@@ -13,36 +13,85 @@ void main(List<String> args) {
1313}
1414
1515class Solution {
16- ////! Success Submitted TC: O(n^2) + Sort TC
16+ ////! Success Submitted TC: O(n^2) + Sort TC (O(NLog(N)))
17+ ////! Enhance
1718 List <List <int >> threeSum (List <int > nums) {
1819 nums.sort ();
1920
20- final Map < String , List <int >> result = {} ;
21+ final List < List <int >> result = [] ;
2122 for (int i = 0 ; i < nums.length - 2 ; i++ ) {
23+ /// this line to break because we will not reach to target at the sorted array
2224 if (nums[i] > 0 ) break ;
23- int target = - 1 * nums[i];
25+ 26+ /// To get rid of the duplicate So we will take the first number appeared in the list
27+ /// So if the number appears for the second time we will continue
28+ if (i != 0 && nums[i - 1 ] == nums[i]) continue ;
29+ 30+ int target = 0 - nums[i];
2431 int left = i + 1 ;
2532 int right = nums.length - 1 ;
2633
2734 while (left < right) {
2835 int sum = nums[right] + nums[left];
2936 if (sum == target) {
30- result.putIfAbsent (
31- "${nums [i ]}${nums [left ]}${nums [right ]}" ,
32- () => [nums[i], nums[left], nums[right]],
33- );
34- right-- ;
37+ result.add ([nums[i], nums[left], nums[right]]);
38+ 39+ /// move to the next
3540 left++ ;
41+ right-- ;
42+ 43+ /// Skip all the left duplicates
44+ /// If the new left is the same as the old then
45+ while (left < right && nums[left] == nums[left - 1 ]) {
46+ left++ ;
47+ }
48+ 49+ /// Skip all the right duplicate
50+ while (left < right &&
51+ right < nums.length - 1 &&
52+ nums[right] == nums[right + 1 ]) {
53+ right-- ;
54+ }
3655 } else if (sum > target) {
3756 right-- ;
3857 } else {
3958 left++ ;
4059 }
4160 }
4261 }
43- return result.values. toList () ;
62+ return result;
4463 }
4564
65+ ////! Success Submitted TC: O(n^2) + Sort TC (O(NLog(N)))
66+ // List<List<int>> threeSum(List<int> nums) {
67+ // nums.sort();
68+ 69+ // final Map<String, List<int>> result = {};
70+ // for (int i = 0; i < nums.length - 2; i++) {
71+ // if (nums[i] > 0) break;
72+ // int target = -1 * nums[i];
73+ // int left = i + 1;
74+ // int right = nums.length - 1;
75+ 76+ // while (left < right) {
77+ // int sum = nums[right] + nums[left];
78+ // if (sum == target) {
79+ // result.putIfAbsent(
80+ // "${nums[i]}${nums[left]}${nums[right]}",
81+ // () => [nums[i], nums[left], nums[right]],
82+ // );
83+ // right--;
84+ // left++;
85+ // } else if (sum > target) {
86+ // right--;
87+ // } else {
88+ // left++;
89+ // }
90+ // }
91+ // }
92+ // return result.values.toList();
93+ // }
94+ 4695 ////! Time Limit Solution --> TC: O(n^3) + Sort TC
4796 //// ! Brute Force .
4897 // List<List<int>> threeSum(List<int> nums) {
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