What is the time complexity of Find All Possible Stable Binary Arrays II?

The optimal solution has a time complexity of O(zero * one).

What is the space complexity of Find All Possible Stable Binary Arrays II?

The space complexity is O(zero * one).

What data structures are used to solve Find All Possible Stable Binary Arrays II?

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

What are the key insights for solving Find All Possible Stable Binary Arrays II?

Insight 1: Dynamic programming is suitable for counting valid arrangements by building up from smaller arrangements. Insight 2: The constraint regarding the limit on the consecutive zeros or ones can be efficiently handled using prefix sums to determine the count of valid arrays.

Which companies ask LeetCode 3130. Find All Possible Stable Binary Arrays II in interviews?

This problem is frequently asked at IBM.

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Find All Possible Stable Binary Arrays II - LeetCode 3130 Solution

Find All Possible Stable Binary Arrays II - Complete Solution Guide

Find All Possible Stable Binary Arrays II is LeetCode problem 3130, 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 3 positive integers zero , one , and limit . A binary array arr is called stable if: The number of occurrences of 0 in arr is exactly zero . The number of occurrences of 1 in arr is exactly one . Each subarray of arr with a size greater than limit must contain both 0 and 1. Return the total number of stable binary arrays. Since the answer may be very large, return it modulo 10 9 + 7 . Example 1: Input: zero = 1, one = 1, limit = 2 Output: 2 Explanation: The two possible stable bina

Detailed Explanation

The problem asks us to find the number of 'stable' binary arrays given constraints on the number of zeros (`zero`), the number of ones (`one`), and a limit (`limit`). A binary array is stable if it contains exactly `zero` zeros and `one` ones, and every subarray of length greater than `limit` must contain both a 0 and a 1. The goal is to return the count of such stable binary arrays modulo 10^9 + 7. Essentially, we need to arrange the zeros and ones in a way that avoids creating long stretches (longer than `limit`) of only zeros or only ones.

Solution Approach

The provided solutions use a dynamic programming approach. The core idea is to build up stable arrays by considering whether the last digit added was a zero or a one. `dp[i][j][0]` represents the number of stable arrays with `i` zeros and `j` ones ending with a zero, and `dp[i][j][1]` represents the number of stable arrays with `i` zeros and `j` ones ending with a one. Prefix sums `pref0` and `pref1` are used to efficiently calculate the sums of previous `dp` values to satisfy the limit constraint. The solution leverages the fact that to append a '0' block, we need to consider all valid configurations with fewer zeros ending with a '1', and similarly, to append a '1' block, we look at configurations with fewer ones ending with a '0'. By iterating through the possibilities and using prefix sums to handle the limit constraint, the number of valid arrangements is computed.

Step-by-Step Algorithm

Step 1: Initialize the DP table `dp[zero+1][one+1][2]` and prefix sum tables `pref0[zero+1][one+1]` and `pref1[zero+1][one+1]` to 0. `dp[0][0][0] = dp[0][0][1] = 1` and `pref0[0][0] = pref1[0][0] = 1` representing the base case (empty array is considered valid for both ending in 0 and ending in 1).
Step 2: Iterate through the DP table from `i = 0` to `zero` and `j = 0` to `one`.
Step 3: If `i > 0`, calculate `dp[i][j][0]` (number of arrays with `i` zeros, `j` ones ending in 0). This is computed by summing up the valid configurations with fewer zeros that end in 1. The number of zeros subtracted ranges from 1 to `limit`. Use prefix sums to efficiently calculate `sum_{k=1..limit} dp[i-k][j][1]`. More specifically, it's computed as `pref1[i-1][j] - pref1[i-limit-1][j]`.
Step 4: If `j > 0`, calculate `dp[i][j][1]` (number of arrays with `i` zeros, `j` ones ending in 1) similarly. This is computed using prefix sums on `pref0` as `pref0[i][j-1] - pref0[i][j-limit-1]`.
Step 5: Update the prefix sum tables `pref0` and `pref1` after calculating `dp[i][j][0]` and `dp[i][j][1]`. `pref0[i][j] = pref0[i][j-1] + dp[i][j][0]` and `pref1[i][j] = pref1[i-1][j] + dp[i][j][1]`.
Step 6: Return `(dp[zero][one][0] + dp[zero][one][1]) % MOD` which represents the total number of stable binary arrays with `zero` zeros and `one` ones, regardless of whether they end in zero or one.

