/
ReconstructOriginalDigitsFromEnglish.java
60 lines (53 loc) · 1.45 KB
/
ReconstructOriginalDigitsFromEnglish.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
// https://leetcode.com/problems/reconstruct-original-digits-from-english/
class Solution {
String[] w = {
"zero",
"one",
"two",
"three",
"four",
"five",
"six",
"seven",
"eight",
"nine"
};
public String originalDigits(String s) {
int[] res = new int[10];
int[] freq = new int[26];
StringBuilder sb = new StringBuilder();
// Process even digits first
for (char c: s.toCharArray()) {
freq[c - 'a'] += 1;
if (c == 'z') {
res[0] += 1;
} else if (c == 'w') {
res[2] += 1;
} else if (c == 'u') {
res[4] += 1;
} else if (c == 'x') {
res[6] += 1;
} else if (c == 'g') {
res[8] += 1;
}
}
for (int i = 0; i < 10; i += 2) {
for (char c: w[i].toCharArray()) {
freq[c - 'a'] -= res[i];
}
}
for (int i = 0; i < 26; i++) {
res[1] = freq['o' - 'a'];
res[3] = freq['t' - 'a'];
res[5] = freq['f' - 'a'];
res[7] = freq['s' - 'a'];
res[9] = (freq['n' - 'a'] - res[7] - res[1])/2;
}
for (int i = 0; i < 10; i++) {
for (int j = 0; j < res[i]; j++) {
sb.append(i);
}
}
return sb.toString();
}
}