added solution to kth missing positive number

Author: Ayon95

problems/binary-search/easy/kth-missing-pos-num/README.md

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+# Kth missing positive number
+
+| #   | Difficulty | Tag(s)                  | Link                                                                       |
+| --- | ---------- | ----------------------- | -------------------------------------------------------------------------- |
+| 42  | easy       | Hash map, binary search | [View problem](https://leetcode.com/problems/kth-missing-positive-number/) |
+
+## Approaches
+
+- linear search using hash map
+- binary search
+
+### Linear search
+
+- use a hash map to store which numbers are present
+- loop through 1 until 2000
+  - problem states that the largest possible number in the input array is 1000
+  - and the largest possible value of k is 1000
+  - so, in the worst case, the 1000th missing number will be 2000
+- whenever we encounter a missing number, we decrement `k`
+- when `k` becomes 0, we have found the kth missing number
+- O(m + n) time complexity
+  - m is the size of the input array
+  - n is 2000
+- O(m) space complexity
+
+### Binary search
+
+- this approach involves finding the count of missing numbers at an index, and comparing it with `k`
+  - how many numbers are missing until the number at that index?
+- we need to find the first index at which `missingNumsCount >= k`
+- use two pointers to track the start and end positions of the current problem space
+- calculate middle index and find `missingNumsCount` at middle index
+- if `missingNumsCount < k`, then the kth missing number must be to the right of `middle`
+- if `missingNumsCount >= k && missingNumsCountJustBeforeMid >= k`, then the kth missing number must be to the left of `middle`
+- O(log n) time complexity
+- O(1) space complexity
+
+![](./kth-missing-pos-number.png)

problems/binary-search/easy/kth-missing-pos-num/kth-missing-pos-number.png

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+/*
+Linear search using hash map
+
+*/
+
+function findKthPositive(arr, k) {
+	const numsPresent = {};
+
+	for (let i = 0; i < arr.length; i++) {
+		numsPresent[arr[i]] = true;
+	}
+
+	for (let i = 1; i <= 2000; i++) {
+		if (!numsPresent[i]) k--;
+		if (k === 0) return i;
+	}
+}
+
+console.log(findKthPositive([2, 3, 4, 7, 11], 5));
+console.log(findKthPositive([1, 2, 3, 4], 2));
+console.log(findKthPositive([2, 3, 4], 1));
+console.log(findKthPositive([5, 6, 7, 8, 9], 9));
+
+/*
+Binary search
+*/
+
+function findKthPositiveV2(arr, k) {
+	let start = 0;
+	let end = arr.length - 1;
+
+	while (start <= end) {
+		const midIndex = Math.floor((end - start) / 2) + start;
+		const missingNumsCountAtMidIndex = arr[midIndex] - (midIndex + 1);
+		const missingNumsCountJustBeforeMidIndex = arr[midIndex - 1] - (midIndex - 1 + 1);
+		if (missingNumsCountAtMidIndex < k) {
+			start = midIndex + 1;
+		} else if (missingNumsCountAtMidIndex >= k && missingNumsCountJustBeforeMidIndex >= k) {
+			end = midIndex - 1;
+		} else {
+			return arr[midIndex] - (missingNumsCountAtMidIndex - k + 1);
+		}
+	}
+
+	return arr.length + k;
+}
+
+console.log('---------');
+console.log(findKthPositiveV2([2, 3, 4, 7, 11], 5));
+console.log(findKthPositiveV2([1, 2, 3, 4], 2));
+console.log(findKthPositiveV2([2, 3, 4], 1));
+console.log(findKthPositiveV2([5, 6, 7, 8, 9], 9));