Java easy with explanation.
Approach\nTo solve this problem, we can use the approach of calculating prefix and suffix sums. For each element in the array, we can calculate the maximum poss
# Approach\nTo solve this problem, we can use the approach of calculating prefix and suffix sums. For each element in the array, we can calculate the maximum possible value of nums[0] | nums[1] | ... | nums[n - 1] by multiplying it by 2^k and bitwise ORing it with the prefix sum of the elements before it and the suffix sum of the elements after it.\n\nHere\'s the step-by-step algorithm:\n\n1. Calculate the prefix sum of the array. For each index i, the prefix sum prefix[i] is equal to the bitwise OR of all elements nums[0] | nums[1] | ... | nums[i].\n\n2. Calculate the suffix sum of the array. For each index i, the suffix sum suffix[i] is equal to the bitwise OR of all elements nums[i] | nums[i+1] | ... | nums[n-1].\n\n3. For each index i in the array, calculate the maximum possible value of nums[0] | nums[1] | ... | nums[n - 1] by multiplying nums[i] by 2^k and bitwise ORing it with prefix[i-1] and suffix[i+1]. Take the maximum value obtained from all indices.\n\n4. Return the maximum value obtained in step 3.\n\n# Complexity\n- Time complexity:\nO(N);\n\n- Space complexity:\nO(N);\n\n# Code\n```\nclass Solution {\n    public long maximumOr(int[] nums, int k) {\n        int n = nums.length;\n        int[] prefix = new int[n];\n        int[] suffix = new int[n];\n        \n        prefix[0] = nums[0];\n        for (int i = 1; i < n; i++) {\n            prefix[i] = prefix[i-1] | nums[i];\n        }\n        \n        suffix[n-1] = nums[n-1];\n        for (int i = n-2; i >= 0; i--) {\n            suffix[i] = suffix[i+1] | nums[i];\n        }\n        \n        long maxOr = 0;\n        for (int i = 0; i < n; i++) {\n            long or = ((long) nums[i]) * (1L << k) | (i > 0 ? prefix[i-1] : 0) | (i < n-1 ? suffix[i+1] : 0);\n            maxOr = Math.max(maxOr, or);\n        }\n        \n        return maxOr;\n    }\n}\n```
['Java']