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Commit 4d40192

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Refactor 3Sum solution to use two-pointer technique and update README for clarity
1 parent 6f72b25 commit 4d40192

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2 files changed

+54
-24
lines changed

2 files changed

+54
-24
lines changed

‎README.md

Lines changed: 2 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -82,6 +82,7 @@
8282
- Map In Dart (Hash Map ---- Linked Hash Map ---- Splay Tree Map)
8383
- Hash Map (Not Ordered) -- Hash Table (Ordered)
8484
- Hash Table
85+
8586
A Hash table is defined as a data structure used to insert, look up, and remove key-value pairs quickly.
8687
It operates on the hashing concept,
8788
where each key is translated by a hash function into a distinct index in an array.
@@ -90,4 +91,4 @@
9091

9192
- Find Sum of 2 Numbers in sorted List
9293
1. Brute Force (Bad)
93-
2. 2 Pointers
94+
2. Two Pointers

‎leetcode/15.3sum.dart

Lines changed: 52 additions & 23 deletions
Original file line numberDiff line numberDiff line change
@@ -6,40 +6,72 @@ void main(List<String> args) {
66
final stopwatch = Stopwatch()..start();
77
runTests();
88

9+
print(Solution().threeSum([-4, -2, -2, -2, 0, 1, 2, 2, 2, 3, 3, 4, 4, 6, 6]));
10+
911
stopwatch.stop();
1012
print('Function Execution Time : ${stopwatch.elapsedMicroseconds} micro s');
1113
}
1214

1315
class Solution {
14-
////! Time Limit Solution --> TC: O(n^3)
15-
//// ! Brute Force .
16+
////! Success Submitted TC: O(n^2) + Sort TC
1617
List<List<int>> threeSum(List<int> nums) {
1718
nums.sort();
1819

1920
final Map<String, List<int>> result = {};
2021
for (int i = 0; i < nums.length - 2; i++) {
2122
if (nums[i] > 0) break;
22-
23-
for (int j = i + 1; j < nums.length - 1; j++) {
24-
if (nums[i] + nums[j] > 0) break;
25-
26-
for (int k = j + 1; k < nums.length; k++) {
27-
int sum = nums[i] + nums[j] + nums[k];
28-
29-
if (sum == 0) {
30-
result.putIfAbsent(
31-
"${nums[i]}${nums[j]}${nums[k]}",
32-
() => [nums[i], nums[j], nums[k]],
33-
);
34-
} else if (sum > 0) {
35-
break;
36-
}
23+
int target = -1 * nums[i];
24+
int left = i + 1;
25+
int right = nums.length - 1;
26+
27+
while (left < right) {
28+
int sum = nums[right] + nums[left];
29+
if (sum == target) {
30+
result.putIfAbsent(
31+
"${nums[i]}${nums[left]}${nums[right]}",
32+
() => [nums[i], nums[left], nums[right]],
33+
);
34+
right--;
35+
left++;
36+
} else if (sum > target) {
37+
right--;
38+
} else {
39+
left++;
3740
}
3841
}
3942
}
4043
return result.values.toList();
4144
}
4245

46+
////! Time Limit Solution --> TC: O(n^3) + Sort TC
47+
//// ! Brute Force .
48+
// List<List<int>> threeSum(List<int> nums) {
49+
// nums.sort();
50+
51+
// final Map<String, List<int>> result = {};
52+
// for (int i = 0; i < nums.length - 2; i++) {
53+
// if (nums[i] > 0) break;
54+
55+
// for (int j = i + 1; j < nums.length - 1; j++) {
56+
// if (nums[i] + nums[j] > 0) break;
57+
58+
// for (int k = j + 1; k < nums.length; k++) {
59+
// int sum = nums[i] + nums[j] + nums[k];
60+
61+
// if (sum == 0) {
62+
// result.putIfAbsent(
63+
// "${nums[i]}${nums[j]}${nums[k]}",
64+
// () => [nums[i], nums[j], nums[k]],
65+
// );
66+
// } else if (sum > 0) {
67+
// break;
68+
// }
69+
// }
70+
// }
71+
// }
72+
// return result.values.toList();
73+
// }
74+
4375
/// Wrong Solution
4476
/// Will make a duplicate
4577
// List<List<int>> threeSum(List<int> nums) {
@@ -166,8 +198,8 @@ void runTests() {
166198
test('All same number except one: [1,1,1,1,1,-2] → [[1,1,-2]]', () {
167199
expect(
168200
s.threeSum([1, 1, 1, 1, 1, -2]),
169-
equals([
170-
[1, 1, -2]
201+
unorderedMatches([
202+
[1, 1, -2]..sort()
171203
]));
172204
});
173205

@@ -194,10 +226,7 @@ void runTests() {
194226
];
195227
final result = s.threeSum(nums);
196228

197-
// Verify all expected triplets are present
198-
for (final triplet in expected) {
199-
expect(result, contains(triplet));
200-
}
229+
expect(result, unorderedMatches(expected));
201230

202231
// Verify no extra triplets are present
203232
expect(result.length, equals(expected.length));

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