Optimized DP Solution Easy to understand
Intuition\n Describe your first thoughts on how to solve this problem. \nWe want to count the number of squares with all elements as 1 in a given matrix. We can
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe want to count the number of squares with all elements as 1 in a given matrix. We can achieve this by iteratively calculating the maximum square size that ends at each cell. To count the squares, we sum up the square sizes.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Initialize a variable count to 0, which will keep track of the count of squares.\n2. Iterate through the matrix:\n* For the first row and the first column, add the cell values to the count. These are squares of size 1 with all elements as 1.\n* For other cells, if the cell value is 1, update the cell value to be the minimum of its top, left, and top-left neighbors plus 1. This represents the maximum square size ending at this cell.\n* Update the count by adding the new square size to it.\n3. Return the final count, which represents the total number of squares with all elements as 1.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(m\xD7n), where m is the number of rows and n is the number of columns in the matrix.\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(1) (constant space), as we are not using any additional data structures that grow with the input size.\n# Code\n```\nclass Solution {\n    public int countSquares(int[][] matrix) {\n        int count = 0;  // Initialize count of squares\n\n        // Iterate through the matrix\n        for (int i = 0; i < matrix.length; i++) {\n            for (int j = 0; j < matrix[0].length; j++) {\n                if (matrix[i][j] > 0) {\n                    // Update the cell value to be the minimum of its top, left, and top-left neighbors plus 1\n                    matrix[i][j] = Math.min(matrix[i-1][j-1], Math.min(matrix[i][j-1], matrix[i-1][j])) + 1;\n                    count += matrix[i][j];  // Add the square size to count\n                }\n            }\n        }\n\n        return count;  // Return the count of squares\n    }\n}\n```
['Java']