You are given an array nums consisting of positive integers.
You are also given an integer array queries of size m. For the ith query, you want to make all of the elements of nums equal to queries[i]. You can perform the following operation on the array any number of times:
1.Return an array answer of size m where answer[i] is the minimum number of operations to make all elements of nums equal to queries[i].
Note that after each query the array is reset to its original state.
\n\n\n
Example 1:
\n\n\nInput: nums = [3,1,6,8], queries = [1,5]\nOutput: [14,10]\nExplanation: For the first query we can do the following operations:\n- Decrease nums[0] 2 times, so that nums = [1,1,6,8].\n- Decrease nums[2] 5 times, so that nums = [1,1,1,8].\n- Decrease nums[3] 7 times, so that nums = [1,1,1,1].\nSo the total number of operations for the first query is 2 + 5 + 7 = 14.\nFor the second query we can do the following operations:\n- Increase nums[0] 2 times, so that nums = [5,1,6,8].\n- Increase nums[1] 4 times, so that nums = [5,5,6,8].\n- Decrease nums[2] 1 time, so that nums = [5,5,5,8].\n- Decrease nums[3] 3 times, so that nums = [5,5,5,5].\nSo the total number of operations for the second query is 2 + 4 + 1 + 3 = 10.\n\n\n
Example 2:
\n\n\nInput: nums = [2,9,6,3], queries = [10]\nOutput: [20]\nExplanation: We can increase each value in the array to 10. The total number of operations will be 8 + 1 + 4 + 7 = 20.\n\n\n
\n
Constraints:
\n\nn == nums.lengthm == queries.length1 <= n, m <= 1051 <= nums[i], queries[i] <= 109Compiled with clang 11 using the latest C++ 20 standard.
Your code is compiled with level two optimization (-O2). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.
Most standard library headers are already included automatically for your convenience.
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