Solution using Reduce
Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem requires counting pairs of elements in an array such that the sum of the el
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires counting pairs of elements in an array such that the sum of the elements in each pair is less than a given target. The initial thought is to use a nested loop to generate all possible pairs and then count the valid ones.\n# Approach\n1. **Pair Generation**: Generate all possible pairs `(nums[i], nums[j])` where `i < j`.\n2. **Reduce Function**: Use the `reduce` function to accumulate the count of valid pairs. The `func` function checks if the sum of the pair is less than the target and increments the count if true.\n3. **Initial Value**: The initial value for the count is set to `0`.\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n-  Time complexity: \\(O(n^2)\\)\n  - Generating all pairs takes \\(O(n^2)\\) time.\n  - The `reduce` function also runs in \\(O(n^2)\\) time as it \n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: \\(O(1)\\)\n  - The generator expression for pairs does not store all pairs in memory at once, keeping space usage minimal.\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nclass Solution:\n    def countPairs(self, nums: List[int], target: int) -> int:\n        length = len(nums)\n        def func(count,pair):\n            if pair[0] + pair[1] < target:\n                return count+1\n            return count\n        pair = ((nums[i],nums[j]) for i in range(0,length) for j in range(i+1,length))\n        return reduce(func,pair,0)\n```
['Python3']