Leetcode 659: Split Array into Consecutive Subsequences
grid47 Exploring patterns and algorithms
Sep 2, 2024
6 min read
Solution to LeetCode 659: Split Array into Consecutive Subsequences Problem
You are given a sorted integer array nums. Determine if it is possible to split nums into one or more subsequences such that each subsequence is a consecutive increasing sequence and each subsequence has a length of 3 or more.
Problem
Approach
Steps
Complexity
Input: The input consists of a sorted integer array nums.
Example: nums = [1, 2, 3, 3, 4, 5]
Constraints:
• 1 <= nums.length <= 10^4
• -1000 <= nums[i] <= 1000
• nums is sorted in non-decreasing order.
Output: Return true if it is possible to split nums into valid subsequences, otherwise return false.
Example: true
Constraints:
• The output is a boolean value indicating whether the split is possible.
Goal: The goal is to check if it is possible to form valid subsequences of length 3 or more, where each subsequence is a consecutive increasing sequence.
Steps:
• 1. Use a map to count occurrences of each number in the array.
• 2. Track the subsequences that can be extended with the current number.
• 3. Ensure that each number is either added to an existing subsequence or starts a new valid subsequence of at least length 3.
Goal: The input array is sorted, and the length of the array is between 1 and 10^4. The integer values are between -1000 and 1000.
Steps:
• 1 <= nums.length <= 10^4
• -1000 <= nums[i] <= 1000
Assumptions:
• The input array is always sorted in non-decreasing order.
• The length of the array is within the specified limits.
• Input: nums = [1, 2, 3, 3, 4, 5]
• Explanation: The array can be split into two subsequences: [1, 2, 3] and [3, 4, 5], both of which are valid.
• Input: nums = [1, 2, 3, 4, 4, 5]
• Explanation: It is impossible to split this array into valid subsequences, as the subsequence starting with the second 4 cannot be extended to form a valid sequence.
Approach: This problem can be solved by tracking the frequency of each element and ensuring that each number is placed in an appropriate subsequence.
Observations:
• We can use a frequency map to keep track of the remaining numbers.
• Subsequences should only be extended if the previous number is part of a valid subsequence.
• We need to carefully manage how we form and extend subsequences to meet the requirement that all subsequences have at least 3 elements.
Steps:
• 1. Use a hash map to track the number of remaining occurrences of each number.
• 2. Use a second map to track how many subsequences can be extended with the current number.
• 3. For each number, try to place it in an existing subsequence or start a new subsequence if possible.
Empty Inputs:
• There will be no empty arrays, as per the constraints.
Large Inputs:
• The solution should handle arrays of size up to 10^4 efficiently.
Special Values:
• Arrays where consecutive numbers repeat can be tricky, but the solution will handle them by extending or starting new subsequences.
Constraints:
• Ensure the number of subsequences is managed correctly to meet the length 3 constraint.
This is the function definition for `isPossible`, which takes a vector of integers `nums` and returns a boolean value indicating whether it's possible to split the array into consecutive subsequences.
2 : Variable Initialization
unordered_map<int, int> left, end;
Two hash maps `left` and `end` are initialized. `left` tracks the count of available elements, and `end` tracks the number of subsequences ending at a specific number.
3 : Loop Initialization
for(intnum: nums)
Iterate over each number in the `nums` array.
4 : Left Count Update
left[num]++;
For each number, increment the count in the `left` map to indicate that this number is available for building subsequences.
5 : Loop Start
for(intnum: nums) { // nums is in increasing order
This is the second loop that iterates through the numbers in `nums`. It processes each number in increasing order to check if it can be added to a subsequence.
6 : Skip Processed Numbers
if(left[num] ==0) continue;
If the number has already been used in a subsequence (i.e., `left[num] == 0`), skip processing it.
7 : Left Count Decrement
left[num]--;
Decrement the count of the current number in `left` because it is being used to form a subsequence.
8 : Check Existing Subsequence
if(end[num -1] >0) {
If there is a subsequence ending with `num - 1`, extend that subsequence by adding the current `num` to it.
9 : Extend Subsequence
end[num -1]--;
Decrement the count of subsequences ending with `num - 1` since we are now using `num` to extend it.
10 : End Subsequence with Current Number
end[num]++;
Increment the count of subsequences ending with `num` after adding it to an existing subsequence.
11 : Check for New Subsequence
} elseif(left[num +1] >0&& left[num +2] >0) {
If no subsequence can be extended with the current number, check if we can start a new subsequence with `num`, `num + 1`, and `num + 2`.
12 : Start New Subsequence
left[num +1]--;
Decrement the count of `num + 1` in `left` as it's being used to start a new subsequence.
13 : New Subsequence Continuation
left[num +2]--;
Decrement the count of `num + 2` in `left` as it's also being used to continue the new subsequence.
14 : End New Subsequence
end [num +2]++;
Increment the count of subsequences ending with `num + 2` to mark the successful creation of a new subsequence.
15 : Return False
} elsereturnfalse;
If neither extending an existing subsequence nor starting a new one is possible, return `false` to indicate that it's not possible to split the array into consecutive subsequences.
16 : Return True
returntrue;
If the loop completes without returning `false`, return `true`, indicating that the array can be split into consecutive subsequences.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is O(n) as we process each element in the array once.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) due to the storage needed for the frequency and subsequence maps.
