This is the first post on a series where I will be resolving some common coding interview questions.

## The question

Given an integer array, output all pairs that sum up to a specific value k.

## My solution

I like to start by creating a test for my solution:

1
2
3

var arr = [1, 4, 2, 6, 8, 3, 9, 0, 7];
var res = findPairs(arr, 7);
// I expect res to be [[1, 6], [4, 3], [0, 7]]

Now, lets think about how to solve this problem. In my test I only expect the pair to appear once, this means that once we find a pair both elements can be removed from the array. With this assumption this is what I would do:

- Get the first element and then compare with all the elements until I find one that sums to k.
- If no match is found then remove the first element.
- If a match is found then add it to the answer and remove both elements from the array.
- Do the same thing for the second element until there are no more elements.

There are two main problems I see with this solution:

- It goes through the elements of the array multiple times, so the efficiency would be O(n^2), which is pretty much as bad as it gets.
- You can’t really remove elements from an array so we would have to settle with assigning null to that array position.

Due to the inefficiency of the previous method I decided to try something different:

- Sort the array (Using merge sort the efficiency would be O(nlogn).
- Start from first element and then start comparing with the last element in reverse order.
- When you find a match you add it to the result.
- When you find that the sum becomes lower than k then you move the left pointer one position to the right.
- Repeat until done.

This is the best solution I could come up with(nlogn) but it turns out this is not the most efficient way to do it. With the help of a hash table you can solve this problem in O(n):

- Start from the left of the array.
- Check if the value we need for it to sum k is in the hash table.
- Add this value to the hash table.

## The best solution

This solution will have to go through all the elements only once and since inserting and reading from a hash table is constant, it gives us an efficiency of O(n)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41

function findPairs(a, k) {
var pairs = []; // We will store the pairs here
var read = {}; // Hashtable with read elements
// Loop once through all the elements in the array
for (var current = 0; current < a.length; current++) {
var expected = k - a[current];
if (read[expected]) {
// If the sum for the current element is found
// then we add the pair to the result and
// delete the element we used from the read
// hashtable
pairs.push([expected, a[current]]);
delete read[expected];
} else {
// If we didn't find the element necessary to
// sum k then we add current element to read
// array
read[a[current]] = true;
}
}
return pairs;
}
function test() {
var arr = [1, 4, 2, 6, 8, 3, 9, 0, 7];
var res = findPairs(arr, 7);
// This is not the most efficient way to compare arrays
// but I didn't want to bloat this exercise with more code
var expected = [[1, 6], [4, 3], [0, 7]];
if (expected.toString() === res.toString()) {
console.log('success');
} else {
console.log('failure');
}
}
test(); // This prints success

application_design

computer_science

algorithms

javascript

programming

]