You are given a 0-indexed array nums containing n integers.
At each second, you perform the following operation on the array:
\n\ni in the range [0, n - 1], replace nums[i] with either nums[i], nums[(i - 1 + n) % n], or nums[(i + 1) % n].Note that all the elements get replaced simultaneously.
\n\nReturn the minimum number of seconds needed to make all elements in the array nums equal.
\n
Example 1:
\n\n\nInput: nums = [1,2,1,2]\nOutput: 1\nExplanation: We can equalize the array in 1 second in the following way:\n- At 1st second, replace values at each index with [nums[3],nums[1],nums[3],nums[3]]. After replacement, nums = [2,2,2,2].\nIt can be proven that 1 second is the minimum amount of seconds needed for equalizing the array.\n\n\n
Example 2:
\n\n\nInput: nums = [2,1,3,3,2]\nOutput: 2\nExplanation: We can equalize the array in 2 seconds in the following way:\n- At 1st second, replace values at each index with [nums[0],nums[2],nums[2],nums[2],nums[3]]. After replacement, nums = [2,3,3,3,3].\n- At 2nd second, replace values at each index with [nums[1],nums[1],nums[2],nums[3],nums[4]]. After replacement, nums = [3,3,3,3,3].\nIt can be proven that 2 seconds is the minimum amount of seconds needed for equalizing the array.\n\n\n
Example 3:
\n\n\nInput: nums = [5,5,5,5]\nOutput: 0\nExplanation: We don't need to perform any operations as all elements in the initial array are the same.\n\n\n
\n
Constraints:
\n\n1 <= n == nums.length <= 1051 <= nums[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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