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by Sumit Chourasia | Nov 03, 2020 | Category :coding | Tags : algorithm array data-structure easy leetcode

Check Array Formation Through Concatenation - Array - Easy - LeetCode

Check Array Formation Through Concatenation - Array - Easy - LeetCode

You are given an array of distinct integers arr and an array of integer arrays pieces, where the integers in pieces are distinct. Your goal is to form arr by concatenating the arrays in pieces in any order. However, you are not allowed to reorder the integers in each array pieces[i].

Return true if it is possible to form the array arr from pieces. Otherwise, return false.

 

Example 1:

Input: arr = [85], pieces = [[85]]
Output: true
Example 2:

Input: arr = [15,88], pieces = [[88],[15]]
Output: true
Explanation: Concatenate [15] then [88]
Example 3:

Input: arr = [49,18,16], pieces = [[16,18,49]]
Output: false
Explanation: Even though the numbers match, we cannot reorder pieces[0].
Example 4:

Input: arr = [91,4,64,78], pieces = [[78],[4,64],[91]]
Output: true
Explanation: Concatenate [91] then [4,64] then [78]
Example 5:

Input: arr = [1,3,5,7], pieces = [[2,4,6,8]]
Output: false
 

Constraints:

1 <= pieces.length <= arr.length <= 100
sum(pieces[i].length) == arr.length
1 <= pieces[i].length <= arr.length
1 <= arr[i], pieces[i][j] <= 100
The integers in arr are distinct.
The integers in pieces are distinct (i.e., If we flatten pieces in a 1D array, all the integers in this array are distinct).

public class Solution {
    public bool CanFormArray(int[] arr, int[][] pieces) {
        var set = new HashSet<string>();
        for(int i=0;i<pieces.Length;i++){
            set.Add(GetKey(pieces[i]));
        }
        
        var strList = new List<int>();
        var str = "";
        for(int i=0;i<arr.Length;i++){
            strList.Add(arr[i]);
            str = GetKey(strList.ToArray());
            if(set.Contains(str)){                
                strList.Clear();
            }            
        }
        
        return strList.Count()==0;
    }
    
    private string GetKey(int[] piece){
        return string.Join(":",piece);
    }
}

Time Complexity: O(m*n)

Space Complexity: O(m)

Where m is the size of the places list and n is the size of the individual place list.

Contributed By: Sumit Chourasia
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