Array Partition I

2n ta butun sonli nums massivi berilgan bo'lsa, bu butun sonlarni n ta juftlik (a1, b1), (a2, b2), ..., (an, bn) ga shunday guruhlangki, barcha i uchun min(ai, bi) yig'indisi maksimal bo'lsin. Maksimal summani qaytaring.

Example 1:

Input: nums = [1,4,3,2] 
Output: 4
Explanation: Barcha mumkin bo'lgan juftliklar (elementlarning tartibini e'tiborsiz qoldiring) quyidagilardir:
1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4
Shunday qilib, maksimal mumkin bo'lgan summa 4 ga teng.

Example 2:

Input: nums = [6,2,6,5,1,2]
Output: 9
Explanation: Optimal juftlik (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9.

Cheklovlar:

  • 1 <= n <= 104

  • nums.length == 2 * n

  • -104 <= nums[i] <= 104

Hint-1

Shubhasiz, brute force bu erda yordam bermaydi. Boshqa narsani o'ylab ko'ring, 1,2,3,4 kabi misollarni oling.

Hint-2

Natijaga erishish uchun qanday qilib juftlik yaratasiz? Biror naqsh bo'lishi kerak.

Hint-3

Siz buni kuzatdingizmi - Minimal element maksimal elementni qurbon qilish natijasida natijaga qo'shiladi.

Hint-4

Hali ham juftliklarni topa olmayapsizmi? Massivni tartiblang va naqshni topishga harakat qiling.

func arrayPairSum(nums []int) int {

	slices.Sort(nums)
    
	s := 0
	for i, v := range nums {
		if i%2 == 0 {
			s += v
		}
	}

	return s
}

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