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2449. Minimum Number of Operations to Make Arrays Similar

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

Solution 1: Odd-Even Classification + Sorting

Notice that, because each operation will only increase or decrease the value of an element by $2$, the parity of the element will not change.

Therefore, we can divide the arrays $nums$ and $target$ into two groups according to their parity, denoted as $a_1$ and $a_2$, and $b_1$ and $b_2$ respectively.

Then, we just need to pair the elements in $a_1$ with the elements in $b_1$, and pair the elements in $a_2$ with the elements in $b_2$, and then perform operations. During the pairing process, we can use a greedy strategy, pairing the smaller elements in $a_i$ with the smaller elements in $b_i$ each time, which can ensure the minimum number of operations. This can be directly implemented through sorting.

Since each operation can reduce the difference of the corresponding elements by $4$, we accumulate the difference of each corresponding position, and finally divide by $4$ to get the answer.

The time complexity is $O(n\times \log n)$, where $n$ is the length of the array $nums$.

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  }
  

• 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;
      }
  };
  

• 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
  
  

• 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
  }
  

• function makeSimilar(nums: number[], target: number[]): number {
      nums.sort((a, b) => a - b);
      target.sort((a, b) => a - b);
  
      const a1: number[] = [];
      const a2: number[] = [];
      const b1: number[] = [];
      const b2: number[] = [];
  
      for (const v of nums) {
          if (v % 2 === 0) {
              a1.push(v);
          } else {
              a2.push(v);
          }
      }
  
      for (const v of target) {
          if (v % 2 === 0) {
              b1.push(v);
          } else {
              b2.push(v);
          }
      }
  
      let ans = 0;
      for (let i = 0; i < a1.length; ++i) {
          ans += Math.abs(a1[i] - b1[i]);
      }
  
      for (let i = 0; i < a2.length; ++i) {
          ans += Math.abs(a2[i] - b2[i]);
      }
  
      return ans / 4;
  }
  
  

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