Algorithms

1. Sort the array in ascending order
2. let int i=0 and int j=n-1
3. if i>=j then search is over
4. calculate sum=array[i]+array[j]
5.   if sum<x, increment i and goto (3)
6.   if sum>x, decrement j and goto (3)
7.   since sum==x, solution found (either stop; or increment i, decrement j and goto 3)

in the name of god<br />
hello and thanks for your answers.<br />
it is the code i wrote in c++:<br />
<br />
int main(){<br />
  //here we declare variables<br />
  cin<<x;<br />
  //here first we sort the array with O(nlg(n))<br />
  mergeSort(Arbegin,Arend);//begin and end of array<br />
  //now we have a sorted array<br />
  int i=0;<br />
  //the order of below algorithm in worst case is O(n) so the order is :O(n)<br />
  while(i<j){<br />
    int sum=A[i]+A[j];<br />
    if(x < sum){<br />
      i++;<br />
    }<br />
    if(x < sum){<br />
      j--;<br />
    }<br />
    if(sum==x){<br />
      a=i;<br />
      b=j;<br />
      break;<br />
    }<br />
  }<br />
}<br />
valhamdolelah.

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goal    # me
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M:
   if there is wall to the left:
       if there is no wall to my diagonal left:
              move there
              rotate 90º left
              M()
       else:
              if there is wall infront:
                       rotate 90º to the right
                       M()
              else:
                       move there.
    else:
        throw exception

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