Return the largest possible value of num after any number of swaps.

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 <= 109

Solution:

Put all even digits into one array, all odd digits into another one, all digits into the third. Sort two arrays, and generate a new number from sorted arrays.

Time complexity: O(logn*loglogn)
Space complexity: O(logn)

// Author: Huahua
class Solution {
public:
  int largestInteger(int num) {
    vector<int> odd, even, org;
    for (int cur = num; cur; cur /= 10) {
      ((cur & 1) ? odd : even).push_back(cur % 10);
      org.push_back(cur % 10);
    }
    sort(begin(odd), end(odd));
    sort(begin(even), end(even));    
    int ans = 0;
    for (int i = 0; i < org.size(); ++i) {
      if (i) ans *= 10;
      auto& cur = (org[org.size() - i - 1] & 1) ? odd : even;
      ans += cur.back();
      cur.pop_back();
    }
    return ans;
  }
};

The post 花花酱 LeetCode 2231. Largest Number After Digit Swaps by Parity appeared first on Huahua's Tech Road.

Example 1:

Input: n = 5
Output: [-7,-1,1,3,4]
Explanation: These arrays also are accepted [-5,-1,1,2,3] , [-3,-1,2,-2,4].

Example 2:

Input: n = 3
Output: [-1,0,1]

Example 3:

Input: n = 1
Output: [0]

Constraints:

• 1 <= n <= 1000

Solution: Generation

Use numbers from {-n/2, … n/2} + {0 if n is odd}

Time complexity: O(n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  vector<int> sumZero(int n) {
    vector<int> ans;
    for (int i = 1; i <= n / 2; ++i) {
      ans.push_back(i);
      ans.push_back(-i);
    }
    if (ans.size() != n) ans.push_back(0);
    return ans;
  }
};

The post 花花酱 LeetCode 1304. Find N Unique Integers Sum up to Zero appeared first on Huahua's Tech Road.