README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/2200-2299/2266.Count%20Number%20of%20Texts/README_EN.md	1856	Weekly Contest 292 Q3	Hash Table Math String Dynamic Programming

2266. Count Number of Texts

中文文档

Description

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'.

Solutions

Solution 1

Python3

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 ch, s in groupby(pressedKeys):
            m = len(list(s))
            ans = ans * (g[m] if ch in "79" else f[m]) % mod
        return ans

Java

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] = 1;
        f[1] = 1;
        f[2] = 2;
        f[3] = 4;
        g[0] = 1;
        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) {
            int j = i;
            char c = pressedKeys.charAt(i);
            for (; 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;
    }
}

Go

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
}