Leetcode 1630: Arithmetic Subarrays
A sequence of numbers is called arithmetic if it consists of at least two elements, and the difference between every two consecutive elements is the same. Given an array of integers
nums and two arrays l and r representing m range queries, determine whether the subarray nums[l[i], ..., r[i]] can be rearranged to form an arithmetic sequence.Problem
Approach
Steps
Complexity
Input: You are given an array of integers `nums`, and two arrays `l` and `r` representing `m` range queries. Each query defines a range in `nums` as `[l[i], r[i]]`.
Example: nums = [1, 2, 5, 3, 7], l = [0, 1, 2], r = [2, 3, 4]
Constraints:
• n == nums.length
• m == l.length
• m == r.length
• 2 <= n <= 500
• 1 <= m <= 500
• 0 <= l[i] < r[i] < n
• -10^5 <= nums[i] <= 10^5
Output: Return a list of booleans, where each boolean represents whether the corresponding subarray can be rearranged into an arithmetic sequence.
Example: [true, false, true]
Constraints:
Goal: For each range in the queries, check if the subarray can be rearranged to form an arithmetic sequence.
Steps:
• For each query, extract the subarray defined by l[i] and r[i].
• Sort the subarray.
• Check if the difference between each consecutive element is constant.
• If it is constant, the subarray can be rearranged to form an arithmetic sequence; otherwise, it cannot.
Goal: The constraints limit the size of the array `nums` and the number of queries, which ensures the problem can be solved within acceptable time limits.
Steps:
• n == nums.length
• m == l.length
• m == r.length
• 2 <= n <= 500
• 1 <= m <= 500
• 0 <= l[i] < r[i] < n
• -10^5 <= nums[i] <= 10^5
Assumptions:
• All input arrays are non-empty.
• The length of `l` and `r` arrays is equal to `m`.
• Input: nums = [1, 2, 5, 3, 7], l = [0, 1, 2], r = [2, 3, 4]
• Explanation: The first query extracts the subarray [1, 2, 5], which can be rearranged as [1, 2, 5], forming an arithmetic sequence.
Approach: To check if a subarray can be rearranged into an arithmetic sequence, sort the subarray and check the differences between consecutive elements.
Observations:
• We can sort the subarray to make checking for differences easier.
• If all differences between consecutive elements are the same, the sequence is arithmetic.
• Sorting the subarray and checking the differences allows us to validate if a rearranged sequence is arithmetic.
Steps:
• For each query, extract the subarray using l[i] and r[i].
• Sort the subarray to bring the elements in order.
• Check the differences between consecutive elements of the sorted subarray.
• If the differences are the same for all consecutive elements, return `true`; otherwise, return `false`.
Empty Inputs:
• Check for cases where the subarray contains only one element.
Large Inputs:
• Ensure the algorithm handles large arrays efficiently.
Special Values:
• Consider cases where elements are negative or the range is small.
Constraints:
• Ensure that input constraints are met and the algorithm can process the largest inputs.
vector<bool> checkArithmeticSubarrays(vector<int>& nums, vector<int>& l, vector<int>& r) {
vector<bool> res;
for (auto i = 0, j = 0; i < l.size(); ++i) {
vector<int> n(begin(nums) + l[i], begin(nums) + r[i] + 1);
sort(begin(n), end(n));
for (j = 2; j < n.size(); ++j)
if (n[j] - n[j - 1] != n[1] - n[0])
break;
res.push_back(j == n.size());
}
return res;
}
1 : Method Definition
vector<bool> checkArithmeticSubarrays(vector<int>& nums, vector<int>& l, vector<int>& r) {
Define the function 'checkArithmeticSubarrays' that takes three parameters: a vector of integers 'nums', and two vectors 'l' and 'r' which specify the start and end indices of subarrays to check.
2 : Variable Initialization
vector<bool> res;
Initialize a vector 'res' to store the boolean results indicating whether each corresponding subarray is an arithmetic subarray.
3 : Loop Constructs
for (auto i = 0, j = 0; i < l.size(); ++i) {
Start a loop to iterate over the indices in vector 'l' and process each subarray defined by 'l[i]' and 'r[i]'.
4 : Subarray Extraction
vector<int> n(begin(nums) + l[i], begin(nums) + r[i] + 1);
Extract the subarray from 'nums' starting at index 'l[i]' and ending at index 'r[i]' (inclusive) into the vector 'n'.
5 : Sorting
sort(begin(n), end(n));
Sort the extracted subarray 'n' to ensure the elements are in ascending order, which is necessary for checking arithmetic progression.
6 : Loop Constructs
for (j = 2; j < n.size(); ++j)
Start a loop to compare consecutive elements in the sorted subarray 'n', beginning at index 2 (the third element).
7 : Conditional Statements
if (n[j] - n[j - 1] != n[1] - n[0])
Check if the difference between the current element and the previous element is not equal to the difference between the first two elements of the sorted subarray. If it is not, the subarray is not arithmetic.
8 : Loop Constructs
break;
If the difference condition is violated, break out of the inner loop as the subarray is not arithmetic.
9 : Vector Operations
res.push_back(j == n.size());
Push a boolean value into 'res' indicating whether the entire subarray 'n' satisfies the arithmetic progression condition (i.e., the loop completed without breaking).
10 : Return Statement
return res;
Return the vector 'res' containing the results for each subarray: 'true' if the subarray is arithmetic, 'false' otherwise.
Best Case: O(n log n) for each query, where n is the length of the subarray.
Average Case: O(n log n) for each query.
Worst Case: O(n log n) for each query.
Description: The sorting operation dominates the time complexity, and it is done for each query.
Best Case: O(1) if no extra space is used.
Worst Case: O(n) for storing the subarray in memory.
Description: The space complexity is determined by the size of the subarray stored for sorting.
| LeetCode Solutions Library / DSA Sheets / Course Catalog |
|---|
comments powered by Disqus