{"id":10093,"date":"2023-10-29T20:08:22","date_gmt":"2023-10-30T03:08:22","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=10093"},"modified":"2023-10-29T20:08:51","modified_gmt":"2023-10-30T03:08:51","slug":"leetcode-2918-minimum-equal-sum-of-two-arrays-after-replacing-zeros","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/uncategorized\/leetcode-2918-minimum-equal-sum-of-two-arrays-after-replacing-zeros\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 2918. Minimum Equal Sum of Two Arrays After Replacing Zeros"},"content":{"rendered":"\n

You are given two arrays nums1<\/code> and nums2<\/code> consisting of positive integers.<\/p>\n\n\n\n

You have to replace all<\/strong> the 0<\/code>‘s in both arrays with strictly<\/strong> positive integers such that the sum of elements of both arrays becomes equal<\/strong>.<\/p>\n\n\n\n

Return the minimum<\/strong> equal sum you can obtain, or <\/em>-1<\/code> if it is impossible<\/em>.<\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

Input:<\/strong> nums1 = [3,2,0,1,0], nums2 = [6,5,0]\nOutput:<\/strong> 12\nExplanation:<\/strong> We can replace 0's in the following way:\n- Replace the two 0's in nums1 with the values 2 and 4. The resulting array is nums1 = [3,2,2,1,4].\n- Replace the 0 in nums2 with the value 1. The resulting array is nums2 = [6,5,1].\nBoth arrays have an equal sum of 12. It can be shown that it is the minimum sum we can obtain.\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> nums1 = [2,0,2,0], nums2 = [1,4]\nOutput:<\/strong> -1\nExplanation:<\/strong> It is impossible to make the sum of both arrays equal.\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= nums1.length, nums2.length <= 105<\/sup><\/code><\/li>
  • 0 <= nums1[i], nums2[i] <= 106<\/sup><\/code><\/li><\/ul>\n\n\n\n
    Solution: A few cases<\/strong><\/h2>\n\n\n\n
    Calculate the sum of number of zeros of each array as (s1, z1), (s2, z2). There are a few cases:<\/p>\n\n\n\n
    1. z1 == z2 == 0, there is no way to change, just check s1 == s2.<\/li>
    2. z1 == 0, z1 != 0 or z2 == 0, z1 != 0. Need to at least increase the sum by number of zeros, check s1 + z1 <= s2 \/ s2 + z2 <= s1<\/li>
    3. z1 != 0, z2 != 0, the min sum is max(s1 + z1, s2 + z2)<\/li><\/ol>\n\n\n\n
      Time complexity: O(n)
      Space complexity: O(1)<\/p>\n\n\n\n
      \n
      C++<\/h2>\n
      \n\n
      \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  long long minSum(vector& nums1, vector& nums2) {\n    auto getSumAndZeros = [](const vector& nums) {\n      long long sum = 0;\n      long long zeros = 0;\n      for (int x : nums) {\n        sum += x;\n        zeros += x == 0;\n      }\n      return pair{sum, zeros};\n    };\n    auto [s1, z1] = getSumAndZeros(nums1);\n    auto [s2, z2] = getSumAndZeros(nums2);\n    if (z1 == 0 && z2 == 0) {\n      return s1 == s2 ? s1 : -1;\n    } else if (z1 == 0) {\n      return s2 + z2 <= s1 ? s1 : -1;\n    } else if (z2 == 0) {\n      return s1 + z1 <= s2 ? s2 : -1;\n    } else {\n      return max(s1 + z1, s2 + z2);\n    }\n    return -1;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
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