Key Insights

Insight 1: Dynamic programming is suitable for counting valid arrangements by building up from smaller arrangements.
Insight 2: The constraint regarding the limit on the consecutive zeros or ones can be efficiently handled using prefix sums to determine the count of valid arrays.
Insight 3: The state of the DP should include the number of zeros and ones already placed, as well as whether the last placed digit was a zero or a one.

Complexity Analysis

Time Complexity: O(zero * one)

Space Complexity: O(zero * one)

Topics

This problem involves: Dynamic Programming, Prefix Sum.

Companies

Asked at: IBM.

Find All Possible Stable Binary Arrays II | LeetCode 3130

Hard

Dynamic Programming Prefix Sum

Time: O(zero * one)Space: O(zero * one)

Solution Code

Available in:

class Solution {
    public int numberOfStableArrays(int zero, int one, int limit) {
        int MOD = (int)1e9 + 7;
        
        // dp[i][j][0]: # of arrays with i zeros, j ones, last block is of 0s
        // dp[i][j][1]: # of arrays with i zeros, j ones, last block is of 1s
        int[][][] dp = new int[zero + 1][one + 1][2];
        
        // pref0[i][j] = sum of dp[i][k][0] for k=0..j
        // This is prefix sum along the row i of dp[][][0]
        int[][] pref0 = new int[zero + 1][one + 1];
        
        // pref1[i][j] = sum of dp[k][j][1] for k=0..i
        // This is prefix sum along the column j of dp[][][1]
        int[][] pref1 = new int[zero + 1][one + 1];

        // Sentinel for an empty prefix. An empty array can be thought of as
        // ending in either 0 or 1, allowing transitions to start a new sequence.
        dp[0][0][0] = 1;
        dp[0][0][1] = 1;
        pref0[0][0] = 1;
        pref1[0][0] = 1;

        for (int i = 0; i <= zero; i++) {
            for (int j = 0; j <= one; j++) {
                if (i == 0 && j == 0) {
                    continue;
                }
                
                // Calculate dp[i][j][0]
                // This is formed by adding a block of 1 to limit zeros to an array ending in 1.
                // Number of ways = sum_{k=1..limit} dp[i-k][j][1]
                if (i > 0) {
                    int upper_sum = pref1[i - 1][j];
                    int lower_sum_idx = i - limit - 1;
                    int lower_sum = (lower_sum_idx >= 0) ? pref1[lower_sum_idx][j] : 0;
                    dp[i][j][0] = (upper_sum - lower_sum + MOD) % MOD;
                }
                
                // Calculate dp[i][j][1]
                // This is formed by adding a block of 1 to limit ones to an array ending in 0.
                // Number of ways = sum_{k=1..limit} dp[i][j-k][0]
                if (j > 0) {
                    int upper_sum = pref0[i][j - 1];
                    int lower_sum_idx = j - limit - 1;
                    int lower_sum = (lower_sum_idx >= 0) ? pref0[i][lower_sum_idx] : 0;
                    dp[i][j][1] = (upper_sum - lower_sum + MOD) % MOD;
                }

                // Update prefix sums
                int current_pref0 = (j > 0) ? pref0[i][j-1] : 0;
                pref0[i][j] = (current_pref0 + dp[i][j][0]) % MOD;
                
                int current_pref1 = (i > 0) ? pref1[i-1][j] : 0;
                pref1[i][j] = (current_pref1 + dp[i][j][1]) % MOD;
            }
        }
        
        return (dp[zero][one][0] + dp[zero][one][1]) % MOD;
    }
}

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Problem Info

Description

Problem 3130: Find All Possible Stable Binary Arrays II - Detailed explanation and solution approaches.

