leetcode

commit b2a0edc84d24edc9f0e722dbb8d926a14d1efdc4
parent 1c99e5756e5dbda6374f3746deb22d8b83c38e29
Author: AndrewLockVI <andrewlaack1@gmail.com>
Date:   Mon,  5 Jun 2023 12:18:56 -0500

Completed number of good pairs using dart

Diffstat:
A
number-of-good-pairs/number-of-good-pairs.dart
 | 
40
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1 file changed, 40 insertions(+), 0 deletions(-)
diff --git a/number-of-good-pairs/number-of-good-pairs.dart b/number-of-good-pairs/number-of-good-pairs.dart
@@ -0,0 +1,40 @@
+//Given a list of numbers return the number of good pairs
+//that can be returned. A good pair is there n[i] == n[j] and
+//i < j. 
+
+//To solve this I created a hashmap with all of the values and
+//the index they occured at. Then from there I iterated through
+//the map and added the total number of pairs that could be created 
+//from each number.
+
+//The time complexity of this is O(n).
+
+//Time: 219ms Beats: 100%
+//Memory: 144.2MB Beats: 8.57%
+
+class Solution {
+  int numIdenticalPairs(List<int> nums) {
+      Map<int,List<int>> vals = {};
+      int count = 0;
+      for(int i in nums){
+          List<int> current = (vals[i] ?? []);
+          current.add(count);
+          vals[i] = current;
+          count += 1;
+      }
+      int options = 0;
+      for(var entry in vals.entries){
+        List<int> curr_list = entry.value;
+        options += sumAll(curr_list.length - 1); 
+      }
+      return options;
+  }
+  int sumAll(int original){
+    if(original <= 1){
+      return original;
+    }
+    int current = sumAll(original - 1);
+    current = current + original;
+    return current;
+  }
+}