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Find the Count of Monotonic Pairs I

Hard

Asked by:

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Topics:

ArraysDynamic Programming

You are given an array of positive integers nums of length n.

We call a pair of non-negative integer arrays (arr1, arr2) monotonic if:

The lengths of both arrays are n.
arr1 is monotonically non-decreasing, in other words, arr1[0] <= arr1[1] <= ... <= arr1[n - 1].
arr2 is monotonically non-increasing, in other words, arr2[0] >= arr2[1] >= ... >= arr2[n - 1].
arr1[i] + arr2[i] == nums[i] for all 0 <= i <= n - 1.

Return the count of monotonic pairs.

Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [2,3,2]

Output: 4

Explanation:

The good pairs are:

([0, 1, 1], [2, 2, 1])
([0, 1, 2], [2, 2, 0])
([0, 2, 2], [2, 1, 0])
([1, 2, 2], [1, 1, 0])

Example 2:

Input: nums = [5,5,5,5]

Output: 126

Constraints:

1 <= n == nums.length <= 2000
1 <= nums[i] <= 50

Try coding it in LeetCode

Find the Count of Monotonic Pairs I

Hard

Asked by:

21 views

Topics:

ArraysDynamic Programming

You are given an array of positive integers nums of length n.

We call a pair of non-negative integer arrays (arr1, arr2) monotonic if:

The lengths of both arrays are n.
arr1 is monotonically non-decreasing, in other words, arr1[0] <= arr1[1] <= ... <= arr1[n - 1].
arr2 is monotonically non-increasing, in other words, arr2[0] >= arr2[1] >= ... >= arr2[n - 1].
arr1[i] + arr2[i] == nums[i] for all 0 <= i <= n - 1.

Return the count of monotonic pairs.

Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [2,3,2]

Output: 4

Explanation:

The good pairs are:

([0, 1, 1], [2, 2, 1])
([0, 1, 2], [2, 2, 0])
([0, 2, 2], [2, 1, 0])
([1, 2, 2], [1, 1, 0])

Example 2:

Input: nums = [5,5,5,5]

Output: 126

Constraints:

1 <= n == nums.length <= 2000
1 <= nums[i] <= 50

Try coding it in LeetCode

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 is the range of integer values within the input array?
Can the input array be empty or null?
Are duplicate numbers allowed in the array, and if so, how should they be handled when determining monotonicity?
Could you define more explicitly what constitutes a 'monotonic pair' in this context? Specifically, should it be strictly increasing/decreasing or non-decreasing/non-increasing?
If no monotonic pairs exist, what should the function return?

Brute Force Solution

Approach

We need to find pairs of numbers where the first number is less than or equal to the second. The brute force method simply checks every single possible pair to see if it meets this condition.

Here's how the algorithm would work step-by-step:

Take the very first number in the list.
Compare that number to every other number in the list, one by one.
If the first number is less than or equal to the second number, count that as a monotonic pair.
Move to the second number in the original list.
Compare that number to every other number in the list, one by one.
Again, if the second number is less than or equal to any other number, count that as a monotonic pair.
Continue this process, taking each number in the list and comparing it to every other number.
Add up all the times you found a pair that fits the condition (the first number is less than or equal to the second number).
The final total is the count of all monotonic pairs.

Code Implementation

def find_monotonic_pairs_brute_force(numbers):
    monotonic_pair_count = 0

    # Iterate through each number in the input list.
    for first_number_index in range(len(numbers)):

        for second_number_index in range(len(numbers)):
            # Check if the pair is monotonic.
            if numbers[first_number_index] <= numbers[second_number_index]:
                # Increment the monotonic pair count.
                monotonic_pair_count += 1

    return monotonic_pair_count

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the n elements in the input array. For each element, it compares it to all other n elements in the array to check for the monotonic pair condition. This results in n comparisons for each of the n elements, leading to a total of approximately n * n operations. Therefore, the time complexity is O(n²).

Space Complexity

O(1) – The described algorithm iterates through the input list using nested loops. It only uses a few integer variables to store indices and the count of monotonic pairs. No additional data structures that scale with the input size N (the number of elements in the list) are created. Therefore, the auxiliary space required is constant.

