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
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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
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function findPairs(a, k) {
var pairs = []; // We will store the pairs here

// Loop once through all the elements in the array
for (var current = 0; current < a.length; current++) {
var expected = k - a[current];

// 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]]);
} else {
// If we didn't find the element necessary to
// array
}
}

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
``````