Java Simple Solution | Using Bit-manipulation | O(N)
\n\n# Complexity\n- Time complexity: O(N)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(1)\n Add your space complexity here, e.g. O(n) \n
\n\n# Complexity\n- Time complexity: O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n    public int maxSubarrays(int[] nums) {\n        int n = nums.length; \n        int cnt = 0;\n        for(int i = 0;i < n;){\n            int and = nums[i];\n            int j;\n            for(j = i;j < n;j++){\n                and &= nums[j];\n                if(and == 0){\n                    break;\n                }\n            }\n            if(and == 0){\n                cnt++;\n            }\n            i = j + 1;\n\n        }\n\n        return Math.max(cnt,1);\n    }\n}\n```
['Java']