Optimal Solution

Approach

The key is to efficiently count the pairs that follow the rules without checking every single possible pair. We can do this by going through the sequence and focusing on how each number relates to the numbers that come after it. By keeping track of the ongoing counts, we can quickly determine how many valid pairs each number contributes to.

Here's how the algorithm would work step-by-step:

Start at the beginning of the sequence of numbers.
For the current number, count how many numbers following it are greater than or equal to it. This gives you the increasing pairs this number can make.
Also, count how many numbers following it are less than or equal to it. This gives you the decreasing pairs this number can make.
Add both of those counts together. This is the total number of monotonic pairs that begin with the current number.
Move on to the next number in the sequence and repeat the process.
Continue until you have checked every number in the sequence.
Finally, add up all the counts you found for each number. The final sum is the total number of monotonic pairs in the entire sequence.

Code Implementation

def find_count_of_monotonic_pairs_i(sequence):
    total_monotonic_pairs = 0
    sequence_length = len(sequence)

    for i in range(sequence_length):
        increasing_pairs_count = 0
        decreasing_pairs_count = 0

        # Iterate through the rest of the sequence
        for j in range(i + 1, sequence_length):
            # Count increasing pairs
            if sequence[j] >= sequence[i]:
                increasing_pairs_count += 1

            # Count decreasing pairs
            if sequence[j] <= sequence[i]:
                decreasing_pairs_count += 1

        # Add the counts to the total
        total_monotonic_pairs += increasing_pairs_count + decreasing_pairs_count

    return total_monotonic_pairs

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the n elements in the input sequence. For each element, it then iterates through the remaining elements after it to count both increasing and decreasing pairs. This inner iteration performs at most n-1 comparisons in the worst case for the first element, and progressively fewer for subsequent elements. Therefore, the total number of operations is proportional to the sum of numbers from 1 to n-1, which can be expressed as n * (n-1) / 2, resulting in a time complexity of O(n²).

Space Complexity

O(1) – The provided explanation calculates monotonic pairs by iterating through the sequence and comparing each number to the subsequent numbers. It only maintains counters for increasing and decreasing pairs for each number, without using any auxiliary data structures that scale with the input size N (the length of the sequence). These counters occupy a fixed amount of space. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Empty input array	Return 0, as no monotonic pairs are possible.
Input array with only one element	Return 0, as a pair requires at least two elements.
Array with all identical elements	Count the number of pairs using n*(n-1)/2 as all pairs are monotonic in this case
Array sorted in strictly increasing order	Count the number of pairs as all are monotonic (increasing).
Array sorted in strictly decreasing order	Count the number of pairs as all are monotonic (decreasing).
Array with positive and negative numbers	The algorithm should handle negative and positive numbers correctly as it is comparison-based.
Large input array exceeding memory limits	Consider using a more memory-efficient data structure or algorithm like an online algorithm if memory is a constraint.
Integer overflow when calculating the count of monotonic pairs for large arrays	Use a larger data type like long to store the count to avoid overflow issues.

Empty input array

How to Handle:

Return 0, as no monotonic pairs are possible.

Input array with only one element

How to Handle:

Return 0, as a pair requires at least two elements.

Array with all identical elements

How to Handle:

Count the number of pairs using n*(n-1)/2 as all pairs are monotonic in this case

Array sorted in strictly increasing order

How to Handle:

Count the number of pairs as all are monotonic (increasing).

Array sorted in strictly decreasing order

How to Handle:

Count the number of pairs as all are monotonic (decreasing).

Array with positive and negative numbers

How to Handle:

The algorithm should handle negative and positive numbers correctly as it is comparison-based.

Large input array exceeding memory limits

How to Handle:

Consider using a more memory-efficient data structure or algorithm like an online algorithm if memory is a constraint.

Integer overflow when calculating the count of monotonic pairs for large arrays

How to Handle:

Use a larger data type like long to store the count to avoid overflow issues.

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 is the range of integer values within the input array?
Can the input array be empty or null?
Are duplicate numbers allowed in the array, and if so, how should they be handled when determining monotonicity?
Could you define more explicitly what constitutes a 'monotonic pair' in this context? Specifically, should it be strictly increasing/decreasing or non-decreasing/non-increasing?
If no monotonic pairs exist, what should the function return?

