Arithmetic Slices II - Subsequence

#define LONG long long
class Solution {
public:
    int findArithmeticSlices(vector<int>& nums) {
        int length = nums.size();
        LONG total = 0;
        vector<map<LONG, int>> count(length);
        for (int current = 1; current < length; current++) {
            for (int prev = 0; prev < current; prev++) {
                LONG difference = (LONG)nums[current] - (LONG)nums[prev];
                int sequenceCount = 0;
                if (count[prev].find(difference) != count[prev].end()) {
                    sequenceCount = count[prev][difference];
                }
                count[current][difference] += sequenceCount + 1;
                total += sequenceCount;
            }
        }

        return (int)total;
    }
};

The solution is a C++ implementation designed to count the number of subsequences in a given array that form an arithmetic progression. Follow these steps to understand how the provided code achieves this:

1. Define a type LONG to represent long long for handling large integers and ensuring there is no integer overflow during calculations.

2. Inside the findArithmeticSlices function, initialize total to track the count of arithmetic subsequences and declare a vector of maps, where each map will store the difference between elements as keys and the count of ways to achieve this difference up to the current index as values.

3. Use a nested loop where the outer loop variable current moves from the second element to the last element of the input vector nums. The inner loop variable prev counts from starting index to one less than current.

4. For each pair of current and prev, compute the difference between the nums[current] and nums[prev]. Look up this difference in the map at the prev index to see how often this difference has occurred before current.

5. If this difference has been seen before prev, the count of these previous occurrences is retrieved and updated in the map at the current index to include these previous subsequences plus the current pair.

6. Add the count of sequences ending at prev and having the same difference to the overall total, which accumulates the total number of valid arithmetic subsequences.

7. Finally, the function returns the total as an integer, giving the count of all arithmetic subsequences present in the array, not including sequences shorter than length 3.

This solution effectively uses dynamic programming with a hashmap to efficiently count all subsequences that are arithmetic, ensuring that even if the input grows large, the approach remains efficient in both time and space.

class Solution {
    public int countArithmeticSubarrays(int[] nums) {
        int length = nums.length;
        long total = 0;
        Map<Integer, Integer>[] maps = new Map[length];
        for (int i = 0; i < length; i++) {
            maps[i] = new HashMap<>(i);
            for (int j = 0; j < i; j++) {
                long difference = (long)nums[i] - (long)nums[j];
                if (difference < Integer.MIN_VALUE || difference > Integer.MAX_VALUE) {
                    continue;
                }
                int diff = (int)difference;
                int count = maps[j].getOrDefault(diff, 0);
                int currentCount = maps[i].getOrDefault(diff, 0);
                maps[i].put(diff, currentCount + count + 1);
                total += count;
            }
        }
        return (int)total;        
    }
}

The problem focuses on counting the number of arithmetic subsequence slices within a given array using Java. The solution effectively uses dynamic programming coupled with hashing to keep track of the progress.

• Initialize array maps of HashMaps to store count of possible sequences ending with a specified difference up to the current index.
• Iterate through each element in the array using two nested loops
  • Calculate the difference between the current pair of indices.
  • Continue to the next iteration if the difference is outside the integer range.
  • Use the difference as a key in the maps for:
    • Keeping track of count of subsequences with this difference up to the current index.
    • Adding this count to the subsequences count of subsequences ending at the next element.
    • Keeping a running total of all valid arithmetic subsequences encountered.
• Return the total count of arithmetic subsequences found.

The emphasis in the solution is on calculating differences between each pair and leveraging hash maps to store the count of each unique difference, which optimizes the process of identifying valid subsequences. This approach allows handling the comparisons effectively, especially for larger datasets. At the end of the iterations, a cast to int for the total count is required to match the function's signature, ensuring the method returns the correct type.