added solution to intersection of two arrays

Author: Ayon95

problems/hash-map/easy/intersection-of-two-arrs/README.md

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+# Intersection of two arrays
+
+| #   | Difficulty | Tag(s)                                         | Link                                                                      |
+| --- | ---------- | ---------------------------------------------- | ------------------------------------------------------------------------- |
+| 40  | Easy       | Hash map, two pointers, sorting, binary search | [View problem](https://leetcode.com/problems/intersection-of-two-arrays/) |
+
+## Approaches
+
+- hash map
+- two pointers
+- binary search
+
+### Hash map
+
+- we have to find the numbers that are present in both arrays
+- we can include in the result array only one occurrence of a common number
+- use a hash map to store unique numbers present in arr1
+- loop through arr2 and check if it contains unique numbers of arr1
+  - whenever we find a match, we add that number to the result arr, and delete it from the hash map
+- O(m + n) time complexity
+- O(k + l) space complexity
+  - k is the count of unique numbers in arr1
+  - l is the size of the result array
+
+### Two pointers
+
+- sort the two arrays in ascending order
+- use a set to store common unique numbers present in both arrays
+- `i` is used to track numbers of arr1
+- `j` is used to track numbers of arr2
+- keep iterating as long as both i and j are within the bounds of their respective arrays
+- if `arr1[i] < arr2[j]`, then increment i so that it points to a bigger number in arr1
+- if `arr2[j] < arr1[i]`, then increment j so that it points to a bigger number in arr2
+- if `arr1[i] === arr2[j]`
+  - add the number to result array
+  - increment both i and j
+- O(mlogm + nlogn) time complexity
+- O(k + l) space complexity
+  - k is the size of the set
+  - l is the size of the result array
+
+### Binary search
+
+- sort the arrays in ascending order
+- use a set to store common unique numbers present in both arrays
+- loop through arr1 and use each number as the target when you perform binary search on arr2
+  - if the target is present in arr2, then we add that number to the set
+- O(mlogm + nlogn) time complexity
+- O(k + l) space complexity

problems/hash-map/easy/intersection-of-two-arrs/intersectionOfTwoArrs.js

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+// Using a hash map
+function intersection(nums1, nums2) {
+	const uniqueNums = {};
+	const result = [];
+
+	for (let i = 0; i < nums1.length; i++) {
+		const num = nums1[i];
+		if (uniqueNums[num] === undefined) {
+			uniqueNums[num] = true;
+		}
+	}
+
+	for (let i = 0; i < nums2.length; i++) {
+		const num = nums2[i];
+		if (uniqueNums[num] !== undefined) {
+			result.push(num);
+			delete uniqueNums[num];
+		}
+	}
+
+	return result;
+}
+
+console.log(intersection([1, 2, 2, 1], [2, 2]));
+console.log(intersection([4, 9, 5], [9, 4, 9, 8, 4]));
+console.log(intersection([6, 2, 5], [1, 2, 10, 6, 11, 5]));
+
+// Two pointers
+function intersectionV2(nums1, nums2) {
+	nums1.sort((a, b) => a - b);
+	nums2.sort((a, b) => a - b);
+
+	const uniqueNums = new Set();
+	let i = 0;
+	let j = 0;
+
+	while (i < nums1.length && j < nums2.length) {
+		const num1 = nums1[i];
+		const num2 = nums2[j];
+
+		if (num1 < num2) i++;
+		else if (num2 < num1) j++;
+		else {
+			uniqueNums.add(num1);
+			i++;
+			j++;
+		}
+	}
+
+	return [...uniqueNums];
+}
+
+console.log('---------');
+console.log(intersectionV2([1, 2, 2, 1], [2, 2]));
+console.log(intersectionV2([4, 9, 5], [9, 4, 9, 8, 4]));
+console.log(intersectionV2([6, 2, 5], [1, 2, 10, 6, 11, 5]));
+
+// Binary search
+function intersectionV3(nums1, nums2) {
+	nums1.sort((a, b) => a - b);
+	nums2.sort((a, b) => a - b);
+
+	const uniqueNums = new Set();
+
+	for (let i = 0; i < nums1.length; i++) {
+		const target = nums1[i];
+		if (numExists(nums2, target)) uniqueNums.add(target);
+	}
+
+	return [...uniqueNums];
+
+	function numExists(arr, target) {
+		let start = 0;
+		let end = arr.length - 1;
+
+		while (start <= end) {
+			const middleIndex = Math.floor((end - start) / 2) + start;
+			if (arr[middleIndex] === target) return true;
+			else if (arr[middleIndex] < target) start = middleIndex + 1;
+			else end = middleIndex - 1;
+		}
+
+		return false;
+	}
+}
+
+console.log('---------');
+console.log(intersectionV3([1, 2, 2, 1], [2, 2]));
+console.log(intersectionV3([4, 9, 5], [9, 4, 9, 8, 4]));
+console.log(intersectionV3([6, 2, 5], [1, 2, 10, 6, 11, 5]));