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Long Pressed Name in C++
Suppose a man is typing some name on keyboard. Sometimes some buttons are long-pressed by mistake. So it may type one or more extra character. So we will take two strings, and check whether the second string is long-pressed name or not. So if the name is “Amit”, and second string is “Ammittt” is longpressed name. But “Ammttt” is not, because character i is not present here.
To solve this, we will follow these steps −
- let j := 0
- for i := 0, i < second.size, increase i by 1 −
- if j < actual_name.size and actual_name[j] = second[i], thenincrease j by 1
- return true when j = actual_name.size, otherwise false
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
bool isLongPressedName(string name, string typed) {
int j = 0;
for(int i = 0; i < typed.size(); i++){
if(j < name.size() && name[j] == typed[i])j++;
}
return j == name.size();
}
};
main(){
Solution ob;
string res = ob.isLongPressedName("Amit", "Ammittt") ? "true" :
"false";
cout << res;
}Input
"Amit" "Ammittt"
Output
true
Updated on: 28-Apr-2020
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