Simple sorting N*log(N), ==> [C]
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```c []\nint com_pare(const void *a , const void *b)\n{\n    return *(int *)a - *(int *)b;\n}\n\nint findValueOfPartition(int* nums, int numsSize){\n\nqsort(nums, numsSize , sizeof(int ) , com_pare);\n\nint min = 1100000000;\n\nfor(int i = 0 ; i < numsSize-1 ; i++)\n{\n    if(abs(nums[i]-nums[i+1]) < min)\n    {\n        min = abs(nums[i]-nums[i+1]);\n    }\n}\n\nreturn min; \n}\n```
['C']