Added code for leetcode daily 15 october 2022

String Compression II.java

@@ -0,0 +1,30 @@
+// We define the state dp[i][k]: the minimum length for s.substring(0, i+1) with at most k deletion.
+// For each char s[i], we can either keep it or delete it.
+// If delete, dp[i][j]=dp[i-1][j-1].
+// If keep, we delete at most j chars in s.substring(0, i+1) that are different from s[i].
+
+class Solution {
+    public int getLengthOfOptimalCompression(String s, int k) {
+        // dp[i][k]: the minimum length for s[:i] with at most k deletion.
+        int n = s.length();
+        int[][] dp = new int[110][110];
+        for (int i = 0; i <= n; i++) for (int j = 0; j <= n; j++) dp[i][j] = 9999;
+		// for (int[] i : dp) Arrays.fill(i, n); // this is a bit slower (100ms)
+        dp[0][0] = 0;
+        for(int i = 1; i <= n; i++) {
+            for(int j = 0; j <= k; j++) {                
+                int cnt = 0, del = 0;
+                for(int l = i; l >= 1; l--) { // keep s[i], concat the same, remove the different
+                    if(s.charAt(l - 1) == s.charAt(i - 1)) cnt++;
+                    else del++;
+                    if(j - del >= 0) 
+                        dp[i][j] = Math.min(dp[i][j], 
+                                            dp[l-1][j-del] + 1 + (cnt >= 100 ? 3 : cnt >= 10 ? 2 : cnt >= 2 ? 1: 0));
+                }
+                if (j > 0) // delete s[i]
+                    dp[i][j] = Math.min(dp[i][j], dp[i-1][j-1]);
+            }
+        }
+        return dp[n][k];
+    }
+}