Leetcode 2811: Check if it is Possible to Split Array
You are given an array nums and an integer m. You need to determine if it’s possible to split the array into n subarrays of size 1, following the rules that each subarray must either have length 1 or have a sum of elements greater than or equal to m.
Problem
Approach
Steps
Complexity
Input: An array nums of length n and an integer m.
Example: Input: nums = [1, 3, 2], m = 5
Constraints:
• 1 <= n == nums.length <= 100
• 1 <= nums[i] <= 100
• 1 <= m <= 200
Output: Return true if it's possible to split the array into n good arrays, otherwise return false.
Example: Output: true
Constraints:
Goal: Determine whether it's possible to split the array into n good arrays based on the described rules.
Steps:
• 1. Start with the entire array.
• 2. Check if it's possible to split the array into two smaller subarrays, where both subarrays are good.
• 3. If a valid split is found, continue the process with the smaller subarrays.
• 4. Return true if it's possible to break down the array into n good subarrays.
Goal: Ensure that the array length and elements meet the constraints.
Steps:
• 1 <= n == nums.length <= 100
• 1 <= nums[i] <= 100
• 1 <= m <= 200
Assumptions:
• The length of the array is at least 1 and no more than 100.
• Each element of the array is between 1 and 100.
• The value of m is between 1 and 200.
• Input: Input: nums = [1, 3, 2], m = 5
• Explanation: By splitting the array into subarrays of length 1, we can achieve a valid result, making the output true.
• Input: Input: nums = [1, 2, 3], m = 7
• Explanation: The array cannot be split into good arrays since no valid subarrays meet the sum >= m condition, so the output is false.
Approach: We need to find a way to split the array into subarrays that meet the good array criteria.
Observations:
• The sum of elements in a subarray plays a crucial role in determining if it's a good array.
• We need to try all possible splits to ensure the array can be divided into valid subarrays.
• We can iterate through the array and look for points where the sum of a split subarray meets the required criteria.
Steps:
• 1. Start by checking if any two consecutive elements meet the criteria for being a good array.
• 2. Recursively try splitting the array into smaller subarrays.
• 3. Return true if a valid split is found, otherwise return false.
Empty Inputs:
• There will always be at least one element in the input array.
Large Inputs:
• Ensure the solution works efficiently for arrays with lengths up to 100.
Special Values:
• If m is large, it may be impossible to split the array in a valid way.
Constraints:
• The solution must be efficient enough to handle the upper limit of input sizes.
bool canSplitArray(vector<int>& nums, int m) {
for (int i = 0; i < nums.size() - 1; ++i)
if (nums[i] + nums[i + 1] >= m)
return true;
return nums.size() < 3;
}
1 : Function Definition
bool canSplitArray(vector<int>& nums, int m) {
The function 'canSplitArray' takes a reference to a vector of integers 'nums' and an integer 'm' as input, and returns a boolean value indicating whether the array can be split based on the condition.
2 : Loop Setup
for (int i = 0; i < nums.size() - 1; ++i)
A loop is initialized to iterate over the elements of the array, from the first element to the second-to-last element, ensuring each pair of consecutive elements is checked.
3 : Condition Check
if (nums[i] + nums[i + 1] >= m)
Within the loop, check if the sum of the current element and the next element in the array is greater than or equal to the given value 'm'.
4 : Return Statement
return true;
If the sum of two consecutive elements meets or exceeds 'm', return 'true' immediately, indicating that the array can be split.
5 : Final Return
return nums.size() < 3;
If no such pair is found, return 'true' if the array contains fewer than 3 elements (since a split is trivially possible with a small array).
Best Case: O(n)
Average Case: O(n^2)
Worst Case: O(n^2)
Description: In the worst case, we may need to check multiple possible splits for each element in the array.
Best Case: O(n)
Worst Case: O(n)
Description: We are storing the intermediate states of the array while attempting to split it into subarrays.
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