Topics

Dynamic Programming Prefix Sum

Companies

IBM

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3130

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

6 steps

1
Step 1: Initialize the DP table `dp[zero+1][one+1][2]` and prefix sum tables `pref0[zero+1][one+1]` and `pref1[zero+1][one+1]` to 0. `dp[0][0][0] = dp[0][0][1] = 1` and `pref0[0][0] = pref1[0][0] = 1` representing the base case (empty array is considered valid for both ending in 0 and ending in 1).
2
Step 2: Iterate through the DP table from `i = 0` to `zero` and `j = 0` to `one`.
3
Step 3: If `i > 0`, calculate `dp[i][j][0]` (number of arrays with `i` zeros, `j` ones ending in 0). This is computed by summing up the valid configurations with fewer zeros that end in 1. The number of zeros subtracted ranges from 1 to `limit`. Use prefix sums to efficiently calculate `sum_{k=1..limit} dp[i-k][j][1]`. More specifically, it's computed as `pref1[i-1][j] - pref1[i-limit-1][j]`.
4
Step 4: If `j > 0`, calculate `dp[i][j][1]` (number of arrays with `i` zeros, `j` ones ending in 1) similarly. This is computed using prefix sums on `pref0` as `pref0[i][j-1] - pref0[i][j-limit-1]`.
5
Step 5: Update the prefix sum tables `pref0` and `pref1` after calculating `dp[i][j][0]` and `dp[i][j][1]`. `pref0[i][j] = pref0[i][j-1] + dp[i][j][0]` and `pref1[i][j] = pref1[i-1][j] + dp[i][j][1]`.
6
Step 6: Return `(dp[zero][one][0] + dp[zero][one][1]) % MOD` which represents the total number of stable binary arrays with `zero` zeros and `one` ones, regardless of whether they end in zero or one.

💡

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(zero * one)

💾

Space Complexity

O(zero * one)

📝

Why?

The algorithm iterates through a 2D DP table of size `(zero + 1) x (one + 1)`. Each cell calculation takes constant time due to the use of prefix sums. Therefore, the time complexity is O(zero * one). The DP table and prefix sum tables each take O(zero * one) space, resulting in a space complexity of O(zero * one).

📖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

Recursive with Memoization

A recursive approach can be used, but it will likely result in time limit exceeded without memoization. Memoization involves storing the results of function calls to avoid redundant calculations. This can turn the exponential solution of pure recursion into a DP solution.

Time: O(zero * one)Space: O(zero * one)

Trade-offs: While conceptually simpler to understand initially, the iterative DP approach is usually preferred for its explicit table structure and often better constant factors.

✨

Key Insights

1Insight 1: Dynamic programming is suitable for counting valid arrangements by building up from smaller arrangements.
2Insight 2: The constraint regarding the limit on the consecutive zeros or ones can be efficiently handled using prefix sums to determine the count of valid arrays.
3Insight 3: The state of the DP should include the number of zeros and ones already placed, as well as whether the last placed digit was a zero or a one.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `zero = 0` and `one = 0`. The algorithm correctly handles this case during initialization with `dp[0][0][0] = dp[0][0][1] = 1`, so that subsequent DP calculations will be initialized correctly for valid cases.
Edge case 2: `limit = 0`. In this case, any subarray of length greater than 0 must contain both 0 and 1, making stable arrays very restrictive. The code handles this since the sum from k=1 to limit becomes an empty sum (or the prefix subtraction handles it correctly).
Edge case 3: When i - limit - 1 < 0, we access pref1[i - limit - 1][j], which is handled by checking `if lower_sum_idx >= 0` before accessing the value, using 0 if the index is negative. This prevents out-of-bounds errors.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not taking modulo at every step. The problem requires calculating the result modulo 10^9 + 7 to avoid integer overflow. Failing to do this at each intermediate step can lead to incorrect results.

✗Common mistake 2: Incorrectly implementing the prefix sum calculation, potentially causing incorrect sums or index out of bound errors. It's important to ensure correct index bounds are checked before calculating sums with prefixes.

✗Common mistake 3: Not initializing the base case `dp[0][0][0]` and `dp[0][0][1]` to 1, leading to an incorrect DP computation. The empty string is a valid base case, and should be accounted for.

✅

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: Consider allocating resources (e.g., CPU cycles, memory) between two types of tasks. The limit constraint can represent the maximum time a single type of task can monopolize resources before switching to another task to ensure fairness and prevent starvation.

🏦Real-world scenario 2: Data compression: In run-length encoding (RLE) or similar techniques, the 'limit' can be used to restrict the maximum length of consecutive sequences of the same symbol to improve compression efficiency and prevent worst-case scenarios for certain data 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

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

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

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

👨‍💻 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: August 10, 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 13

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

@AlgoMasterJul 6

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

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 All Possible Stable Binary Arrays II"?

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

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