algoadvance

You are given an integer array nums that is sorted in non-decreasing order. Determine if it is possible to split nums into one or more subsequences such that both of the following conditions are true:

  1. Each subsequence is a consecutive increasing sequence (i.e., each integer is exactly one more than the previous integer).
  2. Each subsequence has a length of 3 or more.

Return true if you can split the array according to the above conditions, or false otherwise.

Clarifying Questions

  1. Are there any constraints on the size of nums?
    • The problem does not specify constraints, but typical array constraints and common values should be assumed (e.g., size up to 10^5).
  2. What if there are repeating numbers in the array?
    • Since the array is sorted and you are tasked with finding consecutive subsequences, repeating numbers should not disrupt the ability to find or form consecutive subsequences.
  3. Are negative numbers included in the array?
    • The problem statement does not specify, but we should be prepared for both positive and negative integers given the general nature of integers.

Strategy

  1. Use a hashmap (frequency) to keep track of the frequency of each element in nums.
  2. Use another hashmap (appendfreq) to track how many consecutive subsequences can be extended with the current element.
  3. Iterate through each number in nums.
    • Check if the current number can be added to a previously closed subsequence.
    • If not, form a new subsequence if the next two consecutive numbers are available.
  4. Return False if unable to meet the above conditions, otherwise return True.

Code

def isPossible(nums):
    from collections import defaultdict
    
    # Frequency hashmap to count the occurrences of each number
    frequency = defaultdict(int)
    # Appendfreq hashmap to track the possibility of appending the current number to an existing subsequence
    appendfreq = defaultdict(int)
    
    for num in nums:
        frequency[num] += 1
    
    for num in nums:
        if frequency[num] == 0:
            continue
        
        # If we can append num to an existing subsequence ending in num-1
        if appendfreq[num - 1] > 0:
            appendfreq[num - 1] -= 1
            appendfreq[num] += 1
        # If not, try to create a new subsequence num, num+1, num+2
        elif frequency[num + 1] > 0 and frequency[num + 2] > 0:
            frequency[num + 1] -= 1
            frequency[num + 2] -= 1
            appendfreq[num + 2] += 1
        else:
            return False
        
        frequency[num] -= 1
    
    return True

Time Complexity

  1. Time Complexity: O(N)
    • Where N is the length of nums. We make a single pass through nums to build the frequency hashmap and another pass to try and build/append the subsequences.
  2. Space Complexity: O(N)
    • The space complexity is also O(N) due to the usage of the frequency and appendfreq hashmaps.

This approach ensures an efficient check for forming necessary subsequences while adhering to the problem constraints.

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