find pair of numbers in array that add to given sum

Question

Question: Given an unsorted array of positive integers, is it possible to find a pair of integers from that array that sum up to a given sum?

Constraints: This shou

Answer 1

Not guaranteed to be possible; how is the given sum selected?

Example: unsorted array of integers

2, 6, 4, 8, 12, 10

Given sum:

7

??

Answer 2

Naïve double loop printout with O(n x n) performance can be improved to linear O(n) performance using O(n) memory for Hash Table as follows:

void TwoIntegersSum(int[] given, int sum)
{
    Hashtable ht = new Hashtable();
    for (int i = 0; i < given.Length; i++)
        if (ht.Contains(sum - given[i]))
            Console.WriteLine("{0} + {1}", given[i], sum - given[i]);
        else
            ht.Add(given[i], sum - given[i]);
    Console.Read();
}

Answer 3

First you should find reverse array => sum minus actual array then check whether any element from these new array exist in the actual array.

const arr = [0, 1, 2, 6];

const sum = 8;

let isPairExist = arr
  .map(item => sum - item) // [8, 7, 6, 2];
  .find((item, index) => {
    arr.splice(0, 1); // an element should pair with another element
    return arr.find(x => x === item);
  })
  ? true : false;

console.log(isPairExist);

Answer 4

AS @PengOne mentioned it's not possible in general scheme of things. But if you make some restrictions on i/p data.

1. all elements are all + or all -, if not then would need to know range (high, low) and made changes.
2. K, sum of two integers is sparse compared to elements in general.
3. It's okay to destroy i/p array A[N].

Step 1: Move all elements less than SUM to the beginning of array, say in N Passes we have divided array into [0,K] & [K, N-1] such that [0,K] contains elements <= SUM.

Step 2: Since we know bounds (0 to SUM) we can use radix sort.

Step 3: Use binary search on A[K], one good thing is that if we need to find complementary element we need only look half of array A[K]. so in A[k] we iterate over A[ 0 to K/2 + 1] we need to do binary search in A[i to K].

So total appx. time is, N + K + K/2 lg (K) where K is number of elements btw 0 to Sum in i/p A[N]

Note: if you use @PengOne's approach you can do step3 in K. So total time would be N+2K which is definitely O(N)

We do not use any additional memory but destroy the i/p array which is also not bad since it didn't had any ordering to begin with.

Answer 5

First off, sort the array using radix sort. That'll set you back O(kN). Then proceed as @PengOne suggest.

Answer 6

Here's an O(N) algorithm. It relies on an in-place O(N) duplicate removal algorithm, and the existence of a good hash function for the ints in your array.

First, remove all duplicates from the array.

Second, go through the array, and replace each number x with min(x, S-x) where S is the sum you want to reach.

Third, find if there's any duplicates in the array: if 'x' is duplicated, then 'x' and 'S-x' must have occurred in the original array, and you've found your pair.