Solution in O(N) ,  easy approach using hashing and power of 2 method.
Intuition\n Describe your first thoughts on how to solve this problem. \nWell uhm Bitwise OR properties came first to my mind, and generally to form power of 2 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWell uhm Bitwise OR properties came first to my mind, and generally to form power of 2 we can using only itself cuz only when u do 0 | 0 will u get zero, and if it exists then its possible to form thus arriving at conclusion that first power of 2 would be the impossible OR, and the reason y i havent considered non power of 2 numbers is that we can form any other number using pre existing element in array given simple reason being OR properties. \n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Initialize an array hash of size 32 (to cover powers of 2 from 2^0 to 2^31 ( like 32 bits max for int )to zero. This array will track which powers of 2 are present.\n2. Iterate through the input array nums:\nFor each number, check if it is a power of 2 using the condition (nums[i] & (nums[i] - 1)) == 0.\nIf the number is a power of 2, calculate its log base 2 to find its position and increment the corresponding position in the hash array.\n3. Iterate through the hash array to find the first index i where the value is zero. This index corresponds to the smallest missing power of 2\n4. Return 2^i\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nTime complexity: \nO(n), where n is the size of the input array. This is because we iterate through the array once.\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nSpace complexity: \nO(1). The space used by the hash array is constant and does not depend on the input size.\n\n# Code\n```\nint minImpossibleOR(int* nums, int numsSize){\n    int hash[32]={0};\n    int i;\n    for( i=0; i < numsSize; i++){\n        if( ((nums[i] ) & (nums[i]-1) )== 0){\n            hash[(int)log2(nums[i])]++;\n        }\n    }\n    for( i=0; i < 32; i++){\n        if( hash[i] == 0){\n            break;\n        }\n    }\n    return(pow(2,i));\n}\n```
['C']