Java | Sorting | 5 lines | Clean code
Complexity\n- Time complexity: O(n*log(n))\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(log(n)) used by the sorting algorithm\n Add your
# Complexity\n- Time complexity: $$O(n*log(n))$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(log(n))$$ used by the sorting algorithm\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n  public int findValueOfPartition(int[] nums) {\n    int n = nums.length, ans = Integer.MAX_VALUE;\n    Arrays.sort(nums);\n    \n    for (var i=0; i < n-1; i++)\n      ans = Math.min(ans, nums[i+1] - nums[i]);\n    \n    return ans;\n  }\n}\n```\nIf you like my solution, please upvote it!
['Array', 'Sorting', 'Java']