561. Array Partition I
1. Description
Given an integer array nums of 2n integers, group these integers into n pairs (a1, b1), (a2, b2), …, (an, bn) such that the sum of min(ai, bi) for all i is maximized. Return the maximized sum.
2. Example
Example 1:
Input: nums = [1,4,3,2]
Output: 4
Explanation: All possible pairings (ignoring the ordering of elements) are:
- (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
- (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
- (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4 So the maximum possible sum is 4.
Example 2:
Input: nums = [6,2,6,5,1,2]
Output: 9
Explanation: The optimal pairing is (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9.
3. Constraints
- 1 <= n <= $10^4$
- nums.length == 2 * n
- $-10^4$ <= nums[i] <= $10^4$
4. Solutions
My Accepted Solution
n = m_nums.size()
Time complexity: O($nlog_2n$)
Space complexity: O(1)
class Solution
{
public:
// int arrayPairSum(vector<int>& nums)
int arrayPairSum(vector<int> &m_nums)
{
sort(m_nums.begin(), m_nums.end());
int result = 0;
for(int i = 0; i < m_nums.size(); i += 2)
result += m_nums[i];
return result;
}
};
4.1 Bucket Sort
n = m_nums.size()
Time complexity: O(n)
Space complexity: O(n)
// the range of number is [-10000, 10000], it is pretty small, so we could use bucket sort
class Solution
{
public:
// int arrayPairSum(vector<int>& nums)
int arrayPairSum(vector<int> &i_nums)
{
vector<int> buckets(20001); // the range is [-10000, 10000]
for(auto num : i_nums)
buckets[num + 10000]++;
int result = 0;
for(int i = 0; i < buckets.size(); )
{
while(i < buckets.size() && buckets[i] == 0) i++;
// there must be even numbers
// so we just need to judge whether we have the first number in pairs
if(i == buckets.size()) break;
buckets[i]--;
result += (i - 10000);
while(i < buckets.size() && buckets[i] == 0) i++;
buckets[i]--;
}
return result;
}
};