How do you solve Count Number of Texts on LeetCode?

Count Number of Texts (LeetCode #2266) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The number of ways to form messages increases with consecutive identical digits.

What is the brute force solution for Count Number of Texts?

The brute force approach for Count Number of Texts is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible combinations of letters for the given pressed keys. This is straightforward but inefficient, as it may require checking many combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Number of Texts?

Count Number of Texts uses the following concepts: Hash Table, Math, String, Dynamic Programming. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Number of Texts?

Always think about how to break down the problem using dynamic programming. Practice identifying patterns in similar problems. Explain your thought process clearly to the interviewer.

What are common mistakes when solving Count Number of Texts?

Not considering the maximum number of characters that can be represented by a single digit. Failing to handle the modulo operation correctly.

#2266

Count Number of Texts

Medium

Hash TableMathStringDynamic ProgrammingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses dynamic programming to efficiently calculate the number of possible text messages by leveraging previously computed results, avoiding redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP array where dp[i] represents the number of ways to decode the substring pressedKeys[0...i].
2Step 2: Iterate through the pressedKeys and for each digit, look back up to 4 positions (maximum length of the key) to calculate combinations based on the current digit.
3Step 3: Update the DP array based on the number of ways to form messages from previous indices.

solution.py18 lines

1# Full working Python code
2MOD = 10**9 + 7
3
4def countTexts(pressedKeys):
5    n = len(pressedKeys)
6    dp = [0] * (n + 1)
7    dp[0] = 1  # Base case
8
9    for i in range(1, n + 1):
10        count = 0
11        for j in range(1, 5):  # Check up to 4 previous keys
12            if i - j >= 0 and pressedKeys[i - 1] == pressedKeys[i - j]:
13                count += dp[i - j]
14                count %= MOD
15        dp[i] = count
16
17    return dp[n]
18

ℹ

Complexity note: The complexity is O(n) because we only iterate through the string once, and for each character, we check up to 4 previous characters, resulting in linear growth.

1The number of ways to form messages increases with consecutive identical digits.
2Dynamic programming helps in storing intermediate results to avoid redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.