20250303 finished.

src/problem/mod.rs

@@ -514,3 +514,5 @@ mod p2353_design_a_food_rating_system;
 mod p131_palindrome_partitioning;
 
 mod p132_palindrome_partitioning_ii;
+
+mod p1278_palindrome_partitioning_iii;

src/problem/p1278_palindrome_partitioning_iii.rs

@@ -0,0 +1,60 @@
+/**
+ * [1278] Palindrome Partitioning III
+ */
+pub struct Solution {}
+
+// submission codes start here
+
+impl Solution {
+    pub fn palindrome_partition(s: String, k: i32) -> i32 {
+        let s: Vec<char> = s.chars().collect();
+        let length = s.len();
+
+        // cost[i][j]
+        // 将 s[i..=j] 修改为回文串需要的修改次数
+        let mut cost = vec![vec![0; length]; length];
+
+        for span in 2..=length {
+            for i in 0..=length - span {
+                let j = i + span - 1;
+
+                cost[i][j] = cost[i + 1][j - 1] + if s[i] == s[j] { 0 } else { 1 };
+            }
+        }
+
+        let k = k as usize;
+        // dp[i][j]
+        // 将s[..i]分割为j个回文串需要的修改次数
+        let mut dp = vec![vec![i32::MAX; k + 1]; length + 1];
+
+        for i in 1..=length {
+            for j in 1..=i.min(k) {
+                if j == 1 {
+                    dp[i][j] = cost[0][i - 1];
+                } else {
+                    // 这里k从j - 1开始枚举是因为前面需要进行j - 1次分割时
+                    // 字符串的长度必须大于等于j - 1
+                    for k in j - 1..i {
+                        dp[i][j] = dp[i][j].min(dp[k][j - 1] + cost[k][i - 1]);
+                    }
+                }
+            }
+        }
+
+        dp[length][k]
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_1278() {
+        assert_eq!(1, Solution::palindrome_partition("abc".to_owned(), 2));
+        assert_eq!(0, Solution::palindrome_partition("aabbc".to_owned(), 3));
+        assert_eq!(0, Solution::palindrome_partition("leetcode".to_owned(), 8));
+    }
+}