leetcode

Author: ayoubzulfiqar

IntersectionOfTwoArraysII/intersection_of_two_arrays_II.dart

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+/*
+
+
+-* 350. Intersection of Two Arrays II *-
+
+Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order.
+
+
+
+Example 1:
+
+Input: nums1 = [1,2,2,1], nums2 = [2,2]
+Output: [2,2]
+Example 2:
+
+Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
+Output: [4,9]
+Explanation: [9,4] is also accepted.
+
+
+Constraints:
+
+1 <= nums1.length, nums2.length <= 1000
+0 <= nums1[i], nums2[i] <= 1000
+
+
+Follow up:
+
+What if the given array is already sorted? How would you optimize your algorithm?
+What if nums1's size is small compared to nums2's size? Which algorithm is better?
+What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?
+
+*/
+
+import 'dart:collection';
+
+class A {
+  List<int> intersect(List<int> nums1, List<int> nums2) {
+    if (nums1.isEmpty ||
+        nums2.isEmpty ||
+        nums1.length == 0 ||
+        nums2.length == 0) return [];
+    HashMap<int, int> map = HashMap();
+    List<int> result = [];
+    for (int number in nums1)
+      if (map.containsKey(number))
+        map[number] = map[number]! + 1;
+      else
+        map[number] = 1;
+    for (int number in nums2) {
+      if (map.containsKey(number) && map[number]! > 0) {
+        result.add(number);
+        int freq = map[number]!;
+        freq--;
+        map[number] = freq;
+      }
+    }
+    return listToArray(result);
+  }
+
+  List<int> listToArray(List<int> list) {
+    List<int> result = List.filled(list.length, 0);
+    for (int i = 0; i < list.length; i++) {
+      result[i] = list.elementAt(i);
+    }
+    return result;
+  }
+}
+
+class B {
+  List<int> intersect(List<int> nums1, List<int> nums2) {
+    nums1.sort();
+    nums2.sort();
+    List<int> res = [];
+    int left = 0, right = 0;
+    while (left < nums1.length && right < nums2.length) {
+      if (nums1[left] == nums2[right]) {
+        res.add(nums1[left]);
+        left++;
+        right++;
+      } else if (nums1[left] < nums2[right]) {
+        left++;
+      } else
+        right++;
+    }
+    return res;
+  }
+}
+
+class C {
+  List<int> intersect(List<int> nums1, List<int> nums2) {
+    List<int> arr = List.filled(1001, 0);
+    List<int> ans = List.filled(1001, 0);
+    int count = 0;
+    for (int i in nums1) {
+      arr[i]++;
+    }
+    for (int i in nums2) {
+      if (arr[i] > 0) {
+        ans[count++] = i;
+        arr[i]--;
+      }
+    }
+
+    // return List.copyOfRange(ans,0,count);
+    return List.copyRange(ans, 0, [0, count]) as List<int>;
+  }
+}

IntersectionOfTwoArraysII/intersection_of_two_arrays_II.go

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+package main

IntersectionOfTwoArraysII/intersection_of_two_arrays_II.md

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+# 🔥 Intersection of Two Arrays II 🔥 || 3 Approaches || Simple Fast and Easy || with Explanation
+
+## Solution - 1
+
+```dart
+import 'dart:collection';
+
+class Solution {
+  List<int> intersect(List<int> nums1, List<int> nums2) {
+    if (nums1.isEmpty ||
+        nums2.isEmpty ||
+        nums1.length == 0 ||
+        nums2.length == 0) return [];
+    HashMap<int, int> map = HashMap();
+    List<int> result = [];
+    for (int number in nums1)
+      if (map.containsKey(number))
+        map[number] = map[number]! + 1;
+      else
+        map[number] = 1;
+    for (int number in nums2) {
+      if (map.containsKey(number) && map[number]! > 0) {
+        result.add(number);
+        int freq = map[number]!;
+        freq--;
+        map[number] = freq;
+      }
+    }
+    return listToArray(result);
+  }
+
+  List<int> listToArray(List<int> list) {
+    List<int> result = List.filled(list.length, 0);
+    for (int i = 0; i < list.length; i++) {
+      result[i] = list.elementAt(i);
+    }
+    return result;
+  }
+}
+```
+
+## Solution - 2
+
+```dart
+class Solution {
+  List<int> intersect(List<int> nums1, List<int> nums2) {
+    nums1.sort();
+    nums2.sort();
+    List<int> res = [];
+    int left = 0, right = 0;
+    while (left < nums1.length && right < nums2.length) {
+      if (nums1[left] == nums2[right]) {
+        res.add(nums1[left]);
+        left++;
+        right++;
+      } else if (nums1[left] < nums2[right]) {
+        left++;
+      } else
+        right++;
+    }
+    return res;
+  }
+}
+```
+
+## Solution - 3
+
+```dart
+class Solution {
+  List<int> intersect(List<int> nums1, List<int> nums2) {
+    List<int> arr = List.filled(1001, 0);
+    List<int> ans = List.filled(1001, 0);
+    int count = 0;
+    for (int i in nums1) {
+      arr[i]++;
+    }
+    for (int i in nums2) {
+      if (arr[i] > 0) {
+        ans[count++] = i;
+        arr[i]--;
+      }
+    }
+    return List.copyRange(ans, 0, [0, count]) as List<int>;
+  }
+}
+```

README.md

@@ -134,6 +134,7 @@ This repo contain leetcode solution using DART and GO programming language. Most
 - [**344.** Reverse String](ReverseString/reverse_string.dart)
 - [**349.** Intersection of Two Arrays](IntersectionOfTwoArrays/intersection_of_two_arrays.dart)
 - [**223.** Rectangle Area](RectangleArea/rectangle_area.dart)
+- [**350.** Intersection of Two Arrays II](IntersectionOfTwoArraysII/intersection_of_two_arrays_II.dart)
 
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