Skip to content

1775. Equal Sum Arrays With Minimum Number of Operations

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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
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;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
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;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
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
}

Comments