[Java] Clean DP solution with detailed comments.
java\npublic int palindromePartition(String s, int k) {\n\tint n = s.length();\n\t// changes[i][j]: changes needed for s[i...j] to turn into a palindrome\n\tint
```java\npublic int palindromePartition(String s, int k) {\n\tint n = s.length();\n\t// changes[i][j]: changes needed for s[i...j] to turn into a palindrome\n\tint[][] changes = new int[n][n];\n\tfor (int len = 2; len <= n; ++len) {\n\t\tfor (int i = 0, j = i + len - 1; j < n; ++i, ++j) {\n\t\t\tchanges[i][j] = changes[i+1][j-1] + (s.charAt(i) == s.charAt(j) ? 0 : 1);\n\t\t}\n\t}\n\n\t// dp[i][j]: min changes needed for s[0 ... j] to break into i groups with each palindrome.\n\tint[][] dp = new int[k + 1][n];\n\tfor (int i = 0; i < n; ++i) {\n\t\t// Starts with 1 group\n\t\tdp[1][i] = changes[0][i];\n\t}\n\n\tfor (int i = 2; i <= k; ++i) {\n\t\t// Explore dp[i][end], since for i groups, first i - 1 groups take at least i - 1 letters (each group has at least 1 letter), \n\t\t// the last group (i.e., group i) starts at least from index i - 1\n\t\tfor (int end = i - 1; end < n; ++end) {\n\t\t\tint min = n; // Initiated to the largest possible\n\t\t\tfor (int start = i - 1; start <= end; ++start) {\n\t\t\t\tmin = Math.min(min, dp[i - 1][start - 1] + changes[start][end]);\n\t\t\t}\n\t\t\tdp[i][end] = min;\n\t\t}\n\t}\n\treturn dp[k][n - 1];\n}\n```
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