Solved problems

Author: gouthampradhan

README.md

@@ -286,6 +286,7 @@ My accepted leetcode solutions to some of the common interview problems.
 - [Nth Magical Number](problems/src/math/NthMagicalNumber.java) (Hard)
 - [Squirrel Simulation](problems/src/math/SquirrelSimulation.java) (Medium)
 - [Projection Area of 3D Shapes](problems/src/math/ProjectionAreaOf3DShapes.java) (Easy)
+- [Decoded String at Index](problems/src/math/DecodedStringAtIndex.java) (Medium)
 
 #### [Reservoir Sampling](problems/src/reservoir_sampling)
 

problems/src/math/DecodedStringAtIndex.java

@@ -0,0 +1,70 @@
+package math;
+
+/**
+ * Created by gouthamvidyapradhan on 21/07/2019 An encoded string S is given. To find and write the
+ * decoded string to a tape, the encoded string is read one character at a time and the following
+ * steps are taken:
+ *
+ * <p>If the character read is a letter, that letter is written onto the tape. If the character read
+ * is a digit (say d), the entire current tape is repeatedly written d-1 more times in total. Now
+ * for some encoded string S, and an index K, find and return the K-th letter (1 indexed) in the
+ * decoded string.
+ *
+ * <p>Example 1:
+ *
+ * <p>Input: S = "leet2code3", K = 10 Output: "o" Explanation: The decoded string is
+ * "leetleetcodeleetleetcodeleetleetcode". The 10th letter in the string is "o". Example 2:
+ *
+ * <p>Input: S = "ha22", K = 5 Output: "h" Explanation: The decoded string is "hahahaha". The 5th
+ * letter is "h". Example 3:
+ *
+ * <p>Input: S = "a2345678999999999999999", K = 1 Output: "a" Explanation: The decoded string is "a"
+ * repeated 8301530446056247680 times. The 1st letter is "a".
+ *
+ * <p>Note:
+ *
+ * <p>2 <= S.length <= 100 S will only contain lowercase letters and digits 2 through 9. S starts
+ * with a letter. 1 <= K <= 10^9 The decoded string is guaranteed to have less than 2^63 letters.
+ *
+ * <p>Solution: General idea is as shown below example: If S = "leet2" and K = 6 the answer is "e"
+ * which is same as finding answer for K = 2. As soon as the product exceeds the total value of K as
+ * in this case the product of 4 (leet) x 2 is 8 and 8 clearly exceeds 6 therefore we can reduce K
+ * to 8 - 6 = 2 and start from the beginning once again. Repeat the same process until we reach the
+ * answer.
+ */
+public class DecodedStringAtIndex {
+  public static void main(String[] args) {
+    System.out.println(
+        new DecodedStringAtIndex().decodeAtIndex("a2345678999999999999999", 1000000000));
+  }
+
+  public String decodeAtIndex(String S, int K) {
+    long product = 0;
+    char lastC = S.charAt(0);
+    for (int i = 0, l = S.length(); i < l; ) {
+      char c = S.charAt(i);
+      if (Character.isLetter(c)) {
+        lastC = c;
+        product++;
+        i++;
+        if (K == product) break;
+      } else {
+        long temp = (product * Integer.parseInt(String.valueOf(c)));
+        if (temp == K) break;
+        else {
+          if (temp > K) {
+            long x = (K / product);
+            if ((product * x) == K) break;
+            K -= (product * x);
+            i = 0;
+            product = 0;
+          } else {
+            product = temp;
+            i++;
+          }
+        }
+      }
+    }
+    return String.valueOf(lastC);
+  }
+}