# Observe the pattern, concise java code

• My solution is not the best on all posts, just try to share my thinking process

``````public class Solution {
public String originalDigits(String s) {
String[] dict = new String[]{"zero","one","two","three","four","five","six","seven","eight","nine"};
int[] index = new int[]{0,6,7,5,2,4,3,1,8,9};
int[] result = new int[10];
int[] map = new int[26];
for(char a: s.toCharArray()) map[a-'a']++;
StringBuilder res = new StringBuilder();
for(int i =0;i<10  ;i++){
int cur = index[i];
String temp = dict[cur];
int min= Integer.MAX_VALUE;
for(char a: temp.toCharArray()) min = Math.min(min,map[a-'a']);
for(char a: temp.toCharArray()) {map[a-'a'] = map[a-'a']-min;}
result[cur] = min;
}

for(int i =0 ;i<result.length;i++) {
while(result[i]-->0) res.append(String.valueOf(i));
}
return res.toString();
}
}
``````

Explain:
I mainly wanna explain index array part
0:First we start to count how many zero are in the String since character "z" is unique from one to ten
6:same apply to six where character "x" is unique
7:after exclude 6 ,we find that char "s" in seven is unique
5: after exclude 7, we find that char "v" is unique in five
2: char "w" in unique in two, you can put 2 in the front, does not matter
4: after exclude five, char "f" is unique in four
3: after exclude two, we find char "t" is unique in three
1: "o" is unique in one
8: "h" is unique in eight
then 9

rest of the code, I think is pretty straightforward

for index array, there are other combination, you may try yourself

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