Question

1640. Check Array Formation Through Concatenation

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Description

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 = [15,88], pieces = [[88],[15]]
Output: true
Explanation: Concatenate [15] then [88]

Example 2:

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

Example 3:

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

 

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).

Solutions

Solution 1

class Solution:
  def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool:
    i = 0
    while i < len(arr):
      k = 0
      while k < len(pieces) and pieces[k][0] != arr[i]:
        k += 1
      if k == len(pieces):
        return False
      j = 0
      while j < len(pieces[k]) and arr[i] == pieces[k][j]:
        i, j = i + 1, j + 1
    return True

class Solution {
  public boolean canFormArray(int[] arr, int[][] pieces) {
    for (int i = 0; i < arr.length;) {
      int k = 0;
      while (k < pieces.length && pieces[k][0] != arr[i]) {
        ++k;
      }
      if (k == pieces.length) {
        return false;
      }
      int j = 0;
      while (j < pieces[k].length && arr[i] == pieces[k][j]) {
        ++i;
        ++j;
      }
    }
    return true;
  }
}

class Solution {
public:
  bool canFormArray(vector<int>& arr, vector<vector<int>>& pieces) {
    for (int i = 0; i < arr.size();) {
      int k = 0;
      while (k < pieces.size() && pieces[k][0] != arr[i]) {
        ++k;
      }
      if (k == pieces.size()) {
        return false;
      }
      int j = 0;
      while (j < pieces[k].size() && arr[i] == pieces[k][j]) {
        ++i;
        ++j;
      }
    }
    return true;
  }
};

func canFormArray(arr []int, pieces [][]int) bool {
  for i := 0; i < len(arr); {
    k := 0
    for k < len(pieces) && pieces[k][0] != arr[i] {
      k++
    }
    if k == len(pieces) {
      return false
    }
    j := 0
    for j < len(pieces[k]) && arr[i] == pieces[k][j] {
      i, j = i+1, j+1
    }
  }
  return true
}

function canFormArray(arr: number[], pieces: number[][]): boolean {
  const n = arr.length;
  let i = 0;
  while (i < n) {
    const target = arr[i];
    const items = pieces.find(v => v[0] === target);
    if (items == null) {
      return false;
    }
    for (const item of items) {
      if (item !== arr[i]) {
        return false;
      }
      i++;
    }
  }
  return true;
}

use std::collections::HashMap;
impl Solution {
  pub fn can_form_array(arr: Vec<i32>, pieces: Vec<Vec<i32>>) -> bool {
    let n = arr.len();
    let mut map = HashMap::new();
    for (i, v) in pieces.iter().enumerate() {
      map.insert(v[0], i);
    }
    let mut i = 0;
    while i < n {
      match map.get(&arr[i]) {
        None => {
          return false;
        }
        Some(&j) => {
          for &item in pieces[j].iter() {
            if item != arr[i] {
              return false;
            }
            i += 1;
          }
        }
      }
    }
    true
  }
}

/**
 * @param {number[]} arr
 * @param {number[][]} pieces
 * @return {boolean}
 */
var canFormArray = function (arr, pieces) {
  const d = new Map();
  for (const p of pieces) {
    d.set(p[0], p);
  }
  for (let i = 0; i < arr.length; ) {
    if (!d.has(arr[i])) {
      return false;
    }
    const p = d.get(arr[i]);
    for (const v of p) {
      if (arr[i++] != v) {
        return false;
      }
    }
  }
  return true;
};

Solution 2

class Solution:
  def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool:
    d = {p[0]: p for p in pieces}
    i, n = 0, len(arr)
    while i < n:
      if arr[i] not in d:
        return False
      p = d[arr[i]]
      if arr[i : i + len(p)] != p:
        return False
      i += len(p)
    return True

class Solution {
  public boolean canFormArray(int[] arr, int[][] pieces) {
    Map<Integer, int[]> d = new HashMap<>();
    for (var p : pieces) {
      d.put(p[0], p);
    }
    for (int i = 0; i < arr.length;) {
      if (!d.containsKey(arr[i])) {
        return false;
      }
      for (int v : d.get(arr[i])) {
        if (arr[i++] != v) {
          return false;
        }
      }
    }
    return true;
  }
}

class Solution {
public:
  bool canFormArray(vector<int>& arr, vector<vector<int>>& pieces) {
    unordered_map<int, vector<int>> d;
    for (auto& p : pieces) {
      d[p[0]] = p;
    }
    for (int i = 0; i < arr.size();) {
      if (!d.count(arr[i])) {
        return false;
      }
      for (int& v : d[arr[i]]) {
        if (arr[i++] != v) {
          return false;
        }
      }
    }
    return true;
  }
};

func canFormArray(arr []int, pieces [][]int) bool {
  d := map[int][]int{}
  for _, p := range pieces {
    d[p[0]] = p
  }
  for i := 0; i < len(arr); {
    p, ok := d[arr[i]]
    if !ok {
      return false
    }
    for _, v := range p {
      if arr[i] != v {
        return false
      }
      i++
    }
  }
  return true
}