Question

1775. Equal Sum Arrays With Minimum Number of Operations

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Description

You are given two arrays of integers nums1 and nums2, possibly of different lengths. The values in the arrays are between 1 and 6, inclusive.

In one operation, you can change any integer's value in any of the arrays to any value between 1 and 6, inclusive.

Return _the minimum number of operations required to make the sum of values in _nums1_ equal to the sum of values in _nums2_._ Return -1​​​​​ if it is not possible to make the sum of the two arrays equal.

 

Example 1:

Input: nums1 = [1,2,3,4,5,6], nums2 = [1,1,2,2,2,2]
Output: 3
Explanation: You can make the sums of nums1 and nums2 equal with 3 operations. All indices are 0-indexed.
- Change nums2[0] to 6. nums1 = [1,2,3,4,5,6], nums2 = [6,1,2,2,2,2].
- Change nums1[5] to 1. nums1 = [1,2,3,4,5,1], nums2 = [6,1,2,2,2,2].
- Change nums1[2] to 2. nums1 = [1,2,2,4,5,1], nums2 = [6,1,2,2,2,2].

Example 2:

Input: nums1 = [1,1,1,1,1,1,1], nums2 = [6]
Output: -1
Explanation: There is no way to decrease the sum of nums1 or to increase the sum of nums2 to make them equal.

Example 3:

Input: nums1 = [6,6], nums2 = [1]
Output: 3
Explanation: You can make the sums of nums1 and nums2 equal with 3 operations. All indices are 0-indexed. 
- Change nums1[0] to 2. nums1 = [2,6], nums2 = [1].
- Change nums1[1] to 2. nums1 = [2,2], nums2 = [1].
- Change nums2[0] to 4. nums1 = [2,2], nums2 = [4].

 

Constraints:

• 1 <= nums1.length, nums2.length <= 105
• 1 <= nums1[i], nums2[i] <= 6

Solutions

Solution 1

class Solution:
  def minOperations(self, nums1: List[int], nums2: List[int]) -> int:
    s1, s2 = sum(nums1), sum(nums2)
    if s1 == s2:
      return 0
    if s1 > s2:
      return self.minOperations(nums2, nums1)
    arr = [6 - v for v in nums1] + [v - 1 for v in nums2]
    d = s2 - s1
    for i, v in enumerate(sorted(arr, reverse=True), 1):
      d -= v
      if d <= 0:
        return i
    return -1

class Solution {
  public int minOperations(int[] nums1, int[] nums2) {
    int s1 = Arrays.stream(nums1).sum();
    int s2 = Arrays.stream(nums2).sum();
    if (s1 == s2) {
      return 0;
    }
    if (s1 > s2) {
      return minOperations(nums2, nums1);
    }
    int d = s2 - s1;
    int[] arr = new int[nums1.length + nums2.length];
    int k = 0;
    for (int v : nums1) {
      arr[k++] = 6 - v;
    }
    for (int v : nums2) {
      arr[k++] = v - 1;
    }
    Arrays.sort(arr);
    for (int i = 1, j = arr.length - 1; j >= 0; ++i, --j) {
      d -= arr[j];
      if (d <= 0) {
        return i;
      }
    }
    return -1;
  }
}

class Solution {
public:
  int minOperations(vector<int>& nums1, vector<int>& nums2) {
    int s1 = accumulate(nums1.begin(), nums1.end(), 0);
    int s2 = accumulate(nums2.begin(), nums2.end(), 0);
    if (s1 == s2) return 0;
    if (s1 > s2) return minOperations(nums2, nums1);
    int d = s2 - s1;
    int arr[nums1.size() + nums2.size()];
    int k = 0;
    for (int& v : nums1) arr[k++] = 6 - v;
    for (int& v : nums2) arr[k++] = v - 1;
    sort(arr, arr + k, greater<>());
    for (int i = 0; i < k; ++i) {
      d -= arr[i];
      if (d <= 0) return i + 1;
    }
    return -1;
  }
};

func minOperations(nums1 []int, nums2 []int) int {
  s1, s2 := sum(nums1), sum(nums2)
  if s1 == s2 {
    return 0
  }
  if s1 > s2 {
    return minOperations(nums2, nums1)
  }
  d := s2 - s1
  arr := []int{}
  for _, v := range nums1 {
    arr = append(arr, 6-v)
  }
  for _, v := range nums2 {
    arr = append(arr, v-1)
  }
  sort.Sort(sort.Reverse(sort.IntSlice(arr)))
  for i, v := range arr {
    d -= v
    if d <= 0 {
      return i + 1
    }
  }
  return -1
}

func sum(nums []int) (s int) {
  for _, v := range nums {
    s += v
  }
  return
}

Solution 2

class Solution:
  def minOperations(self, nums1: List[int], nums2: List[int]) -> int:
    s1, s2 = sum(nums1), sum(nums2)
    if s1 == s2:
      return 0
    if s1 > s2:
      return self.minOperations(nums2, nums1)
    cnt = Counter([6 - v for v in nums1] + [v - 1 for v in nums2])
    d = s2 - s1
    ans = 0
    for i in range(5, 0, -1):
      while cnt[i] and d > 0:
        d -= i
        cnt[i] -= 1
        ans += 1
    return ans if d <= 0 else -1

class Solution {
  public int minOperations(int[] nums1, int[] nums2) {
    int s1 = Arrays.stream(nums1).sum();
    int s2 = Arrays.stream(nums2).sum();
    if (s1 == s2) {
      return 0;
    }
    if (s1 > s2) {
      return minOperations(nums2, nums1);
    }
    int d = s2 - s1;
    int[] cnt = new int[6];
    for (int v : nums1) {
      ++cnt[6 - v];
    }
    for (int v : nums2) {
      ++cnt[v - 1];
    }
    int ans = 0;
    for (int i = 5; i > 0; --i) {
      while (cnt[i] > 0 && d > 0) {
        d -= i;
        --cnt[i];
        ++ans;
      }
    }
    return d <= 0 ? ans : -1;
  }
}

class Solution {
public:
  int minOperations(vector<int>& nums1, vector<int>& nums2) {
    int s1 = accumulate(nums1.begin(), nums1.end(), 0);
    int s2 = accumulate(nums2.begin(), nums2.end(), 0);
    if (s1 == s2) return 0;
    if (s1 > s2) return minOperations(nums2, nums1);
    int d = s2 - s1;
    int cnt[6] = {0};
    for (int& v : nums1) ++cnt[6 - v];
    for (int& v : nums2) ++cnt[v - 1];
    int ans = 0;
    for (int i = 5; i; --i) {
      while (cnt[i] && d > 0) {
        d -= i;
        --cnt[i];
        ++ans;
      }
    }
    return d <= 0 ? ans : -1;
  }
};

func minOperations(nums1 []int, nums2 []int) (ans int) {
  s1, s2 := sum(nums1), sum(nums2)
  if s1 == s2 {
    return 0
  }
  if s1 > s2 {
    return minOperations(nums2, nums1)
  }
  d := s2 - s1
  cnt := [6]int{}
  for _, v := range nums1 {
    cnt[6-v]++
  }
  for _, v := range nums2 {
    cnt[v-1]++
  }
  for i := 5; i > 0; i-- {
    for cnt[i] > 0 && d > 0 {
      d -= i
      cnt[i]--
      ans++
    }
  }
  if d <= 0 {
    return
  }
  return -1
}

func sum(nums []int) (s int) {
  for _, v := range nums {
    s += v
  }
  return
}