1640. Check Array Formation Through Concatenation

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: Direct Implementation

Traverse arr and find the array item whose first element is equal to the current arr[i] in pieces. If not found, return false directly.

If found, we record the array item as pieces[k], and then continue to traverse arr[i] and pieces[k] until pieces[k] is traversed or the elements are not equal.

The traversal ends and true is returned.

Solution 2

Create a hash table with the key being the first element of each array item in pieces and the value being the array item.

Traverse arr. If the current element does not exist in the hash table, directly return false; otherwise, take out the corresponding array item in the hash table and determine whether it is equal to the element in arr. If not, directly return false.

Otherwise, the traversal ends and true is returned.

Time complexity $O(n)$, space complexity $O(n)$. Where $n$ is the length of arr.

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;
    }
}


// Solution 2
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) {
        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;
    }
};


// Solution 2
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;
    }
};

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


# 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

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
}


// Solution 2
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
}

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;
}

/**
 * @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;
};

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
    }
}

1640 - Check Array Formation Through Concatenation

1640. Check Array Formation Through Concatenation

Description

Solutions

Solution 1: Direct Implementation

Solution 2

All Problems

All Solutions

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1640. Check Array Formation Through Concatenation

Description

Solutions

Solution 1: Direct Implementation

Solution 2

All Problems

All Solutions