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Remove Duplicate Letters in C++
Suppose we have a string consisting of only lowercase letters. We have to remove all duplicate letters such that all letters only occur once. And we have to display the result in the smallest lexicographic sequence. So if the input is like “abccb”, then the result will be “abc”
To solve this, we will follow these steps −
ans := one empty string
Define one stack st
Define an array onStack of size 26
Define one map m
n := size of s
for initializing i := 0, when i < n, increase i by 1 do −
increase m[s[i]] by 1
for initializing i := 0, when i < n, increase i by 1 do −
Define an array x = s of size i
decreae m[x] by 1
if onStack[x - 'a'] is non-zero, then,
Skip to the next iteration, ignore the following part
while st is not empty and x < st.top(), do −
onStack[top of st - 'a'] := false
delete item from st
insert x into st
onStack[x - 'a'] := true
while (st is empty) is false, do −
x := top element of st
delete item from st
ans = ans + x
reverse the array rev
return ans
Example
Let us see the following implementation to get a better understanding −
#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << v[i] << ", ";
}
cout << "]"<<endl;
}
class Solution {
public:
string removeDuplicateLetters(string s) {
string ans = "";
stack <char> st;
vector <int> onStack(26);
map <char, int> m;
int n = s.size();
for(int i = 0; i < n; i++){
m[s[i]]++;
}
for(int i = 0; i < n; i++){
char x = s[i];
m[x]--;
if(onStack[x - 'a'])continue;
while(!st.empty() && x < st.top() && m[st.top()]){
onStack[st.top() - 'a'] = false;
st.pop();
}
st.push(x);
onStack[x - 'a'] = true;
}
while(!st.empty()){
char x = st.top();
st.pop();
ans += x;
}
reverse(ans.begin(), ans.end());
return ans;
}
};
main(){
Solution ob;
cout << (ob.removeDuplicateLetters("abccb"));
}Input
“abccb”
Output
“abc”
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