Decode Messages Sequel - Google Top Interview Questions
Problem Statement :
Given the mapping a = 1, b = 2, ... z = 26, as well as "*" which can be mapped anything from 1 to 9, and an encoded message message (as a string), count the number of ways it can be decoded. Mod the result by 10 ** 9 + 7. Constraints n ≤ 100,000 where n is the length of message Example 1 Input message = "1*" Output 18 Explanation The "*" can represent anything from 1 to 9, so this can be decoded as: ["aa", "ab", "ac", "ad", "ae", "af", "ag", "ah", "ai"] - (1, 1), (1, 2), ..., (1, 9)` ["k", "l", "m", "n", "o", "p", "q", "r", "s"] (11), (12), ..., (19) Example 2 Input message = "22" Output 2 Explanation This can represent "bb" or "v" Example 3 Input message = "*00" Output 0 Explanation There's no valid decoding
Solution :
Solution in C++ :
#define int unsigned long long
int mod = 1000000007;
int helper(string& s, vector<int>& dp, int i) {
if (i == s.length()) return 1;
if (dp[i] != -1) return dp[i] % mod;
if (s[i] == '0') return 0;
int ans = 0;
if (s[i] == '*') {
for (int j = 1; j <= 9; j++) {
ans += helper(s, dp, i + 1) % mod;
ans %= mod;
if (i + 1 < s.length() and s[i + 1] == '*') {
for (int k = 1; k <= 9 and j <= 2; k++) {
if (j * 10 + k <= 26) {
ans += helper(s, dp, i + 2) % mod;
ans %= mod;
}
}
} else if (i + 1 < s.length()) {
if (j * 10 + s[i + 1] - '0' <= 26) {
ans += helper(s, dp, i + 2) % mod;
ans %= mod;
}
}
}
} else {
ans += helper(s, dp, i + 1) % mod;
ans %= mod;
if (i + 1 < s.length() and s[i + 1] == '*') {
for (int k = 1; k <= 9; k++) {
if ((s[i] - '0') * 10 + k <= 26) {
ans += helper(s, dp, i + 2) % mod;
ans %= mod;
}
}
} else if (i + 1 < s.length()) {
if ((s[i] - '0') * 10 + s[i + 1] - '0' <= 26) {
ans += helper(s, dp, i + 2) % mod;
ans %= mod;
}
}
}
return dp[i] = ans;
}
int32_t solve(string mess) {
int n = mess.length();
vector<int> dp(mess.length() + 1, -1);
return helper(mess, dp, 0);
}
Solution in Java :
import java.util.*;
class Solution {
static final long MOD = 1000000007L;
public int solve(String s) {
int N = s.length();
long[] dp = new long[N + 1];
dp[0] = 1L;
for (int i = 1; i <= N; i++) {
dp[i] = dp[i - 1] * one(s.charAt(i - 1));
if (i > 1) {
dp[i] += dp[i - 2] * two(s.charAt(i - 2), s.charAt(i - 1));
}
dp[i] %= MOD;
}
return (int) (dp[N]);
}
public int one(char c) {
if (c == '*')
return 9;
else if (c == '0')
return 0;
else
return 1;
}
public int two(char a, char b) {
if (a == '0')
return 0;
int AL = 1;
int AR = 2;
if (a != '*') {
AL = a - '0';
AR = a - '0';
}
int BL = 1;
int BR = 9;
if (b != '*') {
BL = b - '0';
BR = b - '0';
}
int ans = 0;
for (int i = AL; i <= AR; i++) {
for (int j = BL; j <= BR; j++) {
if (10 * i + j <= 26)
ans++;
}
}
return ans;
}
}
Solution in Python :
class Solution:
def solve(self, message):
n = len(message)
count = 1 # current count
pcount = 1 # previous count
follow = None # char from previous iteration
for i in range(1, n + 1):
c = message[-i]
# count 2 character groups
count2 = 0
if i != 1:
if c == "1" or c == "*":
if follow == "*":
count2 += 9
else:
count2 += 1
if c == "2" or c == "*":
if follow == "*":
count2 += 6
elif ord("0") <= ord(follow) <= ord("6"):
count2 += 1
# count 1 character groups
if c == "*":
count1 = 9
elif c == "0":
count1 = 0
else:
count1 = 1
# sum and mod
ncount = count * count1 + pcount * count2
pcount = count
count = ncount % (10 ** 9 + 7)
follow = c
return count
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