Divide Array Into Increasing Sequences
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ArraysGreedy Algorithms
You are given an integer array nums of length n and an integer k. You want to divide this array into one or more disjoint subsequences such that each subsequence is a strictly increasing sequence.
Return the minimum number of subsequences you need to achieve the above division. If the division is not possible, return -1.
A subsequence of an array is a sequence that can be derived from the array by deleting some or no elements without changing the order of the remaining elements. A sequence seq is strictly increasing if seq[i] < seq[i+1] for all 0 < i < seq.length - 1.
Example 1:
Input: nums = [1,2,3,4] Output: 1 Explanation: The array can be divided into one strictly increasing subsequence.
Example 2:
Input: nums = [1,3,3,3,4] Output: 3 Explanation: The array can be divided into the following subsequences: [1,3,4] (indices = 0, 1, 4) [3] (indices = 2) [3] (indices = 3)
Example 3:
Input: nums = [1,2,1,2,3] Output: 2 Explanation: The array can be divided into the following subsequences: [1, 2, 3] (indices = 0, 1, 4) [1, 2] (indices = 2, 3)
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 105
Solution
Clarifying Questions
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
- What are the constraints on the numbers within the input array, such as the range of possible values and data type (e.g., integers, floats)? Can the array contain negative numbers, zeros, or both?
- How should I handle the case where it's impossible to divide the array into increasing sequences of the specified length? Should I return a boolean value, an empty array, or throw an exception?
- Can you provide more details on what constitutes an 'increasing sequence'? Is strictly increasing required (e.g., `[1, 2, 3]`), or is non-decreasing allowed (e.g., `[1, 2, 2, 3]`?)
- What is the expected output if there are multiple valid ways to divide the array into increasing sequences? Do I need to return all possible solutions, or is any valid solution acceptable?
- What is the minimum length required for an increasing sequence, if not explicitly defined? Is there a maximum limit on how many increasing sequences I should divide the input array into?
Brute Force Solution
Approach
The brute force approach exhaustively explores every possible way to divide the numbers into increasing sequences. It creates all possible groupings and verifies whether each grouping satisfies the increasing sequence requirement. It's like trying out every single combination to find a valid solution.
Here's how the algorithm would work step-by-step:
- Consider the first number and assume it starts a new sequence.
- Then, consider the second number. We can either add it to the existing sequence or start a brand new sequence with it.
- Continue this process for each remaining number. At each number, explore the possibility of adding it to an existing sequence, or starting a completely new sequence.
- Every time you finish placing all numbers into sequences, check if all sequences are in strictly increasing order. Discard any grouping that doesn't satisfy this condition.
- If you find even one way to arrange the numbers into valid increasing sequences after exploring every possibility, then a solution exists. Otherwise, no solution exists.
Code Implementation
def can_divide_into_increasing_sequences_brute_force(numbers):
def is_valid(sequences):
for sequence in sequences:
if len(sequence) > 1:
for i in range(len(sequence) - 1):
if sequence[i] >= sequence[i+1]:
return False
return True
def solve(index, current_sequences):
# Base case: all numbers have been assigned to sequences
if index == len(numbers):
return is_valid(current_sequences)
# Try adding the current number to each existing sequence
for sequence_index in range(len(current_sequences)):
new_sequences = [seq[:] for seq in current_sequences]
new_sequences[sequence_index].append(numbers[index])
# Explore the possibility of adding to sequence
if solve(index + 1, new_sequences):
return True
# Try starting a new sequence with the current number
# This is necessary to explore all combinations
new_sequences = [seq[:] for seq in current_sequences]
new_sequences.append([numbers[index]])
if solve(index + 1, new_sequences):
return True
return False
# Initialize with no sequences
return solve(0, [])Big(O) Analysis
Time Complexity
O(k^n * n) – The brute force approach explores all possible ways to divide the n numbers into sequences. For each number, there are up to k existing sequences (where k could be as large as n), so we have roughly k choices for each of the n numbers. This leads to O(k^n) possibilities. For each of these possibilities, we need to verify if the sequences are strictly increasing, which takes O(n) time in the worst case to iterate through all sequences and their elements. Therefore, the overall time complexity is O(k^n * n). In the worst case, k can be proportional to n, thus the complexity is often considered as O(n^n * n).
Space Complexity
O(N) – The brute force approach explores all possible combinations of placing N numbers into increasing sequences. In the worst-case scenario, the recursion depth could reach N, as each number can potentially start a new sequence, leading to N recursive calls on the call stack. Furthermore, to track each possible grouping of numbers into sequences before validating them, the algorithm implicitly uses space to store these intermediate sequences, also up to N elements across all sequences. Thus, the space complexity is driven by the maximum depth of the recursion and temporary storage of intermediate results, resulting in O(N) space.
