Palindrome Partitioning III
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nPrecomputation:\n\nFirs
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nPrecomputation:\n\nFirst, we\'ll create a 2D array changes[i][j] where changes[i][j] represents the minimum number of changes needed to make the substring s[i:j+1] a palindrome.\nDynamic Programming Table Setup:\n\nWe need another 2D array dp[i][j] where dp[i][j] represents the minimum number of changes needed to partition the substring s[0:i+1] into j+1 palindromic substrings.\nBase Cases:\n\nIf j = 0 (one substring), then dp[i][0] will just be changes[0][i], which is the number of changes needed to make s[0:i+1] a palindrome.\nFilling the DP Table:\n\nFor each substring ending at i and splitting it into j+1 substrings, we will iterate through possible split points p and calculate dp[i][j] using:\ndp[i][j]=min(dp[i][j],dp[p][j\u22121]+changes[p+1][i])\nThis means we\'re adding the cost of making the substring from p+1 to i a palindrome to the minimum cost of partitioning the substring up to p into j palindromic substrings.\nFinal Result:\n\nThe result will be found in dp[n-1][k-1], where n is the length of the string s.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n^2.K)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(n^2 +n\u22C5k)\n\n# Code\n```python []\nclass Solution(object):\n    def palindromePartition(self, s, k):\n        n = len(s)\n        \n        # Step 1: Precompute the number of changes needed to make substrings palindromic\n        changes = [[0] * n for _ in range(n)]\n        \n        for length in range(2, n + 1):  # length of the substring\n            for i in range(n - length + 1):\n                j = i + length - 1\n                changes[i][j] = changes[i + 1][j - 1] + (s[i] != s[j])\n        \n        # Step 2: DP table to find the minimum changes needed for k palindromic partitions\n        dp = [[float(\'inf\')] * k for _ in range(n)]\n        \n        # Fill the dp table\n        for i in range(n):\n            dp[i][0] = changes[0][i]  # One partition\n        \n        for j in range(1, k):\n            for i in range(n):\n                for p in range(i):\n                    dp[i][j] = min(dp[i][j], dp[p][j - 1] + changes[p + 1][i])\n        \n        return dp[n - 1][k - 1]\n\n\n```
['Python']