Easy One line solution
Intuition\nWe can iterate over all possible pairs of indices (i, j) in the list nums. For each pair, we check if nums[i] + nums[j] equals target.\n\n\n# Approac
# Intuition\nWe can iterate over all possible pairs of indices (i, j) in the list nums. For each pair, we check if nums[i] + nums[j] equals target.\n\n\n# Approach\n- **Brute Force Enumeration:**  We can iterate over all possible pairs of indices (i, j) in the list nums. For each pair, we check if nums[i] + nums[j] equals target.\n\n- **Counting Pairs:** We maintain a count of pairs that satisfy the concatenation condition.\n\n- **Efficiency Consideration:** To avoid counting duplicates and improve efficiency, we ensure that i is not equal to j during iteration.\n# Complexity\n- **Time complexity:** O(N^2)\n- **Space complexity:** O(N)\n# Code\n```\nclass Solution:\n    def numOfPairs(self, nums: List[str], target: str) -> int:\n        return sum(1 for i in range(len(nums)) for j in range(len(nums)) if i!=j and nums[i]+nums[j]==target )\n```
['Python3']