# Share my simple and easy O(N) solution

• ``````public class Solution {
public String originalDigits(String s) {
if(s==null || s.length()==0) return "";
int[] count = new int[128];
for(int i=0;i<s.length();i++)  count[s.charAt(i)]++;
int[] num = new int[10];
num[0] = count['z'];
num[2] = count['w'];
num[4] = count['u'];
num[6] = count['x'];
num[8] = count['g'];
num[1] = count['o']-count['z']-count['w']-count['u'];
num[3] = count['h']-count['g'];
num[5] = count['v']-count['s']+count['x'];
num[7] = count['s']-count['x'];
num[9] = count['i']-count['x']-count['g']-count['v']+count['s']-count['x'];
String ret = new String();
for(int i=0;i<10;i++)
for(int j=num[i];j>0;j--) ret += String.valueOf(i);
return ret;
}
}
``````

• I think following is better

``````// even numbers all have unique letter
countNum[0] = countChar['z'];
countNum[2] = countChar['w'];
countNum[4] = countChar['u'];
countNum[6] = countChar['x'];
countNum[8] = countChar['g'] ;

countNum[7] = countChar['s'] - countNum[6];
countNum[5] = countChar['v'] - countNum[7];
countNum[3] = countChar['h'] - countNum[8];
countNum[1] = countChar['o'] - countNum[2] - countNum[4] - countNum[0];
countNum[9] = countChar['i'] - countNum[6] - countNum[5] - countNum[8] ;
``````

• @krisdu
Agree you are better, I will probably ask the eng to rewrote the origin solution since it is not easily readable.

• Thanks for the solution! Here is my C++ code.

``````class Solution {
public:
string originalDigits(string s) {
vector<int> mp(26, 0), nums(10, 0);
for (char c:s) mp[c-'a']++;
nums[0] = mp['z'-'a'];
nums[2] = mp['w'-'a'];
nums[4] = mp['u'-'a'];
nums[1] = mp['o'-'a']-nums[0]-nums[2]-nums[4];
nums[3] = mp['r'-'a']-nums[0]-nums[4];
nums[6] = mp['x'-'a'];
nums[7] = mp['s'-'a']-nums[6];
nums[8] = mp['g'-'a'];
nums[5] = mp['v'-'a']-nums[7];
nums[9] = mp['i'-'a']-nums[5]-nums[6]-nums[8];
string ans;
for (int i = 0; i < 10; i++) {
ans += string(nums[i],'0'+i);
}
return ans;
}
};
``````

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