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

The optimal solution has a time complexity of O(n*max_val).

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

The space complexity is O(max_val).

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

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 II?

Insight 1: The problem can be solved using dynamic programming. We can build up the count of valid arr1 prefixes, considering each element of `nums` one at a time. Insight 2: Since arr1 is non-decreasing and arr2 is non-increasing, the value of arr1[i] is bounded by nums[i]. Therefore, we only need to iterate up to nums[i] to find the number of valid arr1 values at index i.

Which companies ask LeetCode 3251. Find the Count of Monotonic Pairs II in interviews?

This problem is frequently asked at BNY Mellon.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Find the Count of Monotonic Pairs II - LeetCode 3251 Solution

Find the Count of Monotonic Pairs II - Complete Solution Guide

Find the Count of Monotonic Pairs II is LeetCode problem 3251, 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 count the number of 'monotonic pairs' of arrays (arr1, arr2) of length 'n', where 'n' is the length of the given input array 'nums'. A monotonic pair satisfies the following conditions: 1. Both 'arr1' and 'arr2' have the same length 'n' as 'nums'. 2. 'arr1' is monotonically non-decreasing (arr1[i] <= arr1[i+1] for all i). 3. 'arr2' is monotonically non-increasing (arr2[i] >= arr2[i+1] for all i). 4. For each index 'i', the sum of the elements at that index in 'arr1' and 'arr2' must equal the element at the same index in 'nums' (arr1[i] + arr2[i] == nums[i]). We need to return the total count of such valid monotonic pairs, modulo 10^9 + 7.

Solution Approach

The solution uses dynamic programming to calculate the number of monotonic pairs. `dp[v]` represents the number of valid `arr1` prefixes of length `i` where `arr1[i-1]` is equal to `v`. We iterate through the input `nums`. For each element `nums[i]`, we calculate the number of valid `arr1[i]` values. This is done by considering the constraint that `arr1` must be non-decreasing and calculating cumulative sums of the DP table to avoid recomputing counts. We use a prefix sum array to speed up the calculation. We take modulo at each step to prevent integer overflow.

Step-by-Step Algorithm

Step 1: Initialize `dp` array. `dp[v]` stores the number of valid prefixes of `arr1` of length `i`, ending with value `v`. For the first element of `nums`, we initialize `dp[v]` to 1 for all `v` from 0 to `nums[0]`.
Step 2: Iterate through the remaining elements of `nums` (from index 1 to n-1).
Step 3: Calculate `L_prev = max(0, nums[i] - nums[i-1])`. This value is the minimum value arr1[i] can take based on the restriction arr1[i-1] <= arr1[i].
Step 4: Create a `prefix_dp` array to store the cumulative sum of the `dp` array. `prefix_dp[j]` is the sum of `dp[0]` to `dp[j]`. This will help in efficiently computing the number of valid `arr1[i]` values based on the condition arr1[i] >= arr1[i-1]- L_prev.
Step 5: Calculate the next `dp` values, which will be stored in `dp_next` array. For each `v` from 0 to `nums[i]`, calculate `k = v - L_prev`. If `k >= 0`, then `dp_next[v] = prefix_dp[k]`. This step implements the non-decreasing property of arr1.
Step 6: Update the `dp` array with the calculated `dp_next` array. The current DP array `dp` is now updated with dp_next.
Step 7: After iterating through all elements of `nums`, sum all values in the `dp` array and take the modulo to get the final count of monotonic pairs.

Key Insights

Insight 1: The problem can be solved using dynamic programming. We can build up the count of valid arr1 prefixes, considering each element of `nums` one at a time.
Insight 2: Since arr1 is non-decreasing and arr2 is non-increasing, the value of arr1[i] is bounded by nums[i]. Therefore, we only need to iterate up to nums[i] to find the number of valid arr1 values at index i.
Insight 3: The crucial optimization is recognizing that for arr1[i] to be valid, it must be greater than or equal to arr1[i-1]- L_prev, where L_prev = nums[i-1] - nums[i].

Complexity Analysis

Time Complexity: O(n*max_val)

Space Complexity: O(max_val)

Topics

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

Companies

Asked at: BNY Mellon.

