You are given a 0-indexed integer array nums and an integer x.
Find the minimum absolute difference between two elements in the array that are at least x indices apart.
In other words, find two indices i and j such that abs(i - j) >= x and abs(nums[i] - nums[j]) is minimized.
Return an integer denoting the minimum absolute difference between two elements that are at least x indices apart.
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Example 1:
\n\n\nInput: nums = [4,3,2,4], x = 2\nOutput: 0\nExplanation: We can select nums[0] = 4 and nums[3] = 4. \nThey are at least 2 indices apart, and their absolute difference is the minimum, 0. \nIt can be shown that 0 is the optimal answer.\n\n\n
Example 2:
\n\n\nInput: nums = [5,3,2,10,15], x = 1\nOutput: 1\nExplanation: We can select nums[1] = 3 and nums[2] = 2.\nThey are at least 1 index apart, and their absolute difference is the minimum, 1.\nIt can be shown that 1 is the optimal answer.\n\n\n
Example 3:
\n\n\nInput: nums = [1,2,3,4], x = 3\nOutput: 3\nExplanation: We can select nums[0] = 1 and nums[3] = 4.\nThey are at least 3 indices apart, and their absolute difference is the minimum, 3.\nIt can be shown that 3 is the optimal answer.\n\n\n
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Constraints:
\n\n1 <= nums.length <= 1051 <= nums[i] <= 1090 <= x < nums.lengthi < j, as the problem is symmetric.j, we are interested in an index i in the range [0, j - x] that minimizes abs(nums[i] - nums[j]).j, while going from left to right, add nums[j - x] to a set (C++ set, Java TreeSet, and Python sorted set).nums[j - x], we can calculate the closest value to nums[j] in the set using binary search and store the absolute difference. In C++, we can achieve this by using lower_bound and/or upper_bound.Compiled 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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