@@ -10,63 +10,89 @@ void main(List<String> args) {
1010}
1111
1212class Solution {
13- ////! My Solution and solved the problem
14- ////! Hash Map
13+ ////! Enhance My Hash Map solution
14+ ////! AI help
15+ ////! Best Time Solution
1516 int findPairs (List <int > nums, int k) {
16- if (nums.length <= 1 ) return 0 ;
17+ if (nums.isEmpty || k < 0 ) return 0 ;
1718
18- final Map <int , Set <int >> map = {};
1919 int pairsCount = 0 ;
20+ final Map <int , int > map = {};
2021
2122 for (int num in nums) {
22- /// Both Negative Number and positive number are calculated and counted
23- if (map[num ]? .length == 2 ) {
24- continue ;
25- }
23+ map[num ] = (map[num ] ?? 0 ) + 1 ;
24+ }
2625
27- /// Postive number == negative number
26+ final mapKeys = map.keys;
27+ for (int num in mapKeys) {
2828 if (k == 0 ) {
29- if (map[num ] == null ) {
30- map[num ] = {};
31- } else if (map[num ]? .isEmpty == true ) {
32- map[num ]? .add (num );
33- pairsCount++ ;
34- }
35- continue ;
36- }
37- 38- if (map[num ] == null ) {
39- map[num ] = {};
40- }
41- 42- /// Find positive other number i - k = positive number
43- int positiveNumber = num - k;
44- 45- /// if we can add the current num to the positive number set
46- /// then the positive number appeared before and its set isn't contains the current num
47- if (map[positiveNumber] != null &&
48- map[positiveNumber]? .contains (num ) == false &&
49- map[num ]? .contains (positiveNumber) == false ) {
50- map[positiveNumber]? .add (num );
51- map[num ]? .add (positiveNumber);
52- pairsCount++ ;
53- }
54- 55- /// Find negative other number
56- int negativeNumber = num + k;
57- 58- /// if we can add the current num to the negative number set
59- /// then the negative number appeared before and its set isn't contains the current num
60- if (map[negativeNumber] != null &&
61- map[negativeNumber]? .contains (num ) == false &&
62- map[num ]? .contains (negativeNumber) == false ) {
63- map[negativeNumber]? .add (num );
64- map[num ]? .add (negativeNumber);
65- pairsCount++ ;
29+ if (map[num ]! > 1 ) pairsCount++ ;
30+ } else {
31+ int otherNumber = num + k;
32+ if (map.containsKey (otherNumber)) pairsCount++ ;
6633 }
6734 }
6835 return pairsCount;
6936 }
37+ 38+ ////!
39+ ////! My Solution and solved the problem
40+ ////! Hash Map
41+ // int findPairs(List<int> nums, int k) {
42+ // if (nums.length <= 1) return 0;
43+ 44+ // final Map<int, Set<int>> map = {};
45+ // int pairsCount = 0;
46+ 47+ // for (int num in nums) {
48+ // /// Both Negative Number and positive number are calculated and counted
49+ // if (map[num]?.length == 2) {
50+ // continue;
51+ // }
52+ 53+ // /// Postive number == negative number
54+ // if (k == 0) {
55+ // if (map[num] == null) {
56+ // map[num] = {};
57+ // } else if (map[num]?.isEmpty == true) {
58+ // map[num]?.add(num);
59+ // pairsCount++;
60+ // }
61+ // continue;
62+ // }
63+ 64+ // if (map[num] == null) {
65+ // map[num] = {};
66+ // }
67+ 68+ // /// Find positive other number i - k = positive number
69+ // int positiveNumber = num - k;
70+ 71+ // /// if we can add the current num to the positive number set
72+ // /// then the positive number appeared before and its set isn't contains the current num
73+ // if (map[positiveNumber] != null &&
74+ // map[positiveNumber]?.contains(num) == false &&
75+ // map[num]?.contains(positiveNumber) == false) {
76+ // map[positiveNumber]?.add(num);
77+ // map[num]?.add(positiveNumber);
78+ // pairsCount++;
79+ // }
80+ 81+ // /// Find negative other number
82+ // int negativeNumber = num + k;
83+ 84+ // /// if we can add the current num to the negative number set
85+ // /// then the negative number appeared before and its set isn't contains the current num
86+ // if (map[negativeNumber] != null &&
87+ // map[negativeNumber]?.contains(num) == false &&
88+ // map[num]?.contains(negativeNumber) == false) {
89+ // map[negativeNumber]?.add(num);
90+ // map[num]?.add(negativeNumber);
91+ // pairsCount++;
92+ // }
93+ // }
94+ // return pairsCount;
95+ // }
7096}
7197
7298void runTests () {
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