2 sum problem

<span style="font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);">Today I will introduce the classic 2 sum problem.</span>

Problem

Given a array of distinct numbers, write the a program to find if there exists a pair of number whose sum is X.

Variation

Given a array of distinct numbers, write the a program to find all pairs of number whose sum is X.

I will just give a simple description of the algorithm using matlab.

Version 1

N=10;array = rand(1,N);desired_sum=rand(1)+rand(1);for i=1:N    for j = 1:N        if array(i)+array(j) == desired_sum            fprintf('%d %d',array(i),array(j));        end    endend

The complexity of the algorithm is O(n^2).

Version 2

N=10;array = rand(1,N);desired_sum=rand(1)+rand(1);array=sort(array);for i=1:N    % binary_search here return 0 if it cannot find the element    % or return the index of the element.    [k]=binary_search(array,desired_sum-array(i));    if k ~= 0        fprintf('%d %d',array(i),array(k));    endend

The complexity of the algorithm is O(nlog(n)).

Version 3

N=5;array = [1 2 7 3 6];array=sort(array);i = 1;j = N;x = 9;while(i < j)    if (array(i) + array(j) == x)        fprintf('%d %d\n',array(i),array(j));        i = i + 1;        j= j - 1;    end    if array(i) + array(j) > x         j=j-1;    end    if array(i) + array(j) < x         i=i+1;    endend

This algorithm should take about half time of version 2. Let's demonstrate this algorithm really works. For every i, match j to check if they sum to x. if j is to big, j--. If the sum is less than x, we can assert that the we can stop the search for i. Let's search for i+1!!!!

Version 4

The last and the most efficient version is utilizing hashtable, which is a O(n) algorithm. However, it takes much more memory!!!!