Brute Force Solution

Approach

We need to find pairs of numbers where the first number is less than or equal to the second. The brute force method simply checks every single possible pair to see if it meets this condition.

Here's how the algorithm would work step-by-step:

Take the very first number in the list.
Compare that number to every other number in the list, one by one.
If the first number is less than or equal to the second number, count that as a monotonic pair.
Move to the second number in the original list.
Compare that number to every other number in the list, one by one.
Again, if the second number is less than or equal to any other number, count that as a monotonic pair.
Continue this process, taking each number in the list and comparing it to every other number.
Add up all the times you found a pair that fits the condition (the first number is less than or equal to the second number).
The final total is the count of all monotonic pairs.

Code Implementation

def find_monotonic_pairs_brute_force(numbers):
    monotonic_pair_count = 0

    # Iterate through each number in the input list.
    for first_number_index in range(len(numbers)):

        for second_number_index in range(len(numbers)):
            # Check if the pair is monotonic.
            if numbers[first_number_index] <= numbers[second_number_index]:
                # Increment the monotonic pair count.
                monotonic_pair_count += 1

    return monotonic_pair_count

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Start at the beginning of the sequence of numbers.
For the current number, count how many numbers following it are greater than or equal to it. This gives you the increasing pairs this number can make.
Also, count how many numbers following it are less than or equal to it. This gives you the decreasing pairs this number can make.
Add both of those counts together. This is the total number of monotonic pairs that begin with the current number.
Move on to the next number in the sequence and repeat the process.
Continue until you have checked every number in the sequence.
Finally, add up all the counts you found for each number. The final sum is the total number of monotonic pairs in the entire sequence.

Code Implementation

def find_count_of_monotonic_pairs_i(sequence):
    total_monotonic_pairs = 0
    sequence_length = len(sequence)

    for i in range(sequence_length):
        increasing_pairs_count = 0
        decreasing_pairs_count = 0

        # Iterate through the rest of the sequence
        for j in range(i + 1, sequence_length):
            # Count increasing pairs
            if sequence[j] >= sequence[i]:
                increasing_pairs_count += 1

            # Count decreasing pairs
            if sequence[j] <= sequence[i]:
                decreasing_pairs_count += 1

        # Add the counts to the total
        total_monotonic_pairs += increasing_pairs_count + decreasing_pairs_count

    return total_monotonic_pairs

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 0, as no monotonic pairs are possible.
Input array with only one element	Return 0, as a pair requires at least two elements.
Array with all identical elements	Count the number of pairs using n*(n-1)/2 as all pairs are monotonic in this case
Array sorted in strictly increasing order	Count the number of pairs as all are monotonic (increasing).
Array sorted in strictly decreasing order	Count the number of pairs as all are monotonic (decreasing).
Array with positive and negative numbers	The algorithm should handle negative and positive numbers correctly as it is comparison-based.
Large input array exceeding memory limits	Consider using a more memory-efficient data structure or algorithm like an online algorithm if memory is a constraint.
Integer overflow when calculating the count of monotonic pairs for large arrays	Use a larger data type like long to store the count to avoid overflow issues.

Empty input array

How to Handle:

Return 0, as no monotonic pairs are possible.

Input array with only one element

How to Handle:

Return 0, as a pair requires at least two elements.

Array with all identical elements

How to Handle:

Count the number of pairs using n*(n-1)/2 as all pairs are monotonic in this case

Array sorted in strictly increasing order

How to Handle:

Count the number of pairs as all are monotonic (increasing).

Array sorted in strictly decreasing order

How to Handle:

Count the number of pairs as all are monotonic (decreasing).

Array with positive and negative numbers

How to Handle:

The algorithm should handle negative and positive numbers correctly as it is comparison-based.

Large input array exceeding memory limits

How to Handle:

Consider using a more memory-efficient data structure or algorithm like an online algorithm if memory is a constraint.

Integer overflow when calculating the count of monotonic pairs for large arrays

How to Handle:

Use a larger data type like long to store the count to avoid overflow issues.