Java solution | O(n^2) and O(n)
Please Upvote !!!\n#### 1. Nested loop:\n\nclass Solution {\n    public boolean findSubarrays(int[] nums) {\n        for (int i = 0; i < nums.length - 2; i++) {
### ***Please Upvote !!!***\n#### 1. Nested loop:\n```\nclass Solution {\n    public boolean findSubarrays(int[] nums) {\n        for (int i = 0; i < nums.length - 2; i++) {\n            int sum1 = nums[i] + nums[i + 1];\n            for (int j = i + 1; j < nums.length - 1; j++) {\n                int sum2 = nums[j] + nums[j + 1];\n                if (sum1 == sum2) return true;\n            }\n        }\n\n        return false;\n    }\n}\n\n// TC: O(n ^ 2), SC: O(1)\n```\n#### 2. Single loop (using hashset):\n```\nclass Solution {\n    public boolean findSubarrays(int[] nums) {\n        Set<Integer> set = new HashSet<>();\n        for (int i = 0; i < nums.length - 1; i++) {\n            int sum = nums[i] + nums[i + 1];\n            if (set.contains(sum)) return true;\n            set.add(sum);\n        }\n\n        return false;\n    }\n}\n\n// TC: O(n), SC: O(n)\n```
['Java']