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Find Permutation in C++
Suppose we have a secret signature consisting of character 'D' and 'I'. 'D' denotes the decreasing relationship between two numbers, 'I' denotes increasing relationship between two numbers. And the secret signature was constructed by a special integer array, which contains uniquely all the different number from 1 to n.
For example, the secret signature "DI" can be constructed from an array like [2,1,3] or [3,1,2], but not be constructed using array like [3,2,4] or [2,1,3,4], which are both illegal constructing special string that can't represent the "DI" secret signature.
Now we have to find the lexicographically smallest permutation of [1, 2, ... n] that could refer to the given secret signature in the input.
So, if the input is like "DI", then the output will be [2,1,3], As we know [3,1,2] can also construct the secret signature "DI", but since we want to find the one with the smallest lexicographical permutation, we need to output [2,1,3]
To solve this, we will follow these steps −
Define one stack st
cnt := 2
Define an array ret
for initialize i := 1, when i <= size of s, update (increase i by 1), do −
if s[i - 1] is same as 'D', then −
insert i into st
Otherwise
insert i at the end of ret
while (not st is empty), do −
insert top element of st at the end of ret
delete element from st
insert size of s into st
while (not st is empty), do −
insert top element of st at the end of ret
delete element from st
return ret
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:
vector<int< findPermutation(string s) {
stack <int< st;
int cnt = 2;
vector <int< ret;
for(int i = 1; i <= s.size(); i++){
if(s[i - 1] == 'D'){
st.push(i);
}
else{
ret.push_back(i);
while(!st.empty()){
ret.push_back(st.top());
st.pop();
}
}
}
st.push(s.size() + 1);
while(!st.empty()){
ret.push_back(st.top());
st.pop();
}
return ret;
}
};
main(){
Solution ob;
print_vector(ob.findPermutation("DIID"));
}Input
"DIID"
Output
[2, 1, 3, 5, 4, ]
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