Find the Count of Monotonic Pairs II | LeetCode 3251

Hard

Array Math Dynamic Programming Combinatorics Prefix Sum

Time: O(n*max_val)Space: O(max_val)

Solution Code

Available in:

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

        if (n == 1) {
            return nums[0] + 1;
        }

        int max_val = 1000;
        int[] dp = new int[max_val + 1];

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

        for (int i = 1; i < n; i++) {
            int L_prev = Math.max(0, nums[i] - nums[i - 1]);

            int[] prefix_dp = new int[max_val + 1];
            prefix_dp[0] = dp[0];
            for (int j = 1; j <= max_val; j++) {
                prefix_dp[j] = (prefix_dp[j - 1] + dp[j]) % MOD;
            }

            int[] dp_next = new int[max_val + 1];
            for (int v = 0; v <= nums[i]; v++) {
                int k = v - L_prev;
                if (k >= 0) {
                    dp_next[v] = prefix_dp[k];
                }
            }

            dp = dp_next;
        }

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

        return total_count;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 3251: Find the Count of Monotonic Pairs II - Detailed explanation and solution approaches.

Topics

Array Math Dynamic Programming Combinatorics Prefix Sum

Companies

BNY Mellon

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3251

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

7 steps

1
Step 1: Initialize `dp` array. `dp[v]` stores the number of valid prefixes of `arr1` of length `i`, ending with value `v`. For the first element of `nums`, we initialize `dp[v]` to 1 for all `v` from 0 to `nums[0]`.
2
Step 2: Iterate through the remaining elements of `nums` (from index 1 to n-1).
3
Step 3: Calculate `L_prev = max(0, nums[i] - nums[i-1])`. This value is the minimum value arr1[i] can take based on the restriction arr1[i-1] <= arr1[i].
4
Step 4: Create a `prefix_dp` array to store the cumulative sum of the `dp` array. `prefix_dp[j]` is the sum of `dp[0]` to `dp[j]`. This will help in efficiently computing the number of valid `arr1[i]` values based on the condition arr1[i] >= arr1[i-1]- L_prev.
5
Step 5: Calculate the next `dp` values, which will be stored in `dp_next` array. For each `v` from 0 to `nums[i]`, calculate `k = v - L_prev`. If `k >= 0`, then `dp_next[v] = prefix_dp[k]`. This step implements the non-decreasing property of arr1.
6
Step 6: Update the `dp` array with the calculated `dp_next` array. The current DP array `dp` is now updated with dp_next.
7
Step 7: After iterating through all elements of `nums`, sum all values in the `dp` array and take the modulo to get the final count of monotonic pairs.

💡

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)

💾

Space Complexity

O(max_val)

📝

Why?

The outer loop iterates `n` times, and the inner loop iterates up to `max_val` times. The prefix sum calculation also takes O(max_val) time inside the outer loop. The DP array uses O(max_val) space to store the intermediate results.

📖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

One might try to derive a closed-form formula using combinatorics. However, due to the monotonically increasing and decreasing constraints, deriving a general formula is extremely difficult and not practical for implementation.

Time: Unknown (Likely exponential if trying to brute-force combinations)Space: Unknown (Likely exponential)

Trade-offs: This approach is not feasible given the constraints and complexity of the problem.

✨

Key Insights

1Insight 1: The problem can be solved using dynamic programming. We can build up the count of valid arr1 prefixes, considering each element of `nums` one at a time.
2Insight 2: Since arr1 is non-decreasing and arr2 is non-increasing, the value of arr1[i] is bounded by nums[i]. Therefore, we only need to iterate up to nums[i] to find the number of valid arr1 values at index i.
3Insight 3: The crucial optimization is recognizing that for arr1[i] to be valid, it must be greater than or equal to arr1[i-1]- L_prev, where L_prev = nums[i-1] - 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` has only one element. The code handles this by returning `nums[0] + 1` because arr1 can take values from 0 to nums[0] inclusively, resulting in nums[0] + 1 monotonic pairs.
Edge case 2: `nums[i]` is smaller than `nums[i-1]`. L_prev handles this condition. It can be a negative number and using the max function it will always be zero if it's a negative value. Thus, it won't affect the overall result.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow. The number of monotonic pairs can be very large, so it's crucial to take the modulo at each step to prevent integer overflow.

✗Common mistake 2: Incorrectly calculating `L_prev`. It's important to use `max(0, nums[i] - nums[i-1])` because L_prev has to be non-negative. The lower bound for array values cannot be negative.

✅

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 over time, where `nums` represents the total resources available at each time step. `arr1` could represent the cumulative resources allocated, which must be non-decreasing. `arr2` represents the remaining resources, which must be non-increasing. The problem is analogous to finding different valid allocation strategies.

🏦Real-world scenario 2: Inventory management. Consider `nums` as the total inventory requirement at each period. `arr1` might represent the amount of products manufactured and delivered, and `arr2` can denote the available product at the warehouse at each time period. Finding valid pairs represents managing inventory to fulfill demand following specific rules.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

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

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 12, 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

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

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

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

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.

🤖 Need help with this problem?