n): subarray_sum += arr[i] if subarray_sum < 0: subarray_sum = 0 local_start = i + 1 elif subarray_sum > max_subarray_sum: max_subarray_sum = subarray_sum start = local_start finish = i # There is at-least one # non-negative number if finish != -1: return max_subarray_sum, start, finish # Special Case: When all numbers in arr[] are negative max_subarray_sum = arr[0] start = finish = 0 # Find the maximum element in array for i in range(1, n): if arr[i] > max_subarray_sum: max_subarray_sum = arr[i] start = finish = i return max_subarray_sum, start, finish  This is fast and works well. However, I would like to find a pair of subarrays with maximal sum. Take this example input. arr = [3, 3, 3, -8, 3, 3, 3]  The maximal subarray is the entire array with sum 10. But if I am allowed to take two subarrays they can be [3, 3, 3] and [3, 3, 3] which has sum 18. Is there a fast algorithm to compute the maximal pair of subarrays? I am assuming the two subarrays will not overlap.

Answer: Kadane's algorithm is used to find the maximal subarray sum in an array of integers. However, to find the maximal pair of non-overlapping subarrays with maximal sum, a different approach is needed. One way to solve this is by using the concept of prefix and suffix sums. Here's a high-level overview of the algorithm:  Compute the prefix sum and suffix sum arrays for the given array. Use Kadane's algorithm to find the maximum sum subarray in the prefix sum array, which gives the maximal subarray ending at each position. Use Kadane's algorithm in reverse to find the maximum sum subarray in the suffix sum array, which gives the maximal subarray starting at each position. Iterate through the array and find the pair of non-overlapping subarrays with maximal sum by combining the results from steps 2 and 3.  This approach has a time complexity of O(n) (and space complexity of O(n)) and can efficiently find the maximal pair of subarrays with maximal sum, where n is the size of the input array. def max_subarray(arr, rev=False): best_sum = [(arr[0], 0, 0)] current_sum = (arr[0], 0, 0) for j in range(1, len(arr)): current_sum = max( (arr[j], j, j), (current_sum[0] + arr[j], current_sum[1], j) ) best_sum.append(max(best_sum[-1], current_sum)) return [ (t[0], len(arr) - t[2] - 1, len(arr) - t[1] - 1) for t in reversed(best_sum) ] if rev else best_sum def max_pair_subarray(arr): max_sub = max_subarray(arr) max_sub_rev = max_subarray(list(reversed(arr)), rev=True) max_sum = (max_sub[0][0] + max_sub_rev[1][0], max_sub[0], max_sub_rev[1]) for j in range(1, len(arr) - 2): max_sum = max( max_sum, (max_sub[j][0] + max_sub_rev[j + 1][0], max_sub[j], max_sub_rev[j + 1]), ) return ( max_sum[0], (max_sum[1][1], max_sum[1][2]), (max_sum[2][1], max_sum[2][2]), ) max_sum, (start_idx_1, end_idx_1), (start_idx_2, end_idx_2) = max_pair_subarray(arr)