343: Space 100%, Solution with step by step explanation
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nTo solve this problem, we can use dynamic programming. We will create a d
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nTo solve this problem, we can use dynamic programming. We will create a dp array of size n+1 where dp[i] will represent the maximum product that can be obtained by breaking the integer i into the sum of k positive integers, where k>=2.\n\nWe will then iterate from 2 to n and for each i, we will break it into the sum of j and i-j, where j is in the range 1 to i/2. We will then calculate the product of these two integers and take the maximum of all such products. This maximum value will be stored in dp[i].\n\nFinally, we will return dp[n], which will represent the maximum product that can be obtained by breaking n into the sum of k positive integers, where k>=2.\n\nThe time complexity of this algorithm is O(n^2) and the space complexity is O(n).\n\n# Complexity\n- Time complexity:\n42.66%\n\n- Space complexity:\n100%\n\n# Code\n```\nclass Solution:\n    def integerBreak(self, n: int) -> int:\n        # Initialize dp array with 0\'s\n        dp = [0]*(n+1)\n        \n        # Base case\n        dp[2] = 1\n        \n        # Iterate from 3 to n\n        for i in range(3, n+1):\n            # Iterate from 1 to i//2\n            for j in range(1, i//2+1):\n                # Calculate the product of j and i-j\n                prod = j*(i-j)\n                \n                # Update the maximum product in dp[i]\n                dp[i] = max(dp[i], max(prod, j*dp[i-j]))\n        \n        # Return dp[n]\n        return dp[n]\n\n```
['Math', 'Dynamic Programming', 'Python', 'Python3']