What is the time complexity of Count Ways To Build Good Strings?

The optimal solution has a time complexity of O(high).

What is the space complexity of Count Ways To Build Good Strings?

The space complexity is O(high).

What data structures are used to solve Count Ways To Build Good Strings?

This problem uses Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Count Ways To Build Good Strings?

Insight 1: Dynamic Programming is suitable because the number of good strings of length `i` can be built using the number of good strings of length `i - zero` and `i - one`. Insight 2: The problem can be solved iteratively by building up the number of good strings for each length from 0 to `high`.

Which companies ask LeetCode 2466. Count Ways To Build Good Strings in interviews?

This problem is frequently asked at Citadel.

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Count Ways To Build Good Strings - LeetCode 2466 Solution

Count Ways To Build Good Strings - Complete Solution Guide

Count Ways To Build Good Strings is LeetCode problem 2466, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given the integers zero , one , low , and high , we can construct a string by starting with an empty string, and then at each step perform either of the following: Append the character '0' zero times. Append the character '1' one times. This can be performed any number of times. A good string is a string constructed by the above process having a length between low and high ( inclusive ). Return the number of different good strings that can be constructed satisfying these properties. Since the an

Detailed Explanation

The problem asks us to count the number of "good" strings that can be built using two operations: appending '0' a certain number of times (given by `zero`) or appending '1' a certain number of times (given by `one`). A "good" string is defined as having a length between `low` and `high` (inclusive). The result needs to be returned modulo 10^9 + 7 to avoid integer overflow.

Solution Approach

The solution uses dynamic programming to solve this problem. An array `dp` of size `high + 1` is created to store the number of good strings of each length from 0 to `high`. `dp[i]` represents the number of good strings of length `i`. The base case is `dp[0] = 1` (an empty string). The algorithm then iterates from 1 to `high`, calculating `dp[i]` by adding the number of good strings of length `i - zero` (if `i >= zero`) and the number of good strings of length `i - one` (if `i >= one`). Finally, the algorithm sums the values in the `dp` array from index `low` to `high` (inclusive) to get the total number of good strings of lengths within the desired range. The modulo operation is applied at each step to prevent overflow.

Step-by-Step Algorithm

Step 1: Initialize a DP array `dp` of size `high + 1` with all elements set to 0.
Step 2: Set the base case: `dp[0] = 1` (representing an empty string).
Step 3: Iterate from `i = 1` to `high`:
Step 4: If `i >= zero`, update `dp[i] = (dp[i] + dp[i - zero]) % MOD`.
Step 5: If `i >= one`, update `dp[i] = (dp[i] + dp[i - one]) % MOD`.
Step 6: Initialize a variable `sum` to 0.
Step 7: Iterate from `i = low` to `high`:
Step 8: Update `sum = (sum + dp[i]) % MOD`.
Step 9: Return `sum`.

Key Insights

Insight 1: Dynamic Programming is suitable because the number of good strings of length `i` can be built using the number of good strings of length `i - zero` and `i - one`.
Insight 2: The problem can be solved iteratively by building up the number of good strings for each length from 0 to `high`.
Insight 3: Modulo operation is crucial at each step to prevent integer overflow, ensuring the result remains within the specified range.

Complexity Analysis

Time Complexity: O(high)

Space Complexity: O(high)

Topics

This problem involves: Dynamic Programming.

Companies

Asked at: Citadel.

Count Ways To Build Good Strings | LeetCode 2466

Medium

Dynamic Programming

Time: O(high)Space: O(high)

Solution Code

Available in:

class Solution {
    public int countGoodStrings(int low, int high, int zero, int one) {
        int MOD = 1_000_000_007;
        
        int[] dp = new int[high + 1];
        dp[0] = 1;
        
        for (int i = 1; i <= high; i++) {
            if (i >= zero) {
                dp[i] = (dp[i] + dp[i - zero]) % MOD;
            }
            if (i >= one) {
                dp[i] = (dp[i] + dp[i - one]) % MOD;
            }
        }
        
        int sum = 0;
        for (int i = low; i <= high; i++) {
            sum = (sum + dp[i]) % MOD;
        }
        
        return sum;
    }
}

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Problem Info

Description

Problem 2466: Count Ways To Build Good Strings - Detailed explanation and solution approaches.

Topics

Dynamic Programming

Companies

Citadel

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2466

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 a DP array `dp` of size `high + 1` with all elements set to 0.
2
Step 2: Set the base case: `dp[0] = 1` (representing an empty string).
3
Step 3: Iterate from `i = 1` to `high`:
4
Step 4: If `i >= zero`, update `dp[i] = (dp[i] + dp[i - zero]) % MOD`.
5
Step 5: If `i >= one`, update `dp[i] = (dp[i] + dp[i - one]) % MOD`.
6
Step 6: Initialize a variable `sum` to 0.
7
Step 7: Iterate from `i = low` to `high`:
8
Step 8: Update `sum = (sum + dp[i]) % MOD`.
9
Step 9: Return `sum`.

💡

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(high)

💾

Space Complexity

O(high)

📝

Why?

The time complexity is derived from the for loops that iterate up to `high`. The space complexity comes from the `dp` array that stores the number of ways to construct strings of lengths up to `high`.

📖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

Recursion with Memoization

A recursive approach with memoization can also solve this problem. It involves defining a recursive function that calculates the number of good strings of a given length and storing the results in a memo table to avoid redundant calculations.

Time: O(high)Space: O(high)

Trade-offs: Recursion with memoization has similar time and space complexity as the iterative DP approach. However, it might be slightly less efficient due to function call overhead. The iterative DP is generally preferred in terms of performance and avoiding potential stack overflow issues in cases of very large values for `high`.

✨

Key Insights

1Insight 1: Dynamic Programming is suitable because the number of good strings of length `i` can be built using the number of good strings of length `i - zero` and `i - one`.
2Insight 2: The problem can be solved iteratively by building up the number of good strings for each length from 0 to `high`.
3Insight 3: Modulo operation is crucial at each step to prevent integer overflow, ensuring the result remains within the specified range.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `low == high == 1`, `zero == 1`, `one == 1`. The solution should return 2 ('0' and '1'). The algorithm correctly handles this.
Edge case 2: `zero == one`. The algorithm should still work correctly because it adds the contributions from both appending zero and appending one.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not applying the modulo operation at each step, leading to integer overflow and incorrect results.

✗Common mistake 2: Incorrectly initializing the base case of the DP array, which can lead to incorrect calculations.

✅

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: Bioinformatics - Counting the number of DNA sequences within a certain length range that can be formed by appending specific base pairs ('A', 'T', 'C', 'G') a given number of times. Similar to building strings with '0' and '1', we could build DNA sequences with different base pair lengths.

🏦Real-world scenario 2: Network Routing - Determining the number of valid network paths with a length between `low` and `high` hops, where each hop can be of a certain fixed cost or bandwidth. Each '0' could represent a hop of cost `zero`, and each '1' a hop of cost `one`. The algorithm counts valid paths based on length constraints.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

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

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: October 5, 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

❓

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."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Count Ways To Build Good Strings"?

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

🤖 Need help with this problem?