What is the time complexity of Count Number of Texts?

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

What data structures are used to solve Count Number of Texts?

This problem uses Hash Table, Math, String, Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Count Number of Texts?

Insight 1: Dynamic Programming is suitable because the number of possible texts for a given prefix of `pressedKeys` can be calculated based on the results for shorter prefixes. Insight 2: Recognize repeating digit sequences. The problem boils down to counting the combinations within each sequence of identical digits.

Which companies ask LeetCode 2266. Count Number of Texts in interviews?

This problem is frequently asked at Goldman Sachs.

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Count Number of Texts - LeetCode 2266 Solution

Count Number of Texts - Complete Solution Guide

Count Number of Texts is LeetCode problem 2266, 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

Alice is texting Bob using her phone. The mapping of digits to letters is shown in the figure below. In order to add a letter, Alice has to press the key of the corresponding digit i times, where i is the position of the letter in the key. For example, to add the letter 's' , Alice has to press '7' four times. Similarly, to add the letter 'k' , Alice has to press '5' twice. Note that the digits '0' and '1' do not map to any letters, so Alice does not use them. However, due to an error in transmi

Detailed Explanation

The problem asks us to count the number of possible text messages Alice could have sent, given a string of pressed keys Bob received. Each digit from '2' to '9' on the phone keypad corresponds to a set of letters. Pressing a digit multiple times cycles through these letters. For example, '2' can be 'a', 'b', or 'c'. The digits '7' and '9' have four possible letters each. We are given the sequence of pressed keys and must determine how many different combinations of letters could have produced that sequence, returning the result modulo 10^9 + 7.

Solution Approach

The provided code uses dynamic programming to solve this problem. The `dp[i]` value represents the number of possible text messages that can be formed from the first `i` characters of `pressedKeys`. The solution iterates through the `pressedKeys` string, and for each position `i`, it considers the possibilities of pressing the same key 1, 2, 3, or 4 times (only for '7' and '9'). If the previous keys are the same, it adds the corresponding `dp` values to the current `dp[i]` value, ensuring the result is taken modulo 10^9 + 7.

Step-by-Step Algorithm

Step 1: Initialize a dynamic programming array `dp` of size `n + 1`, where `n` is the length of `pressedKeys`. `dp[0]` is initialized to 1, representing an empty string.
Step 2: Iterate from `i = 1` to `n`. `dp[i]` is initially set to `dp[i-1]`, representing the case where the current key is pressed only once.
Step 3: If the current key is the same as the previous key (i.e., `pressedKeys[i-1] == pressedKeys[i-2]`), add `dp[i-2]` to `dp[i]`. This represents the case where the current key is pressed twice.
Step 4: If the current key is the same as the previous two keys (i.e., `pressedKeys[i-1] == pressedKeys[i-2] == pressedKeys[i-3]`), add `dp[i-3]` to `dp[i]`. This represents the case where the current key is pressed three times.
Step 5: If the current key is '7' or '9' and is the same as the previous three keys (i.e., `pressedKeys[i-1] == pressedKeys[i-2] == pressedKeys[i-3] == pressedKeys[i-4]`), add `dp[i-4]` to `dp[i]`. This represents the case where the current key is pressed four times.
Step 6: After each addition, take the modulo `10^9 + 7` to prevent integer overflow.
Step 7: Finally, return `dp[n]`, which represents the total number of possible text messages.

Key Insights

Insight 1: Dynamic Programming is suitable because the number of possible texts for a given prefix of `pressedKeys` can be calculated based on the results for shorter prefixes.
Insight 2: Recognize repeating digit sequences. The problem boils down to counting the combinations within each sequence of identical digits.
Insight 3: Handle the '7' and '9' keys specially, as they map to four letters instead of three.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, Math, String, Dynamic Programming.

Companies

Asked at: Goldman Sachs.

