Maximize Subarray Sum After Removing All Occurrences of One Element

• Calculate the maximum subarray sum for the original nums array using Kadane's algorithm and store it as the initial global_max_sum.
• Find all unique elements in nums and store them in a set.
• For each unique element x:
  • Check if removing all occurrences of x would make the array empty. If so, continue to the next unique element.
  • Create a new temporary array by filtering out all occurrences of x from nums.
  • Calculate the maximum subarray sum of the temporary array using Kadane's algorithm.
  • Update global_max_sum = max(global_max_sum, new_max_sum).
• Return global_max_sum.

The algorithm proceeds as follows:

1. First, handle the base case of not performing any operation. Calculate the maximum subarray sum of the original nums array using Kadane's algorithm. This value serves as our initial answer.
2. Find all unique elements in nums. A HashSet is suitable for this, allowing O(n) discovery.
3. Iterate through each unique element x found in the previous step.
4. For each x, check if removing it would result in an empty array. If count(x) == nums.length, we cannot remove x, so we skip it.
5. If removing x is a valid operation, create a new list or array that contains all elements from nums except for x. This takes O(n) time.
6. Run Kadane's algorithm on this new array to find its maximum subarray sum. Kadane's algorithm finds the maximum sum of a contiguous subarray in O(n) time.
7. Compare this sum with the overall maximum sum found so far and update it if the new sum is greater.
8. After checking all unique elements, the value stored as the overall maximum is the answer.

Here is a code snippet for this approach:

1import java.util.*;2 3class Solution {4    public long maxSubarraySum(int[] nums) {5        // Calculate max subarray sum for the original array6        long maxSoFar = kadane(nums);7 8        Set<Integer> uniqueNums = new HashSet<>();9        Map<Integer, Integer> counts = new HashMap<>();10        for (int num : nums) {11            uniqueNums.add(num);12            counts.put(num, counts.getOrDefault(num, 0) + 1);13        }14 15        for (int x : uniqueNums) {16            // Check if removing x makes the array empty17            if (counts.get(x) == nums.length) {18                continue;19            }20 21            List<Integer> tempList = new ArrayList<>();22            for (int num : nums) {23                if (num != x) {24                    tempList.add(num);25                }26            }27            28            int[] tempArray = tempList.stream().mapToInt(i -> i).toArray();29            maxSoFar = Math.max(maxSoFar, kadane(tempArray));30        }31 32        return maxSoFar;33    }34 35    private long kadane(int[] arr) {36        if (arr.length == 0) {37            return Long.MIN_VALUE; // Or some other indicator for empty array38        }39        long maxCurrent = arr[0];40        long maxGlobal = arr[0];41        for (int i = 1; i < arr.length; i++) {42            maxCurrent = Math.max(arr[i], maxCurrent + arr[i]);43            if (maxCurrent > maxGlobal) {44                maxGlobal = maxCurrent;45            }46        }47        return maxGlobal;48    }49}