What is the time complexity of Find the Count of Monotonic Pairs I?

The optimal solution has a time complexity of O(n*MAX_VAL^2).

What is the space complexity of Find the Count of Monotonic Pairs I?

The space complexity is O(MAX_VAL).

What data structures are used to solve Find the Count of Monotonic Pairs I?

This problem uses Array, Math, Dynamic Programming, Combinatorics, Prefix Sum. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Find the Count of Monotonic Pairs I?

Insight 1: The core idea is to use dynamic programming to count valid `arr1` prefixes. `arr2` is implicitly determined by `arr1` because `arr2[i] = nums[i] - arr1[i]`. Insight 2: The monotonicity constraints are critical. They enable the use of prefix sums for efficient computation of the number of valid `arr1` values at each step.

Which companies ask LeetCode 3250. Find the Count of Monotonic Pairs I in interviews?

This problem is frequently asked at Arcesium.

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Find the Count of Monotonic Pairs I - LeetCode 3250 Solution

Q: How to solve Find the Count of Monotonic Pairs I?

The provided solution uses dynamic programming. `dp[j]` represents the number of valid monotonic prefixes `arr1[0...i]` such that `arr1[i] = j`. The solution iterates through the `nums` array. For each `nums[i]`, it calculates the `dp` array based on the previous `dp` values and the monotonicity constraints. It uses prefix sums to efficiently calculate the number of valid values for `arr1[i]`.

Find the Count of Monotonic Pairs I - Complete Solution Guide

Find the Count of Monotonic Pairs I is LeetCode problem 3250, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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, retur

Detailed Explanation

The problem asks us to find the number of 'monotonic pairs' of arrays (arr1, arr2) that satisfy specific conditions based on a given input array `nums`. Both arr1 and arr2 must have the same length as `nums`. `arr1` must be monotonically non-decreasing (each element is greater than or equal to the previous one), and `arr2` must be monotonically non-increasing (each element is less than or equal to the previous one). Crucially, for each index `i`, the sum of `arr1[i]` and `arr2[i]` must equal `nums[i]`. The answer should be returned modulo 10^9 + 7 to handle potentially large results. The constraints indicate `nums` can have up to 2000 elements, and each element in `nums` is a positive integer between 1 and 50.

Solution Approach

The provided solution uses dynamic programming. `dp[j]` represents the number of valid monotonic prefixes `arr1[0...i]` such that `arr1[i] = j`. The solution iterates through the `nums` array. For each `nums[i]`, it calculates the `dp` array based on the previous `dp` values and the monotonicity constraints. It uses prefix sums to efficiently calculate the number of valid values for `arr1[i]`.

Step-by-Step Algorithm

Step 1: Initialize `dp` array. For the first element of `nums`, `dp[j] = 1` for all `j` such that `0 <= j <= nums[0]`. This means arr1[0] can be any value from 0 to nums[0].
Step 2: Iterate through the remaining elements of `nums` (from `i = 1` to `n - 1`).
Step 3: For each element `nums[i]`, create a `new_dp` array to store the updated counts.
Step 4: Calculate a `prefix_sum` array based on the previous `dp` array. This prefix sum array will help calculate how many possible values `arr1[i-1]` could have had given what `arr1[i]` is.
Step 5: Iterate through all possible values of `arr1[i]` (from `j = 0` to `nums[i]`).
Step 6: For each `j`, calculate the maximum possible value `limit` for `arr1[i-1]` such that the monotonicity condition `arr1[i-1] <= arr1[i]` is satisfied, and `arr1[i-1] + arr2[i-1] == nums[i-1]` remains true, given that `arr1[i] = j`. The `upper_bound_k` variable represents the maximum possible value that `arr1[i-1]` can take to satisfy the constraints.
Step 7: Set `new_dp[j]` to `prefix_sum[limit]`. This means that for `arr1[i] = j`, there are `prefix_sum[limit]` ways to construct `arr1` from the beginning up to index `i` given the rules of the problem.
Step 8: Update `dp` to `new_dp` for the next iteration.
Step 9: After iterating through all elements of `nums`, sum the values in the final `dp` array to get the total number of valid monotonic pairs, taking the modulo at each step to prevent overflow.

Key Insights

Insight 1: The core idea is to use dynamic programming to count valid `arr1` prefixes. `arr2` is implicitly determined by `arr1` because `arr2[i] = nums[i] - arr1[i]`.
Insight 2: The monotonicity constraints are critical. They enable the use of prefix sums for efficient computation of the number of valid `arr1` values at each step.
Insight 3: The solution needs to carefully handle the constraints, especially the monotonically increasing constraint on arr1 and the condition `arr1[i] + arr2[i] == nums[i]`

Complexity Analysis

Time Complexity: O(n*MAX_VAL^2)

Space Complexity: O(MAX_VAL)

Topics

This problem involves: Array, Math, Dynamic Programming, Combinatorics, Prefix Sum.

Companies

Asked at: Arcesium.