Count Number of Texts | LeetCode 2266

Medium

Hash Table Math String Dynamic Programming

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int countTexts(String pressedKeys) {
        int MOD = (int)1e9 + 7;
        int n = pressedKeys.length();
        int[] dp = new int[n + 1];
        dp[0] = 1;

        for (int i = 1; i <= n; i++) {
            // 1-press option
            dp[i] = dp[i-1];

            // 2-press option
            if (i >= 2 && pressedKeys.charAt(i-1) == pressedKeys.charAt(i-2)) {
                dp[i] = (dp[i] + dp[i-2]) % MOD;
                
                // 3-press option
                if (i >= 3 && pressedKeys.charAt(i-2) == pressedKeys.charAt(i-3)) {
                    dp[i] = (dp[i] + dp[i-3]) % MOD;
                    
                    // 4-press option (only for '7' and '9')
                    if (pressedKeys.charAt(i-1) == '7' || pressedKeys.charAt(i-1) == '9') {
                        if (i >= 4 && pressedKeys.charAt(i-3) == pressedKeys.charAt(i-4)) {
                            dp[i] = (dp[i] + dp[i-4]) % MOD;
                        }
                    }
        
                }
            }
        }
        
        return dp[n];
    }
}

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Problem Info

Description

Problem 2266: Count Number of Texts - Detailed explanation and solution approaches.

Topics

Hash Table Math String Dynamic Programming

Companies

Goldman Sachs

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2266

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 a dynamic programming array `dp` of size `n + 1`, where `n` is the length of `pressedKeys`. `dp[0]` is initialized to 1, representing an empty string.
2
Step 2: Iterate from `i = 1` to `n`. `dp[i]` is initially set to `dp[i-1]`, representing the case where the current key is pressed only once.
3
Step 3: If the current key is the same as the previous key (i.e., `pressedKeys[i-1] == pressedKeys[i-2]`), add `dp[i-2]` to `dp[i]`. This represents the case where the current key is pressed twice.
4
Step 4: If the current key is the same as the previous two keys (i.e., `pressedKeys[i-1] == pressedKeys[i-2] == pressedKeys[i-3]`), add `dp[i-3]` to `dp[i]`. This represents the case where the current key is pressed three times.
5
Step 5: If the current key is '7' or '9' and is the same as the previous three keys (i.e., `pressedKeys[i-1] == pressedKeys[i-2] == pressedKeys[i-3] == pressedKeys[i-4]`), add `dp[i-4]` to `dp[i]`. This represents the case where the current key is pressed four times.
6
Step 6: After each addition, take the modulo `10^9 + 7` to prevent integer overflow.
7
Step 7: Finally, return `dp[n]`, which represents the total number of possible text messages.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because the outer loop iterates 'n' times and the inner conditional checks execute in constant time, regardless of 'n'. The space complexity is O(n) due to the dp array used to store 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

Recursion with Memoization

A recursive approach with memoization can also be used to solve this problem. The recursive function would calculate the number of possible texts for a given substring of `pressedKeys` and store the results in a memoization table to avoid redundant calculations.

Time: O(n)Space: O(n)

Trade-offs: Recursion with memoization might be slightly less efficient than the iterative dynamic programming approach due to function call overhead. However, it can be more intuitive for some people to understand.

Optimized DP with O(1) space

Since the dp array only uses the previous 4 values, we can optimize the space complexity to O(1) by storing the last 4 values in variables instead of an array.

Time: O(n)Space: O(1)

Trade-offs: Slightly more complex to write and understand but uses significantly less space.

✨

Key Insights

1Insight 1: Dynamic Programming is suitable because the number of possible texts for a given prefix of `pressedKeys` can be calculated based on the results for shorter prefixes.
2Insight 2: Recognize repeating digit sequences. The problem boils down to counting the combinations within each sequence of identical digits.
3Insight 3: Handle the '7' and '9' keys specially, as they map to four letters instead of three.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: `pressedKeys` is a single character. The code handles this correctly because the base case `dp[0] = 1` and `dp[1] = dp[0] = 1` ensures the correct result.
Edge case 2: `pressedKeys` consists of only '7's or '9's. The code correctly handles this by checking for the possibility of 4 presses when the digit is '7' or '9'.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not taking the modulo at each step. This can lead to integer overflow, resulting in an incorrect answer.

✗Common mistake 2: Incorrectly handling the '7' and '9' cases. It's important to ensure that the code only considers the possibility of 4 presses when the digit is '7' or '9' and that all the preceding digits are the same.

✗Common mistake 3: Not handling the edge case when n = 1 properly.

✅

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: Autocomplete and predictive text features in mobile devices use similar dynamic programming techniques to suggest possible words based on the keys pressed.

🏦Real-world scenario 2: Speech recognition systems can employ similar algorithms to decode sequences of phonemes into possible words or sentences.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

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: September 15, 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 13

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

@InterviewProJul 6

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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 Number of Texts"?

For this medium-level Hash Table 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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