1545. Find Kth Bit in Nth Binary String: AC

Author: tan-wei

src/solution/mod.rs

@@ -1165,3 +1165,4 @@ mod s1540_can_convert_string_in_k_moves;
 mod s1541_minimum_insertions_to_balance_a_parentheses_string;
 mod s1542_find_longest_awesome_substring;
 mod s1544_make_the_string_great;
+mod s1545_find_kth_bit_in_nth_binary_string;

src/solution/s1545_find_kth_bit_in_nth_binary_string.rs

@@ -0,0 +1,100 @@
+/**
+ * [1545] Find Kth Bit in Nth Binary String
+ *
+ * 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).
+ * For example, the first four strings in the above sequence are:
+ *
+ * 	S1 = "0"
+ * 	S2 = "011"
+ * 	S3 = "0111001"
+ * 	S4 = "011100110110001"
+ *
+ * Return the k^th bit in Sn. It is guaranteed that k is valid for the given n.
+ *  
+ * Example 1:
+ *
+ * Input: n = 3, k = 1
+ * Output: "0"
+ * Explanation: S3 is "<u>0</u>111001".
+ * The 1^st bit is "0".
+ *
+ * Example 2:
+ *
+ * Input: n = 4, k = 11
+ * Output: "1"
+ * Explanation: S4 is "0111001101<u>1</u>0001".
+ * The 11^th bit is "1".
+ *
+ *  
+ * Constraints:
+ *
+ * 	1 <= n <= 20
+ * 	1 <= k <= 2^n - 1
+ *
+ */
+pub struct Solution {}
+
+// problem: https://leetcode.com/problems/find-kth-bit-in-nth-binary-string/
+// discuss: https://leetcode.com/problems/find-kth-bit-in-nth-binary-string/discuss/?currentPage=1&orderBy=most_votes&query=
+
+// submission codes start here
+
+impl Solution {
+    pub fn find_kth_bit(n: i32, k: i32) -> char {
+        let n = n as usize;
+        let k = k as usize;
+        let mut arr = vec![vec![]; n];
+
+        arr[0].push(false);
+        for i in 1..n {
+            let len = arr[i - 1].len();
+            for j in 0..len {
+                let v = arr[i - 1][j];
+                arr[i].push(v);
+            }
+            arr[i].push(true);
+            for j in (0..len).rev() {
+                let v = !arr[i - 1][j];
+                arr[i].push(v);
+            }
+        }
+
+        if arr[n - 1][k - 1] {
+            '1'
+        } else {
+            '0'
+        }
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_1545_example_1() {
+        let n = 3;
+        let k = 1;
+
+        let result = '0';
+
+        assert_eq!(Solution::find_kth_bit(n, k), result);
+    }
+
+    #[test]
+    fn test_1545_example_2() {
+        let n = 4;
+        let k = 11;
+
+        let result = '1';
+
+        assert_eq!(Solution::find_kth_bit(n, k), result);
+    }
+}