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Solve
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Thoughts (Read these aloud)
- We need to find two numbers that sum to the target value.
- We can use two nested loops to check all possible pairs.
- For each number, we'll check if it sums with any other number to reach the target.
Solution
Solution
1function twoSum(nums, target) {
2 // Check all possible pairs of numbers
3 for (let i = 0; i < nums.length; i++) {
4 for (let j = i + 1; j < nums.length; j++) {
5 // If we find a pair that sums to target
6 if (nums[i] + nums[j] === target) {
7 return [i, j];
8 }
9 }
10 }
11 // No solution found
12 return [];
13}Complexity
- Time Complexity: O(n²)
- Space Complexity: O(1)
What I Changed (Read these aloud)
- The current solution uses nested loops, resulting in O(n²) time complexity.
- We can optimize this by using a hash map to store previously seen numbers.
- This reduces time complexity to O(n) with O(n) space trade-off.
- The hash map lets us find complements in constant time O(1).
Solution
Solution
1function twoSum(nums, target) {
2 // Value -> Index mapping
3 const seen = {};
4
5 for (let i = 0; i < nums.length; i++) {
6 const complement = target - nums[i];
7 // Check if we've seen the complement before
8 if (complement in seen) {
9 return [seen[complement], i];
10 }
11 // Store current number and its index
12 seen[nums[i]] = i;
13 }
14 // No solution found
15 return [];
16}Complexity
- Time Complexity: O(n) - single pass through the array
- Space Complexity: O(n) - storing up to n elements in hash map
- Trade-off: Using more space to achieve better time complexity
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