Initial version of the README.MD

README.md

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+---
+license: llama2
+language:
+- en
+tags:
+- Python
+- Leetcode
+- Problem Solving
+- CP
+---
+
+```
+Below is an instruction that describes a task, paired with an input that provides further context. Write a response that appropriately completes the request.
+
+### Instruction:
+Solve the problem using Python.
+
+### Input:
+Consider a matrix M with dimensions width * height, such that every cell has value 0 or 1, and any square sub-matrix of M of size sideLength * sideLength has at most maxOnes ones.
+
+Return the maximum possible number of ones that the matrix M can have.
+
+ 
+
+Example 1:
+
+Input: width = 3, height = 3, sideLength = 2, maxOnes = 1
+Output: 4
+Explanation:
+In a 3*3 matrix, no 2*2 sub-matrix can have more than 1 one.
+The best solution that has 4 ones is:
+[1,0,1]
+[0,0,0]
+[1,0,1]
+
+Example 2:
+
+Input: width = 3, height = 3, sideLength = 2, maxOnes = 2
+Output: 6
+Explanation:
+[1,0,1]
+[1,0,1]
+[1,0,1]
+
+ 
+
+Constraints:
+
+    1 <= width, height <= 100
+    1 <= sideLength <= width, height
+    0 <= maxOnes <= sideLength * sideLength
+
+
+
+### Response:
+Sure, here you have a possible implementation to solve the problem.
+
+'''python
+class Solution:
+    def maximumNumberOfOnes(self, C, R, K, maxOnes):
+        # every K*K square has at most maxOnes ones
+        count = [0] * (K*K)
+        for r in range(R):
+            for c in range(C):
+                code = (r%K) * K + c%K
+                count[code] += 1
+        count.sort()
+        ans = 0
+        for _ in range(maxOnes):
+            ans += count.pop()
+        return ans
+'''
+```