Simple easy python solution (with detailed explanation)
Intuition\nGet all the possible selections and go for the one with the highest satisfaction (i.e the one which covers most rows)\n# Approach\n1. I created an ar
# Intuition\nGet all the possible selections and go for the one with the highest satisfaction (i.e the one which covers most rows)\n# Approach\n1. I created an array "a" and filled it up with numbers upto the number of columns.\n2. I had a helper function in which I either chose a number from a or ignored it, I continued either till I reached the end of the array or when the length of the choosing array has met the condition of becoming equal to the required number of selection.\n3. After this I created a dictionary to get, all the positions that are occupied by one in a row. so basically d[i] represents all the position in row i which are having one\'s\n4. After this I compared it with the subsets I have. If at all a subset is the superset of a row that means we can count it covered.\n5. I just returned the maximum number of rows that are covered.\n# Complexity\n- Time complexity:\nO(n c  se) *m *n \n- Space complexity:\nO(n c se)+m*n\n# Code\n```python3 []\nclass Solution:\n    def maximumRows(self, matrix: List[List[int]], se: int) -> int:\n        n=len(matrix[0])\n        a=[]\n        for i in range(n):\n            a.append(i)\n        aa=[]\n        b=[]\n        def s(i,aa):\n            if len(aa)>=se:\n                if len(aa)==se:\n                    b.append(set(aa))\n                return\n            if i==len(a):\n                return\n            s(i+1,aa+[a[i]])\n            s(i+1,aa)\n        s(0,aa)\n        d=dict()\n        for i in range(len(matrix)):\n            d[i]=set()\n            for j in range(n):\n                if matrix[i][j]==1:\n                    d[i].add(j)\n        m=0\n        for i in b:\n            t=0\n            for j in range(len(matrix)):\n                if d[j].issubset(i):\n                    t=t+1\n            m=max(t,m)\n        \n\n        return m\n        \n        \n```
['Python3']