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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