## Microsoft Interview Question

Software Engineer in Testsusing only one variable i don't think anyone can do it. For iteration to the length of string we need atleast one variable. so here is my thinking.

```
a = input string
s = 0;
e = n-1;
i = 0;
while(i<n){
if 'A' <= a[i] <= 'Z' ans i > s then
swap(a[s], a[i]) //swap using third variable
s++;
else if 'a'<=a[i]<='z' and i < e then
swap(a[i], a[e]);
e--;
else i++;
}
```

int l = a.length();

int s = 0;

int e = l-1;

int i =0;

while(i<l){

if('A' <= a[i] && a[i] <= 'Z' && i>s){

swap(a[i], a[s]);

s++;

}else if('a' <= a[i] && a[i] <= 'z' and i<e){

swap(a[i], a[e]);

e--;

}else i++;

}

```
/* A generic code implementing Dutch Flag Problem solution. Let me know if there are any bugs. Thanks*/
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
/*!
* \fn void DFPBuckets(vector<T> &a, T bktBound1, T bktBound2)
* \brief Rearrange the elements of an array into buckets
* \param a A vector containing the elements that need to be
* \param bktBound1 the value that separates lower buckets from mid bucket
* \param bktBound2 the value that separates mid buckets from higher bucket
*/
template<typename T>
void DFPBuckets(vector<T> &a, T bktBound1, T bktBound2){
int sz = a.size();
int low=0,mid=0,high=sz-1;
while(mid<=high){
if(bktBound1 >= a[mid]){ //case 1st bucket
swap(a[low],a[mid]);//cout<<"swap: "<< a[low] << " , " << a[mid]<<endl;
++low; ++mid;
}else if(bktBound2<=a[mid]){ //case 3rd bucket
while(high>mid && a[high]>bktBound2)
--high;
swap(a[mid],a[high]);//cout<<"swap: "<<a[mid]<<" , "<<a[high]<<endl;
--high;
}else//case 2nd bucket
++mid;
}
}
/*!
* \fn printVector(vector<T> &a);
* \brief print contents of a vector
*/
template<typename T>
void printVector(vector<T> &a){
int sz = a.size();
for(int i=0;i<sz;++i)cout<<", "<<a[i];
}
int main(){
vector<char> arrayC {'A','b','3','C','1','d','c','4','2','B','a'};
vector<double> arrayF {11.2, 0.1, 15.4, 0.0, 3.2, 2.8, 6.2};
cout<<"Input: ";printVector(arrayC);cout<<"\t\t\t\t";
DFPBuckets(arrayC, '9','Z');
cout<<"Output: ";printVector(arrayC);cout<<endl;
cout<<"Input: ";printVector(arrayF);cout<<"\t\t\t";
DFPBuckets(arrayF, 5.0,10.0);
cout<<"Output: ";printVector(arrayF);cout<<endl;
}
```

void swap(char *a,int i,int j )

{

char temp= a[i];

a[i]=a[j];

a[j]=temp;

}

void move()

{

int s = 0,i = 0;

int e = size - 1;

while(i < e)

{

if(str[i]>= 'a' && str[i]<= 'z' && i>=s )

{swap(str,i++,s++); //i should be incremented for o(n) complexity

}

else if(str[i]>='A' && str[i] <='Z' && i < e)

{swap(str,i,e--);

}

else

{i++;

}

}

}

//small correction req.

while(i <=e) //required as not working in case of {0, 'A', 'c', 'a'

Just another version of Dutch flag problem. Look it up on google.

- Rayden November 18, 2010