Check Array Formation Through Concatenation

class Solution {
public:
    bool canAssembleArray(vector<int>& mainArray, vector<vector<int>>& subArrays) {
        int length = mainArray.size();
        unordered_map<int, vector<int>> arrayMap;
        // Populate the map with subArrays
        for (auto section: subArrays){
            arrayMap[section[0]] = section;
        }

        int index = 0;
        while (index < length) {
            // Check if current main array element is a start of some subArray
            if (arrayMap.count(mainArray[index]) == 0) {
                return false;
            }
            // Validate the sequence
            auto currentSection = arrayMap[mainArray[index]];
            for (int element : currentSection) {
                if (element != mainArray[index]) {
                    return false;
                }
                index++;
            }
        }
        return true;
    }
};

This C++ solution addresses the problem of checking if an array can be formed by concatenating subsequences of another set of arrays. The solution involves a few key operations which are detailed below:

1. Initialize an unordered_map to associate the first element of each subarray with the subarray itself. This facilitates quick access and checking during the assembly process.

2. The function iterates through the mainArray, using index to keep track of the current position.

3. At each step, check if the current element in mainArray is the starting element of any subarray by querying the map.

4. If the element is found, verify that subsequent elements in mainArray match the elements in the corresponding subarray from the map. If any element does not match, the function immediately returns false.

5. If the mainArray successfully passes through all checks against the contents of the map without discrepancies, true is returned, indicating the array formation is possible through concatenation of the provided subarrays.

By handling the mapping and sequential verification efficiently, this approach ensures a streamlined check with a focus on early exits upon detecting mismatches, enhancing the performance potential for larger datasets.

class Solution {
    public boolean canAssembleArray(int[] mainArray, int[][] subArrays) {
        int length = mainArray.length;
        Map<Integer, int[]> arrayMap = new HashMap<>();
        for (int[] subArray : subArrays) {
            arrayMap.put(subArray[0], subArray);
        }

        int index = 0;
        while (index < length) {
            if (!arrayMap.containsKey(mainArray[index])) {
                return false;
            }
            int[] currentArray = arrayMap.get(mainArray[index]);
            for (int element : currentArray) {
                if (element != mainArray[index]) {
                    return false;
                }
                index++;
            }
        }
        return true;
    }
}

This Java solution introduces a method called canAssembleArray designed to determine if an integer array (mainArray) can be formed by concatenating other subarrays (subArrays) in a specific sequence. The method accomplishes the following:

1. Initializes length to the length of mainArray and generates an empty HashMap, arrayMap, which maps the first element of each subarray to the corresponding subarray.

2. Iterates through the subArrays. Each subarray gets placed in arrayMap with its first element as the key and the subarray itself as the value.

3. Establishes an index (index) to track positions in mainArray. The loop continues until index matches the length of mainArray.

4. Checks if the current element of mainArray (accessed via index) is a key in arrayMap. If not, the function immediately returns false, indicating that the required subarray to continue the sequence is missing.

5. Retrieves the subarray (currentArray) using the value from arrayMap and iterates through it:

   • If any element of currentArray doesn't match the corresponding element in mainArray, it returns false.
   • Increments index with each successfully matched element.

6. Once all elements have been validated and the loop concludes without mismatches, the method returns true, signifying that mainArray can indeed be formed by concatenating the subarrays in a specific sequence.

class Solution:
    def formArray(self, array: List[int], parts: List[List[int]]) -> bool:
        length = len(array)
        # create dictionary
        hash_map = {part[0]: part for part in parts}

        index = 0
        while index < length:
            # search for the part
            if array[index] not in hash_map:
                return False
            # validate the piece
            current_piece = hash_map[array[index]]
            for val in current_piece:
                if val != array[index]:
                    return False
                index += 1

        return True

The Python solution presented addresses the problem of checking if an array can be formed by concatenating subarrays that are given in a specific sequence. Here’s a breakdown of how the solution works:

• A dictionary (hash_map) is created with the first element of each subarray as the key and the subarray itself as the value. This setup speeds up the search process, allowing quick checks if a part of the main array matches the beginning of any given subarray.
• Iteration through the main array (array) is conducted using a while loop. The variable index is used to keep track of the current position in the main array.
• If the current element in the main array isn't a key in hash_map, it indicates that this element can't start any of the given subarrays, and False is returned.
• If the current element is in the hash_map, the corresponding subarray (current_piece) is retrieved and checked against the main array. Each element in this subarray must match the sequential elements in the main array for the solution to remain viable.
• If all elements of the current subarray match correctly, the index is incremented for the length of the subarray, moving the validation forward.
• If the entire main array is successfully validated with the subarrays, the function returns True.

This solution is effective because it utilizes a hashmap for constant-time look-up operations, reducing the overall complexity when verifying matching arrays and ensuring efficient performance. This mechanism specifically enhances the solution by minimizing the need for nested iterations over both input arrays.