Find the Count of Monotonic Pairs I | LeetCode 3250

Hard

Array Math Dynamic Programming Combinatorics Prefix Sum

Time: O(n*MAX_VAL^2)Space: O(MAX_VAL)

Solution Code

Available in:

class Solution {
    public int countOfPairs(int[] nums) {
        int n = nums.length;
        int MOD = (int)1e9 + 7;

        int MAX_VAL = 51;

        int[] dp = new int[MAX_VAL];

        for (int j = 0; j <= nums[0]; j++) {
            dp[j] = 1;
        }

        for (int i = 1; i < n; i++) {
            int[] prefix_sum = new int[MAX_VAL];
            int current_sum = 0;
            for (int j = 0; j < MAX_VAL; j++) {
                current_sum = (current_sum + dp[j]) % MOD;
                prefix_sum[j] = current_sum;
            }

            int[] new_dp = new int[MAX_VAL];

            int num_i = nums[i];
            int num_prev = nums[i - 1];
            int diff = num_i - num_prev;

            for (int j = 0; j <= num_i; j++) {
                int upper_bound_k = j - Math.max(0, diff);
                int limit = Math.min(num_prev, upper_bound_k);

                if (limit >= 0) {
                    new_dp[j] = prefix_sum[limit];
                }
            }

            dp = new_dp;
        }

        int total_count = 0;
        for (int count : dp) {
            total_count = (total_count + count) % MOD;
        }

        return total_count;
    }
}

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Problem Info

Description

Problem 3250: Find the Count of Monotonic Pairs I - Detailed explanation and solution approaches.

Topics

Array Math Dynamic Programming Combinatorics Prefix Sum

Companies

Arcesium

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3250

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Initialize `dp` array. For the first element of `nums`, `dp[j] = 1` for all `j` such that `0 <= j <= nums[0]`. This means arr1[0] can be any value from 0 to nums[0].
2
Step 2: Iterate through the remaining elements of `nums` (from `i = 1` to `n - 1`).
3
Step 3: For each element `nums[i]`, create a `new_dp` array to store the updated counts.
4
Step 4: Calculate a `prefix_sum` array based on the previous `dp` array. This prefix sum array will help calculate how many possible values `arr1[i-1]` could have had given what `arr1[i]` is.
5
Step 5: Iterate through all possible values of `arr1[i]` (from `j = 0` to `nums[i]`).
6
Step 6: For each `j`, calculate the maximum possible value `limit` for `arr1[i-1]` such that the monotonicity condition `arr1[i-1] <= arr1[i]` is satisfied, and `arr1[i-1] + arr2[i-1] == nums[i-1]` remains true, given that `arr1[i] = j`. The `upper_bound_k` variable represents the maximum possible value that `arr1[i-1]` can take to satisfy the constraints.
7
Step 7: Set `new_dp[j]` to `prefix_sum[limit]`. This means that for `arr1[i] = j`, there are `prefix_sum[limit]` ways to construct `arr1` from the beginning up to index `i` given the rules of the problem.
8
Step 8: Update `dp` to `new_dp` for the next iteration.
9
Step 9: After iterating through all elements of `nums`, sum the values in the final `dp` array to get the total number of valid monotonic pairs, taking the modulo at each step to prevent overflow.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*MAX_VAL^2)

💾

Space Complexity

O(MAX_VAL)

📝

Why?

The time complexity is dominated by the nested loops calculating the `dp` values for each element in `nums`. The space complexity is determined by the size of the `dp` arrays, which are proportional to MAX_VAL.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Combinatorial Approach

A purely combinatorial solution involves finding a closed-form expression or using combinatorial identities to directly calculate the number of monotonic pairs. This is generally very difficult to derive.

Time: Potentially O(n) with precomputed factorials/inverses, but derivation is complex.Space: Potentially O(n) for storing precomputed values.

Trade-offs: If a simple closed-form expression can be found, this could be more efficient. However, it's unlikely to exist for this problem.

✨

Key Insights

1Insight 1: The core idea is to use dynamic programming to count valid `arr1` prefixes. `arr2` is implicitly determined by `arr1` because `arr2[i] = nums[i] - arr1[i]`.
2Insight 2: The monotonicity constraints are critical. They enable the use of prefix sums for efficient computation of the number of valid `arr1` values at each step.
3Insight 3: The solution needs to carefully handle the constraints, especially the monotonically increasing constraint on arr1 and the condition `arr1[i] + arr2[i] == nums[i]`

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `nums` array with a single element. The solution correctly initializes the `dp` array and returns the sum, which is `nums[0] + 1`.
Edge case 2: `nums` array with all elements being the same. The algorithm correctly calculates the number of monotonic pairs based on combinations.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow. Failing to take the modulo `% MOD` at each step can lead to integer overflow, resulting in an incorrect answer.

✗Common mistake 2: Incorrectly implementing the monotonicity condition. A subtle error in calculating the upper bound `limit` for `arr1[i-1]` can lead to undercounting or overcounting the valid monotonic pairs.

✗Common mistake 3: Not understanding prefix sums. Forgetting that to sum from index 0 to `limit` in `prefix_sum`, you can directly access `prefix_sum[limit]` avoids an inner loop.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation. Imagine allocating resources (e.g., bandwidth, memory) to different processes over time. The `nums` array could represent the total available resources at each time point. `arr1` could represent resources allocated to high-priority processes, which tend to increase monotonically, and `arr2` could represent resources allocated to low-priority processes, which tend to decrease monotonically. The algorithm helps count the possible allocation strategies that adhere to these constraints.

🏦Real-world scenario 2: Time series analysis. Consider analyzing stock prices or sensor data where you need to find patterns where one set of values increases over time while another set decreases, and their sum relates to a reference value. This algorithm helps count the number of possible ways to achieve those patterns.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 11, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Find the Count of Monotonic Pairs I"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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