Merge pull request #90 from panchal20bhavin/main

Add solution for Day 19 problem in Java

solutions/day19/README.md

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+# Find Kth Bit in Nth Binary String
+
+## Problem Statement
+Given two positive integers n and k, the binary string Sn is formed as follows:
+
+    S1 = "0"
+    Si = Si - 1 + "1" + reverse(invert(Si - 1)) for i > 1
+
+Where + denotes the concatenation operation, reverse(x) returns the reversed string x, and invert(x) inverts all the bits in x (0 changes to 1 and 1 changes to 0).
+
+## Explanation
+For example, the first four strings in the above sequence are:
+
+    S1 = "0"
+    S2 = "011"
+    S3 = "0111001"
+    S4 = "011100110110001"
+
+Return the kth bit in Sn. It is guaranteed that k is valid for the given n.
+
+## Example
+Example 1:
+
+Input: n = 3, k = 1
+Output: "0"
+Explanation: S3 is "0111001".
+The 1st bit is "0".
+
+Example 2:
+
+Input: n = 4, k = 11
+Output: "1"
+Explanation: S4 is "011100110110001".
+The 11th bit is "1".
+
+### JAVA Approach Explanation
+
+1. Base case: When n = 1, the binary string is "0"
+2. Find the length of the current string Sn, which is 2^n - 1
+3. Find the middle position
+4. If k is the middle position, return '1'
+5. If k is in the first half, find the bit in Sn-1
+6. If k is in the second half, find the bit in Sn-1 and invert it
+
+### Time Complexity ⏱️
+- The overall **time complexity** is `O(n)` because we recursively reduce the problem size with each step by one level.
+
+### Space Complexity 💾
+- The space complexity is `O(n)` due to the recursion stack.

solutions/day19/solutions/day19/solution_java.java

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+package solutions.day19;
+
+public class solution_java {
+	public char findKthBit(int n, int k) {
+		if (n == 1)
+			return '0';
+
+		int length = (1 << n) - 1;
+
+		int mid = length / 2 + 1;
+
+		if (k == mid)
+			return '1';
+
+		if (k < mid)
+			return findKthBit(n - 1, k);
+
+		return findKthBit(n - 1, length - k + 1) == '0' ? '1' : '0';
+	}
+
+}