[Python🤫🐍🐍🐍]  Simple Clean Hashmap Counter Solution
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nUse a HM that tells you
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nUse a HM that tells you how much to add to the counter for each row when you reach (i,j)\nUse a second HM that tells you how much to remove from the counter when you reach (i,j)\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n^2 + q*n)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(n^2)\n\n# Code\n```\nclass Solution:\n    def rangeAddQueries(self, n: int, queries: List[List[int]]) -> List[List[int]]:\n        \n        out = [ [0]*n for i in range(n)]\n\n        #When we reach hms with key [i,j] we increment counter by hms[i,j]\n        hms = defaultdict(int)\n        #When we reach hme with key [i,j] we increment counter by hme[i,j]\n        hme = defaultdict(int)\n        \n        for q in queries:\n            \n            for i in range(q[0],q[2]+1):\n                hms[(i,q[1])] += 1\n                hme[(i,q[3]+1)] += 1\n\n        for i in range(n):\n            count = 0\n            for j in range(n):\n                count += hms[(i,j)]\n                count -= hme[(i,j)]\n                out[i][j] = count\n        return out\n\n```
['Hash Table', 'Python3']