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
Solution
equal-sum-arrays-with-minimum-number-of-operations.py
class Solution:
def minOperations(self, nums1: List[int], nums2: List[int]) -> int:
n1, n2 = len(nums1), len(nums2)
if n1 * 6 < n2 or n1 > n2 * 6: return -1
sum1, sum2 = sum(nums1), sum(nums2)
if sum1 > sum2: return self.minOperations(nums2, nums1)
nums1.sort()
nums2.sort()
i, j ,res = 0, n2 - 1, 0
while sum2 > sum1:
if (j < 0 or i < n1) and 6 - nums1[i] > nums2[j] - 1:
sum1 += 6 - nums1[i]
i += 1
else:
sum2 -= nums2[j] - 1
j -= 1
res += 1
return res