Optimal Solution
Approach
The goal is to split a collection of numbers into the smallest possible number of increasing sequences. We achieve this by figuring out how many times each number appears and then arranging the numbers smartly to minimize the sequences needed. This is a clever way to quickly find the solution without checking every possible arrangement.
Here's how the algorithm would work step-by-step:
- First, count how many times each number appears in the collection.
- Next, examine the counts to determine how many sequences will be needed in total, because it's at least as many sequences as the number that appears most often.
- Try to form the increasing sequences. Start with the smallest number and put it into each sequence until you've used up all of that number.
- Move on to the next smallest number and do the same thing, making sure each number is larger than the one before it in its sequence.
- If at any point you can't place a number because it's not bigger than the last number in a sequence, that means your initial number of sequences was too small. If this happens, it's impossible to divide up the initial collection of numbers into the desired number of increasing sequences.
- If you can successfully place all the numbers following the procedure above, you've successfully divided up the collection into the minimum number of increasing sequences.
Code Implementation
def divide_array_into_sequences(numbers, sequence_length):
number_counts = {}
for number in numbers:
number_counts[number] = number_counts.get(number, 0) + 1
if not number_counts:
return True
max_frequency = max(number_counts.values())
if max_frequency > sequence_length:
return False
sequences = [[] for _ in range(sequence_length)]
sorted_numbers = sorted(number_counts.keys())
# Iterate through numbers to build sequences
for number in sorted_numbers:
for _ in range(number_counts[number]):
placed = False
for i in range(sequence_length):
# Place number only if its greater than last in sequence
if not sequences[i] or number > sequences[i][-1]:
sequences[i].append(number)
placed = True
break
# Cannot build required sequences
if not placed:
return False
return TrueBig(O) Analysis
Time Complexity
O(n log n) – Counting the frequency of each number involves iterating through the input array of size n, which takes O(n) time. Sorting the distinct numbers (keys of the frequency map) takes O(k log k) where k is the number of distinct elements, and k <= n, so this is at most O(n log n). The subsequent steps, involve iterating through the sorted numbers and attempting to place them in sequences. While the placement logic involves checking the current sequences, the dominant cost comes from sorting the numbers, hence the overall time complexity is O(n log n).
Space Complexity
O(N) – The algorithm primarily uses a frequency map to count the occurrences of each number, where N is the number of elements in the input array. This map stores at most N unique numbers and their corresponding counts, resulting in O(N) space. While the sequences themselves are implicitly created, the dominant space complexity comes from this frequency map. Therefore, the auxiliary space complexity is O(N).
Edge Cases
| Case | How to Handle |
|---|---|
| Null or empty input array | Return an empty list or throw an IllegalArgumentException, depending on the problem specification, as there are no elements to divide. |
| Array with only one element | Return false or an empty list as a single element cannot form an increasing sequence of length k>1. |
| k is greater than the array length | Return false or an empty list because you cannot divide the array into sequences longer than the array's size. |
| Array with all identical elements and k > 1 | Return false because it's impossible to create strictly increasing sequences from identical elements. |
| Array is already sorted in decreasing order | The algorithm should correctly identify and return false because increasing sequences cannot be formed in this scenario. |
| Array contains negative numbers, zeros, and positive numbers | The sorting step will handle this correctly by arranging elements in ascending order regardless of sign. |
| k = 1 (sequences of length 1) | Return True since the input array trivially satisfies the condition, or return the entire array as valid subsequences. |
| Integer overflow when calculating counts of elements if using large inputs and frequency maps | Use a data type with sufficient capacity (e.g., long) or check for overflows during the count calculation. |
Null or empty input array
How to Handle:
Return an empty list or throw an IllegalArgumentException, depending on the problem specification, as there are no elements to divide.
Array with only one element
How to Handle:
Return false or an empty list as a single element cannot form an increasing sequence of length k>1.
k is greater than the array length
How to Handle:
Return false or an empty list because you cannot divide the array into sequences longer than the array's size.
Array with all identical elements and k > 1
How to Handle:
Return false because it's impossible to create strictly increasing sequences from identical elements.
Array is already sorted in decreasing order
How to Handle:
The algorithm should correctly identify and return false because increasing sequences cannot be formed in this scenario.
Array contains negative numbers, zeros, and positive numbers
How to Handle:
The sorting step will handle this correctly by arranging elements in ascending order regardless of sign.
k = 1 (sequences of length 1)
How to Handle:
Return True since the input array trivially satisfies the condition, or return the entire array as valid subsequences.
Integer overflow when calculating counts of elements if using large inputs and frequency maps
How to Handle:
Use a data type with sufficient capacity (e.g., long) or check for overflows during the count calculation.
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