Update README_EN.md

solution/1500-1599/1531.String Compression II/README.md

@@ -65,7 +65,54 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int getLengthOfOptimalCompression(String s, int k) {
+        // dp[i][k] := the length of the optimal compression of s[i..n) with at most
+        // k deletion
+        dp = new int[s.length()][k + 1];
+        Arrays.stream(dp).forEach(A -> Arrays.fill(A, K_MAX));
+        return compression(s, 0, k);
+    }
+
+    private static final int K_MAX = 101;
+    private int[][] dp;
+
+    private int compression(final String s, int i, int k) {
+        if (k < 0) {
+            return K_MAX;
+        }
+        if (i == s.length() || s.length() - i <= k) {
+            return 0;
+        }
+        if (dp[i][k] != K_MAX) {
+            return dp[i][k];
+        }
+        int maxFreq = 0;
+        int[] count = new int[128];
+        // Make letters in s[i..j] be the same.
+        // Keep the letter that has the maximum frequency in this range and remove
+        // the other letters.
+        for (int j = i; j < s.length(); ++j) {
+            maxFreq = Math.max(maxFreq, ++count[s.charAt(j)]);
+            dp[i][k] = Math.min(dp[i][k], getLength(maxFreq) + compression(s, j + 1, k - (j - i + 1 - maxFreq)));
+        }
+        return dp[i][k];
+    }
+
+    // Returns the length to compress `maxFreq`.
+    private int getLength(int maxFreq) {
+        if (maxFreq == 1) {
+            return 1; // c
+        }
+        if (maxFreq < 10) {
+            return 2; // [1-9]c
+        }
+        if (maxFreq < 100) {
+            return 3; // [1-9][0-9]c
+        }
+        return 4;   // [1-9][0-9][0-9]c
+    }
+}
 ```
 
 ### **...**

solution/1500-1599/1531.String Compression II/README_EN.md

@@ -58,7 +58,54 @@
 ### **Java**
 
 ```java
-
+class Solution {
+    public int getLengthOfOptimalCompression(String s, int k) {
+        // dp[i][k] := the length of the optimal compression of s[i..n) with at most
+        // k deletion
+        dp = new int[s.length()][k + 1];
+        Arrays.stream(dp).forEach(A -> Arrays.fill(A, K_MAX));
+        return compression(s, 0, k);
+    }
+
+    private static final int K_MAX = 101;
+    private int[][] dp;
+
+    private int compression(final String s, int i, int k) {
+        if (k < 0) {
+            return K_MAX;
+        }
+        if (i == s.length() || s.length() - i <= k) {
+            return 0;
+        }
+        if (dp[i][k] != K_MAX) {
+            return dp[i][k];
+        }
+        int maxFreq = 0;
+        int[] count = new int[128];
+        // Make letters in s[i..j] be the same.
+        // Keep the letter that has the maximum frequency in this range and remove
+        // the other letters.
+        for (int j = i; j < s.length(); ++j) {
+            maxFreq = Math.max(maxFreq, ++count[s.charAt(j)]);
+            dp[i][k] = Math.min(dp[i][k], getLength(maxFreq) + compression(s, j + 1, k - (j - i + 1 - maxFreq)));
+        }
+        return dp[i][k];
+    }
+
+    // Returns the length to compress `maxFreq`.
+    private int getLength(int maxFreq) {
+        if (maxFreq == 1) {
+            return 1; // c
+        }
+        if (maxFreq < 10) {
+            return 2; // [1-9]c
+        }
+        if (maxFreq < 100) {
+            return 3; // [1-9][0-9]c
+        }
+        return 4;   // [1-9][0-9][0-9]c
+    }
+}
 ```
 
 ### **...**