**Physical Address**

304 North Cardinal St.

Dorchester Center, MA 02124

You are given a positive integer `num`

. You may swap any two digits of `num`

that have the same **parity** (i.e. both odd digits or both even digits).

Return* the largest possible value of *

`num`

**Example 1:**

Input: num = 1234 Output: 3412 Explanation: Swap the digit 3 with the digit 1, this results in the number 3214. Swap the digit 2 with the digit 4, this results in the number 3412. Note that there may be other sequences of swaps but it can be shown that 3412 is the largest possible number. Also note that we may not swap the digit 4 with the digit 1 since they are of different parities.

**Example 2:**

Input: num = 65875 Output: 87655 Explanation: Swap the digit 8 with the digit 6, this results in the number 85675. Swap the first digit 5 with the digit 7, this results in the number 87655. Note that there may be other sequences of swaps but it can be shown that 87655 is the largest possible number.

**Constraints:**

`1 <= num <= 10`

^{9}

```
class Solution {
public:
int largestInteger(int num) {
priority_queue<int> p; // priority queue to store odd digits in descending order
priority_queue<int> q; // priority queue to store even digits in descending order
string nums=to_string(num); // converting num to a string for easy access of the digits
int n=nums.size(); // n stores the number of digits in num
for(int i=0;i<n;i++){
int digit=nums[i]-'0';
if((digit)%2) // if digit is odd, push it into priority queue p
p.push(digit);
else
q.push(digit); // if digit is even, push it into priority queue q
}
int answer=0;
for(int i=0; i<n; i++){
answer=answer*10;
if((nums[i]-'0')%2) // if the digit is odd, add the largest odd digit of p into the answer
{answer+=p.top();p.pop();}
else
{answer+=q.top();q.pop();} // if the digit is even, add the largest even digit of q into the answer
}
return answer;
}
};
```

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