Leetcode 1764: Form Array by Concatenating Subarrays of Another Array
Given an array of subarrays
groups and a single array nums, your task is to determine whether it is possible to extract each subarray from groups as a contiguous segment from nums. The subarrays must appear in the same order as in groups, and they should be disjoint, meaning no element from nums can belong to more than one subarray. Return true if this is possible, and false otherwise.Problem
Approach
Steps
Complexity
Input: You are given a 2D array `groups` where each subarray is a sequence of integers. Additionally, you are given an integer array `nums`.
Example: Input: groups = [[2, 3], [1, 2, 3]], nums = [2, 3, 1, 2, 3, 4]
Constraints:
• 1 <= groups.length <= 1000
• 1 <= groups[i].length <= 1000
• 1 <= sum(groups[i].length) <= 1000
• 1 <= nums.length <= 1000
• nums and group elements are between -10^7 and 10^7
Output: Return true if you can find each subarray from `groups` as disjoint subarrays within `nums` while maintaining the order, otherwise return false.
Example: Output: true
Constraints:
Goal: The goal is to check whether each subarray from `groups` can be found in `nums` in the given order and that the subarrays are disjoint.
Steps:
• 1. Start iterating through `nums` while keeping track of the current subarray being matched from `groups`.
• 2. For each subarray, check if it appears in `nums` as a contiguous block.
• 3. Once a match is found, ensure no elements from `nums` are reused for the next subarray by continuing the search beyond the matched segment.
• 4. Return true if all subarrays are found, otherwise return false.
Goal: The solution must handle edge cases, such as when there are no possible matches or the input sizes are near their limits.
Steps:
• The solution must handle cases where `nums` or the subarrays in `groups` are large efficiently.
Assumptions:
• The arrays `nums` and `groups` consist of integers only.
• Subarrays in `groups` must appear in the same order in `nums`.
• Input: Input: groups = [[2, 3], [1, 2, 3]], nums = [2, 3, 1, 2, 3, 4]
• Explanation: The first subarray [2, 3] can be found at the start of `nums`. The second subarray [1, 2, 3] follows immediately after the first. Thus, the answer is true.
• Input: Input: groups = [[10, -2], [1, 2, 3, 4]], nums = [1, 2, 3, 4, 10, -2]
• Explanation: The subarrays are not in the same order in `nums`. The second subarray [1, 2, 3, 4] appears before [10, -2], so the answer is false.
• Input: Input: groups = [[3, 4], [5, 6]], nums = [3, 5, 4, 6]
• Explanation: The subarrays are disjoint, but they do not appear in the same order. Therefore, the answer is false.
Approach: The solution will iterate through `nums` and try to match each subarray from `groups` in sequence. We will attempt to find each subarray as a contiguous block, ensuring that no element is shared between subarrays.
Observations:
• The subarrays must be found in order, and no overlap should occur between the subarrays in `nums`.
• An efficient approach is to use a greedy matching strategy to find each subarray and continue searching from the end of the last match.
Steps:
• 1. Start iterating through `nums` with an index to check for the first subarray.
• 2. For each subarray in `groups`, find its exact match in `nums` while ensuring no overlap with previous matches.
• 3. Once a subarray is found, move the index in `nums` past that subarray and continue checking for the next subarray in `groups`.
• 4. Return true if all subarrays are found, otherwise return false.
Empty Inputs:
• If `nums` is empty or if the total sum of lengths of subarrays in `groups` exceeds the length of `nums`, return false.
Large Inputs:
• The solution must handle cases where the length of `nums` and the total number of elements in `groups` are close to their upper limits efficiently.
Special Values:
• If any group is empty or a group doesn't have a matching subsequence in `nums`, return false.
Constraints:
• The solution should be optimized to handle up to 1000 subarrays in `groups` and 1000 elements in `nums`.
bool canChoose(vector<vector<int>>& group, vector<int>& nums) {
int numsIdx = 0, grpIdx = 0;
while(numsIdx < nums.size() && grpIdx < group.size()) {
int matchCnt = 0;
while(numsIdx + matchCnt < nums.size() &&
matchCnt < group[grpIdx].size() &&
nums[numsIdx + matchCnt] == group[grpIdx][matchCnt])
matchCnt++;
if(matchCnt == group[grpIdx].size()) {
grpIdx++;
numsIdx += matchCnt;
} else numsIdx++;
}
return grpIdx == group.size();
}
1 : Function Declaration
bool canChoose(vector<vector<int>>& group, vector<int>& nums) {
Declares the function for determining if groups can be matched in the given array `nums`.
2 : Variable Initialization
int numsIdx = 0, grpIdx = 0;
Initializes indices for tracking the current position in `nums` and `group`.
3 : While Loop
while(numsIdx < nums.size() && grpIdx < group.size()) {
Iterates over `nums` and `group` as long as indices are within bounds.
4 : Variable Initialization
int matchCnt = 0;
Initializes a counter to track the number of matching elements for the current group.
5 : While Loop
while(numsIdx + matchCnt < nums.size() &&
Checks for matching elements between the current group and the array `nums`.
6 : Condition Check
matchCnt < group[grpIdx].size() &&
Ensures the match count does not exceed the size of the current group.
7 : Condition Check
nums[numsIdx + matchCnt] == group[grpIdx][matchCnt])
Checks if the current elements in `nums` and `group` match.
8 : Increment
matchCnt++;
Increments the match count when elements match.
9 : Block End
Empty line for separating blocks for readability.
10 : Condition Check
if(matchCnt == group[grpIdx].size()) {
Checks if all elements in the current group have been matched.
11 : Index Update
grpIdx++;
Moves to the next group after successfully matching the current group.
12 : Index Update
numsIdx += matchCnt;
Skips past the matched elements in `nums`.
13 : Else Block
} else numsIdx++;
Moves to the next element in `nums` if the current group does not match.
14 : Return Statement
return grpIdx == group.size();
Returns `true` if all groups were matched successfully; otherwise, `false`.
Best Case: O(n)
Average Case: O(n^2)
Worst Case: O(n^3)
Description: In the worst case, each subarray needs to be matched against a portion of `nums`, leading to a time complexity of O(n^3) when considering the sum of subarray lengths.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is constant since we are only using a few integer variables to track indices and matches.
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