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src/com/dsa/binarysearch/KthMissingNumber.java

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+package com.dsa.binarysearch;
+
+import java.io.BufferedReader;
+import java.io.FileReader;
+import java.io.IOException;
+import java.util.Arrays;
+
+public class KthMissingNumber {
+    public static void main(String[] args) {
+        try (BufferedReader reader = new BufferedReader(new FileReader("input.txt"))) {
+            int n = Integer.parseInt(reader.readLine().trim());
+            String[] numbers = reader.readLine().trim().split(",");
+            int k = Integer.parseInt(reader.readLine().trim());
+            int[] arr = new int[n];
+            for (int i = 0; i < n; i++) {
+                arr[i] = Integer.parseInt(numbers[i].trim());
+            }
+            KthMissingNumber solution = new KthMissingNumber();
+            System.out.println("Input Array: " + Arrays.toString(arr) + ", K: " + k);
+            System.out.println("Kth Missing Number: " + solution.getMissingNumber(arr, k));
+        } catch (IOException e) {
+            throw new RuntimeException(e);
+        }
+    }
+
+    public int getMissingNumber(int[] nums, int k) {
+        int left = 0;
+        int right = nums.length - 1;
+
+        while (left <= right) {
+            int mid = (left + right) / 2;
+            int missingCount = nums[mid] - mid - 1;
+            if (missingCount < k) {
+                left = mid + 1;
+            } else {
+                right = mid - 1;
+            }
+        }
+        /*
+        After the loop, the left pointer indicates the position where the kth missing number would fit.
+        Eg: 2, 3, 4, 7, 11 and k = 5
+        Missing numbers are 1, 5, 6, 8, 9, 10, ...
+        At the end of the loop, left = 4 (pointing to 11)
+        and right = 3 (pointing to 7)
+        The missing number would be between 7 and 11.
+        So to get the 5th missing number, we calculate it as:
+        nums[high] + k - missingCount at high
+        missingCount at index 3 (value 7) = 7 - 3 - 1 = 3
+        So, 7 + 5 - 3 = 9
+        we can rewrite nums[high] + k - (nums[high] - high - 1)
+        which simplifies to nums[high] + k - nums[high] + high + 1
+        which further simplifies to high + k + 1
+        Since high = left - 1, we can rewrite it as:
+        (left - 1) + k + 1 = left + k
+         */
+        return left + k;
+    }
+}