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Solution.java
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Solution.java
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package checkarrayformationconcatenation.bruteforce;
/**
* 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).
*/
class Solution {
/**
* Runtime: 0 ms, faster than 100.00% of Java online submissions for Check Array Formation Through Concatenation.
* Memory Usage: 38.4 MB, less than 36.68% of Java online submissions for Check Array Formation Through Concatenation.
*
* @param arr
* @param pieces
* @return
*/
boolean canFormArray(int[] arr, int[][] pieces) {
int i = 0;
int pos = -1;
while (i < arr.length) {
int curr = arr[i];
// find curr in a piece in pieces
for (int j = 0; j < pieces.length; j++) {
// find first matching piece in pieces
if (pieces[j][0] == curr) {
pos = j;
break;
}
}
// curr arr[i] not found anywhere
if (pos == -1)
return false;
// check if subsequent pieces match subsequent elements in arr
for (int x : pieces[pos]) {
if (arr[i] != x)
return false;
i++;
}
}
return true;
}
}