LeetCode每日一题(2449. Minimum Number of Operations to Make Arrays Similar)

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.

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1. 因为加和减是对称的， 所以我们只需要统计一种即可
2. num 与 target 都进行排序， 排序后， 对于 num[i], 一定是变成 target[i]的成本最低
3. 注意奇偶的问题， 因为无论加减操作数都是 2， 所以奇数只能变成奇数，偶数只能变成偶数

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impl Solution {
    pub fn make_similar(mut nums: Vec<i32>, mut target: Vec<i32>) -> i64 {
        nums.sort();
        let mut num_odds = Vec::new();
        let mut num_evens = Vec::new();
        for n in nums {
            if n % 2 == 0 {
                num_evens.push(n);
                continue;
            }
            num_odds.push(n);
        }
        target.sort();
        let mut target_odds = Vec::new();
        let mut target_evens = Vec::new();
        for n in target {
            if n % 2 == 0 {
                target_evens.push(n);
                continue;
            }
            target_odds.push(n);
        }
        let mut ans = 0;
        for (s, t) in num_odds.into_iter().zip(target_odds) {
            ans += if s > t { (s - t) as i64 / 2 } else { 0 };
        }
        for (s, t) in num_evens.into_iter().zip(target_evens) {
            ans += if s > t { (s - t) as i64 / 2 } else { 0 };
        }
        ans
    }
}