LeetCode #2266 — MEDIUM

Count Number of Texts

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

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 transmission, Bob did not receive Alice's text message but received a string of pressed keys instead.

For example, when Alice sent the message "bob", Bob received the string "2266622".

Given a string pressedKeys representing the string received by Bob, return the total number of possible text messages Alice could have sent.

Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: pressedKeys = "22233"
Output: 8
Explanation:
The possible text messages Alice could have sent are:
"aaadd", "abdd", "badd", "cdd", "aaae", "abe", "bae", and "ce".
Since there are 8 possible messages, we return 8.

Example 2:

Input: pressedKeys = "222222222222222222222222222222222222"
Output: 82876089
Explanation:
There are 2082876103 possible text messages Alice could have sent.
Since we need to return the answer modulo 10⁹ + 7, we return 2082876103 % (10⁹ + 7) = 82876089.

Constraints:

1 <= pressedKeys.length <= 10⁵
pressedKeys only consists of digits from '2' - '9'.

Step 01

Brute Force Baseline

Problem summary: 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 transmission, Bob did not receive Alice's text message but received a string of pressed keys instead. For example, when Alice sent the message "bob", Bob received the string "2266622". Given a string pressedKeys representing the string received by Bob, return the total number of possible text messages Alice could have sent. Since the answer may be very large, return it modulo 109 + 7.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Math · Dynamic Programming

Example 1

"22233"

Example 2

"222222222222222222222222222222222222"

Core Insight

What unlocks the optimal approach

For a substring consisting of the same digit, how can we count the number of texts it could have originally represented?
How can dynamic programming help us calculate the required answer?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2266: Count Number of Texts
class Solution {
    private static final int N = 100010;
    private static final int MOD = (int) 1e9 + 7;
    private static long[] f = new long[N];
    private static long[] g = new long[N];
    static {
        f[0] = f[1] = 1;
        f[2] = 2;
        f[3] = 4;
        g[0] = g[1] = 1;
        g[2] = 2;
        g[3] = 4;
        for (int i = 4; i < N; ++i) {
            f[i] = (f[i - 1] + f[i - 2] + f[i - 3]) % MOD;
            g[i] = (g[i - 1] + g[i - 2] + g[i - 3] + g[i - 4]) % MOD;
        }
    }

    public int countTexts(String pressedKeys) {
        long ans = 1;
        for (int i = 0, n = pressedKeys.length(); i < n; ++i) {
            char c = pressedKeys.charAt(i);
            int j = i;
            while (j + 1 < n && pressedKeys.charAt(j + 1) == c) {
                ++j;
            }
            int cnt = j - i + 1;
            ans = c == '7' || c == '9' ? ans * g[cnt] : ans * f[cnt];
            ans %= MOD;
            i = j;
        }
        return (int) ans;
    }
}

// Accepted solution for LeetCode #2266: Count Number of Texts
const mod int = 1e9 + 7
const n int = 1e5 + 10

var f = [n]int{1, 1, 2, 4}
var g = f

func init() {
	for i := 4; i < n; i++ {
		f[i] = (f[i-1] + f[i-2] + f[i-3]) % mod
		g[i] = (g[i-1] + g[i-2] + g[i-3] + g[i-4]) % mod
	}
}

func countTexts(pressedKeys string) int {
	ans := 1
	for i, j, n := 0, 0, len(pressedKeys); i < n; i++ {
		c := pressedKeys[i]
		j = i
		for j+1 < n && pressedKeys[j+1] == c {
			j++
		}
		cnt := j - i + 1
		if c == '7' || c == '9' {
			ans = ans * g[cnt] % mod
		} else {
			ans = ans * f[cnt] % mod
		}
		i = j
	}
	return ans
}

# Accepted solution for LeetCode #2266: Count Number of Texts
mod = 10**9 + 7
f = [1, 1, 2, 4]
g = [1, 1, 2, 4]
for _ in range(100000):
    f.append((f[-1] + f[-2] + f[-3]) % mod)
    g.append((g[-1] + g[-2] + g[-3] + g[-4]) % mod)


class Solution:
    def countTexts(self, pressedKeys: str) -> int:
        ans = 1
        for c, s in groupby(pressedKeys):
            m = len(list(s))
            ans = ans * (g[m] if c in "79" else f[m]) % mod
        return ans

// Accepted solution for LeetCode #2266: Count Number of Texts
/**
 * [2266] Count Number of Texts
 *
 * Alice is texting Bob using her phone. The mapping of digits to letters is shown in the figure below.
 * <img alt="" src="https://assets.leetcode.com/uploads/2022/03/15/1200px-telephone-keypad2svg.png" style="width: 200px; height: 162px;" />
 * 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 transmission, Bob did not receive Alice's text message but received a string of pressed keys instead.
 *
 * 	For example, when Alice sent the message "bob", Bob received the string "2266622".
 *
 * Given a string pressedKeys representing the string received by Bob, return the total number of possible text messages Alice could have sent.
 * Since the answer may be very large, return it modulo 10^9 + 7.
 *  
 * Example 1:
 *
 * Input: pressedKeys = "22233"
 * Output: 8
 * Explanation:
 * The possible text messages Alice could have sent are:
 * "aaadd", "abdd", "badd", "cdd", "aaae", "abe", "bae", and "ce".
 * Since there are 8 possible messages, we return 8.
 *
 * Example 2:
 *
 * Input: pressedKeys = "222222222222222222222222222222222222"
 * Output: 82876089
 * Explanation:
 * There are 2082876103 possible text messages Alice could have sent.
 * Since we need to return the answer modulo 10^9 + 7, we return 2082876103 % (10^9 + 7) = 82876089.
 *
 *  
 * Constraints:
 *
 * 	1 <= pressedKeys.length <= 10^5
 * 	pressedKeys only consists of digits from '2' - '9'.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/count-number-of-texts/
// discuss: https://leetcode.com/problems/count-number-of-texts/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn count_texts(pressed_keys: String) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2266_example_1() {
        let pressed_keys = "22233".to_string();

        let result = 8;

        assert_eq!(Solution::count_texts(pressed_keys), result);
    }

    #[test]
    #[ignore]
    fn test_2266_example_2() {
        let pressed_keys = "222222222222222222222222222222222222".to_string();

        let result = 82876089;

        assert_eq!(Solution::count_texts(pressed_keys), result);
    }
}

// Accepted solution for LeetCode #2266: Count Number of Texts
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2266: Count Number of Texts
// class Solution {
//     private static final int N = 100010;
//     private static final int MOD = (int) 1e9 + 7;
//     private static long[] f = new long[N];
//     private static long[] g = new long[N];
//     static {
//         f[0] = f[1] = 1;
//         f[2] = 2;
//         f[3] = 4;
//         g[0] = g[1] = 1;
//         g[2] = 2;
//         g[3] = 4;
//         for (int i = 4; i < N; ++i) {
//             f[i] = (f[i - 1] + f[i - 2] + f[i - 3]) % MOD;
//             g[i] = (g[i - 1] + g[i - 2] + g[i - 3] + g[i - 4]) % MOD;
//         }
//     }
// 
//     public int countTexts(String pressedKeys) {
//         long ans = 1;
//         for (int i = 0, n = pressedKeys.length(); i < n; ++i) {
//             char c = pressedKeys.charAt(i);
//             int j = i;
//             while (j + 1 < n && pressedKeys.charAt(j + 1) == c) {
//                 ++j;
//             }
//             int cnt = j - i + 1;
//             ans = c == '7' || c == '9' ? ans * g[cnt] : ans * f[cnt];
//             ans %= MOD;
//             i = j;
//         }
//         return (int) ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.