What if you could find matching pairs instantly without checking every possibility?
Why Two Sum Problem Classic Hash Solution in DSA C#?
Imagine you have a list of prices for items in a store, and you want to find two items that add up to a gift card amount. You try checking every pair one by one.
Checking every pair means looking at each item with every other item. This takes a lot of time if the list is big, and you might miss pairs or repeat checks, making it slow and tiring.
The classic hash solution uses a simple helper (a dictionary) to remember what numbers you've seen. It quickly tells you if the matching number to reach the target is already found, so you don't waste time checking all pairs.
for (int i = 0; i < nums.Length; i++) { for (int j = i + 1; j < nums.Length; j++) { if (nums[i] + nums[j] == target) return new int[] { i, j }; } }
var map = new Dictionary<int, int>(); for (int i = 0; i < nums.Length; i++) { int complement = target - nums[i]; if (map.ContainsKey(complement)) return new int[] { map[complement], i }; map[nums[i]] = i; }
This method lets you find pairs that add up to a target quickly, even in very large lists, saving time and effort.
When shopping online, you want to quickly find two products that together use your full gift card without going over. This solution helps the website find those pairs instantly.
Manual pair checking is slow and repeats work.
Using a dictionary remembers seen numbers for quick lookup.
This approach finds pairs in one pass, making it fast and efficient.
Practice
Two Sum Problem Classic Hash Solution?Solution
Step 1: Understand the role of hash map
The hash map stores numbers and their indices for quick lookup.Step 2: Use hash map to find complement
For each number, check if (target - number) exists in the map to find the pair.Final Answer:
To quickly check if the complement of the current number exists -> Option DQuick Check:
Hash map = fast complement check [OK]
- Thinking hash map sorts the array
- Trying to store sums instead of indices
- Ignoring the complement concept
Solution
Step 1: Identify correct dictionary declaration
In C#,Dictionary<int, int>is used for key-value pairs.Step 2: Check syntax correctness
Dictionary map = new Dictionary(); correctly declares and initializes the dictionary.Final Answer:
Dictionary map = new Dictionary(); -> Option AQuick Check:
C# dictionary syntax = Dictionary map = new Dictionary(); [OK]
- Using List instead of Dictionary
- Using Java's HashMap syntax in C#
- Not specifying generic types
[2, 7, 11, 15] and target 9?
public int[] TwoSum(int[] nums, int target) {
Dictionary map = new Dictionary();
for (int i = 0; i < nums.Length; i++) {
int complement = target - nums[i];
if (map.ContainsKey(complement)) {
return new int[] { map[complement], i };
}
map[nums[i]] = i;
}
return new int[] { -1, -1 };
}Solution
Step 1: Trace the loop and map updates
At i=0, map is empty; add (2,0). At i=1, complement=9-7=2 found in map at index 0.Step 2: Return indices of complement and current
Return [0,1] as the pair sums to 9.Final Answer:
[0, 1] -> Option BQuick Check:
nums[0]+nums[1] = 2+7 = 9 [OK]
- Returning indices in wrong order
- Not updating map correctly
- Returning [-1, -1] incorrectly
public int[] TwoSum(int[] nums, int target) {
Dictionary map = new Dictionary();
for (int i = 0; i <= nums.Length; i++) {
int complement = target - nums[i];
if (map.ContainsKey(complement)) {
return new int[] { map[complement], i };
}
map[nums[i]] = i;
}
return new int[] { -1, -1 };
}Solution
Step 1: Check loop boundary
The loop uses i <= nums.Length which causes index out of range error.Step 2: Correct loop condition
Change to i < nums.Length to avoid accessing invalid index.Final Answer:
Loop condition should be < instead of <= -> Option CQuick Check:
Loop boundary off by one causes error [OK]
- Using <= instead of < in loop
- Confusing data structures
- Ignoring index out of range exceptions
[3, 3, 4, 5] and target 6, what will the Two Sum hash solution return and why?Solution
Step 1: Trace the map and complements
At i=0, map empty, add (3,0). At i=1, complement=6-3=3 found at index 0.Step 2: Return indices of matching pair
Return [0,1] because nums[0]+nums[1]=3+3=6.Final Answer:
[0, 1] because the first two 3's sum to 6 -> Option AQuick Check:
Duplicate values handled correctly [OK]
- Ignoring duplicates in map
- Assuming no pair exists with same number twice
- Returning wrong indices
Ready to prove it?
Pass the test and this pattern moves unknown
