Check Array Formation Through Concatenation - Array - Easy - LeetCode
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Check Array Formation Through Concatenation - Array - Easy - LeetCode

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  • 1The goal is to form the array 'arr' by concatenating arrays from 'pieces' without reordering elements within each piece.
  • 2The function returns true if 'arr' can be formed; otherwise, it returns false based on the provided constraints.
  • 3The time complexity of the solution is O(m*n), where m is the number of pieces and n is the size of individual pieces.

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Key Insight
AskGif

"The goal is to form the array 'arr' by concatenating arrays from 'pieces' without reordering elements within each piece."

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.

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sumitc91

Published on 3 November 2020 · 1 min read · 270 words

Part of AskGif Blog · coding

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