String t is generated by random shuffling string s and then add one more letter at a random position.
Return the letter that was added to t.
Solution
When I first looked at the question, I immediately realized that using a dictionary to solve it would be a good idea. The keys correspond to characters, while the values represent their frequencies.
C# Solution
Solution1
public class Solution {
public char FindTheDifference(string s, string t) {
Dictionary<char,int> dict = new Dictionary<char,int>();
//1.count number of occurrences of each value
foreach(var tmp in s)
{
if (!dict.ContainsKey(tmp))
{
dict.Add(tmp, 1);
}
else
{
dict[tmp] = dict[tmp]+1;
}
}
//2. find the added char
foreach(var tmp in t){
if (!dict.ContainsKey(tmp)){
return tmp;
}
var num = dict[tmp]-1;
if(num<0){
return tmp;
}
dict[tmp] = num;
}
return '\0'; // Default value used to ensure every possible code path returns a value.
}
}
If we can’t find the char in the dictionary or its appearance count is not enough, we will return the char.
Java Solution
Solution1
class Solution {
public char findTheDifference(String s, String t) {
Map<Character, Integer> dict = new HashMap<Character, Integer>();
//1.count number of occurrences of each value
for (char tmp: s.toCharArray())
{
if (!dict.containsKey(tmp))
{
dict.put(tmp, 1);
}
else
{
dict.put(tmp, dict.get(tmp)+1);
}
}
//2. find the added char
for (char tmp: t.toCharArray()) {
if (!dict.containsKey(tmp)){
return tmp;
}
int num = dict.get(tmp)-1;
if(num<0){
return tmp;
}
dict.put(tmp,num);
}
return '\0'; // Default value used to ensure every possible code path returns a value.
}
}
uh… that’s too slow, maybe we need to find another way… (just faster than 20%)
Conclusion
We will be back, with faster code…
🧡If my solution helps, that is my honor!
🧡You can support me by sharing my posts, thanks a lot
✅If you got any problem about the explanation, please feel free to let me know
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