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2449. Minimum Number of Operations to Make Arrays Similar

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Description

You are given two positive integer arrays nums and target, of the same length.

In one operation, you can choose any two distinct indices i and j where 0 <= i, j < nums.length and:

• set nums[i] = nums[i] + 2 and
• set nums[j] = nums[j] - 2.

Two arrays are considered to be similar if the frequency of each element is the same.

Return the minimum number of operations required to make nums similar to target. The test cases are generated such that nums can always be similar to target.

 

Example 1:

Input: nums = [8,12,6], target = [2,14,10]
Output: 2
Explanation: It is possible to make nums similar to target in two operations:
- Choose i = 0 and j = 2, nums = [10,12,4].
- Choose i = 1 and j = 2, nums = [10,14,2].
It can be shown that 2 is the minimum number of operations needed.

Example 2:

Input: nums = [1,2,5], target = [4,1,3]
Output: 1
Explanation: We can make nums similar to target in one operation:
- Choose i = 1 and j = 2, nums = [1,4,3].

Example 3:

Input: nums = [1,1,1,1,1], target = [1,1,1,1,1]
Output: 0
Explanation: The array nums is already similiar to target.

 

Constraints:

• n == nums.length == target.length
• 1 <= n <= 105
• 1 <= nums[i], target[i] <= 106
• It is possible to make nums similar to target.

Solutions

Python3

class Solution:
    def makeSimilar(self, nums: List[int], target: List[int]) -> int:
        nums.sort(key=lambda x: (x & 1, x))
        target.sort(key=lambda x: (x & 1, x))
        return sum(abs(a - b) for a, b in zip(nums, target)) // 4

Java

class Solution {
    public long makeSimilar(int[] nums, int[] target) {
        Arrays.sort(nums);
        Arrays.sort(target);
        List<Integer> a1 = new ArrayList<>();
        List<Integer> a2 = new ArrayList<>();
        List<Integer> b1 = new ArrayList<>();
        List<Integer> b2 = new ArrayList<>();
        for (int v : nums) {
            if (v % 2 == 0) {
                a1.add(v);
            } else {
                a2.add(v);
            }
        }
        for (int v : target) {
            if (v % 2 == 0) {
                b1.add(v);
            } else {
                b2.add(v);
            }
        }
        long ans = 0;
        for (int i = 0; i < a1.size(); ++i) {
            ans += Math.abs(a1.get(i) - b1.get(i));
        }
        for (int i = 0; i < a2.size(); ++i) {
            ans += Math.abs(a2.get(i) - b2.get(i));
        }
        return ans / 4;
    }
}

C++

class Solution {
public:
    long long makeSimilar(vector<int>& nums, vector<int>& target) {
        sort(nums.begin(), nums.end());
        sort(target.begin(), target.end());
        vector<int> a1;
        vector<int> a2;
        vector<int> b1;
        vector<int> b2;
        for (int v : nums) {
            if (v & 1)
                a1.emplace_back(v);
            else
                a2.emplace_back(v);
        }
        for (int v : target) {
            if (v & 1)
                b1.emplace_back(v);
            else
                b2.emplace_back(v);
        }
        long long ans = 0;
        for (int i = 0; i < a1.size(); ++i) ans += abs(a1[i] - b1[i]);
        for (int i = 0; i < a2.size(); ++i) ans += abs(a2[i] - b2[i]);
        return ans / 4;
    }
};

Go

func makeSimilar(nums []int, target []int) int64 {
	sort.Ints(nums)
	sort.Ints(target)
	a1, a2, b1, b2 := []int{}, []int{}, []int{}, []int{}
	for _, v := range nums {
		if v%2 == 0 {
			a1 = append(a1, v)
		} else {
			a2 = append(a2, v)
		}
	}
	for _, v := range target {
		if v%2 == 0 {
			b1 = append(b1, v)
		} else {
			b2 = append(b2, v)
		}
	}
	ans := 0
	for i := 0; i < len(a1); i++ {
		ans += abs(a1[i] - b1[i])
	}
	for i := 0; i < len(a2); i++ {
		ans += abs(a2[i] - b2[i])
	}
	return int64(ans / 4)
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

TypeScript

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