### hollow's blog

By hollow, 8 years ago, ,

In many problems i see the tag "two pointers". I assume its a technique. I dont have an idea on it can anybody give any type of reference. Thanks.

• +10

 » 8 years ago, # | ← Rev. 5 →   +10 I will try to describe this method on example of a problem. Given two arrays (A and B) sorted in ascending order, and an integer x. we need to find i and j, such that a[i] + b[j] is equal to X. i and j our pointers, at first step i is points to the first element of a, and j points to the last element of b. i = 0; j = b.size() - 1; move first pointer one by one through the array a, and correct position of the second pointer if needed while (i < a.size()) { while(a[i] + b[j] > X && j > 0) j--; if (a[i] + b[j] == X) writeAnswer(i, j); i++; } Also, there was the same question in Russian, maybe it can helps. link to google-translated version http://codeforces.ru/blog/entry/4136
•  » » 8 years ago, # ^ |   +3 In addition to post of Alias:The feature of this method is that solutions are O(n) (n=b.size() using his terminology) despite of possibility of decreasing second pointer with O(n) while 1 iteration of outer cycle made. It's right because every pointer will change only O(n) times.
•  » » » 3 years ago, # ^ |   0 amm... should it not be O(a.size()+b.size()) in worst case like if x=101 , a=1 2 5 7 8 100 , b= 1 2 3 4 47
•  » » » » 2 years ago, # ^ |   0 Do you need to be that harsh with him?
•  » » » » » 18 months ago, # ^ |   0 ^__^ :p Being Harsh
•  » » 7 years ago, # ^ |   0 and what if we have to find all the pairs that sums up to X ?
•  » » » 7 years ago, # ^ | ← Rev. 3 →   0 the same approach works as well, just continue moving your pointers and count the number of pairs.actually the "writeAnswer" function in Alias code will be called for all such pairs
•  » » » » 7 years ago, # ^ |   0 I guess in that case we need to do this : writeAnswer(i++,j--), right ?
•  » » » » » 7 years ago, # ^ | ← Rev. 3 →   0 no. this code works fine: #include using namespace std; int n,a[100001],b[100001]; void writeAnswer(int ii,int jj){ cout<>n>>X; for(int i=0;i>a[i]; } for(int i=0;i>b[i]; } int i=0,j=n-1; while (i < n) { while(a[i] + b[j] > X && j > 0) j--; if (a[i] + b[j] == X) writeAnswer(i, j); i++; } while(1); //pause } this code may not work when the array values is not unique, anyway the main point is to understand two pointers technique
•  » » » » » » 7 years ago, # ^ |   0 i guess writeAnswer(i++,j--); thingy would also work provided all elements in the array are unique.
•  » » » » » » » 7 years ago, # ^ |   0 I don't think so, try it up and see why :)
•  » » » » » » » » 7 years ago, # ^ | ← Rev. 2 →   0 can you think of a test case where my idea fails. Because for almost all inputs that i have, its working fine !
•  » » » » » » » » » 7 years ago, # ^ | ← Rev. 4 →   0 5 6 1 2 3 4 5 1 2 3 4 5 
•  » » » » » » » » » 7 years ago, # ^ |   0 well, it gave me as expected, it gave me two pairs : 1,5 2,4hence passed !
•  » » » » » » » » » 7 years ago, # ^ |   0 correct output: 1 5 2 4 3 3 4 2 5 1 
•  » » » » » » » » » 3 years ago, # ^ |   0 This case worked with me output 1,5 2,4 3,3 4,2 5,1I think you mean this case which failed with me 5 6 1 2 3 4 5 1 2 3 5 5
•  » » » » » » 4 years ago, # ^ | ← Rev. 2 →   0 why this code will not work if array elements not unique
•  » » 7 years ago, # ^ |   0 i came up with this problem : http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=515it reminded me of almost the same algorithm as used here except it needs perhaps more than two pointers. can you help me in figuring out how to approach above mentioned problem given in the link.
•  » » » 7 years ago, # ^ |   0 it is just a bruteforce. There is total 2^12 subsets, try all.
•  » » » » 7 years ago, # ^ |   0 is there any more efficient algorithm if input size is much larger than this complete search ?
•  » » » » » 7 years ago, # ^ | ← Rev. 2 →   0 answer can be vey big, so there is no big difference. there is O(sum * n * n + total_size_of_answer) dfs like approach. vertex is a pair (sum, last_used_number). in this graph you have to find all paths between two vertices, but the anwser can be very big as it mentioned above.
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 Is there any condition for arrays 'a' and 'b', like all the elements will be unique or others? If not, then how your given code will give correct ans under the following condition: a.size() = 6; X = 6; a[] = {1, 2, 3, 4, 5, 6}; b[] = {1, 2, 3, 5, 5, 6}; 
•  » » » 4 years ago, # ^ |   +8 Why don't you try it for yourself?If you mean "why the possible answer i=0, j=3 is not displayed", well, be more specific next time, and it's because the original problem is to find any one pair of indices. What if the problem asks to find all such pairs? The maximum running time is O(mn) instead of O(m+n), e.g., when X=2 and all elements of a and b are equal to 1. So we can actually use the straightforward algorithm (for i, for j, check a[i]+b[j]) instead of the two pointers method. The above bound can be improved to O(s) where s is the number of pairs in the answer. You can still patch the algorithm (while equal, go up and down) to make it output all s pairs in O(s). If you need the number of pairs and not the pairs themselves, it can also be done in O(m+n) by maintaining highest and lowest possible values of j instead of a single variable.
•  » » » » 4 years ago, # ^ | ← Rev. 2 →   0 h.m.m...m..I was wrong!!! Thanks Gassa !!!
•  » » 18 months ago, # ^ |   0 And since X - a[ i ] is independent of the second pointer j, your two-pointers example can be rewritten as follows. #include using namespace std; int main() { int m, n, X; cin >> m >> n >> X; int a[ m + 1 ], b[ n + 1 ]; for( int i = 1; i <= m; cin >> a[ i++ ] ); for( int j = 1; j <= n; cin >> b[ j++ ] ); for( int i = 1, j = n; i <= m; i++ ) { int Y = X - a[ i ]; while( j >= 1 and b[ j ] > Y ) j--; if( j >= 1 and b[ j ] == Y ) cout << i << ' ' << j << '\n'; } } 
 » 18 months ago, # |   0 you can see this: https://tp-iiita.quora.com/The-Two-Pointer-Algorithm
•  » » 17 months ago, # ^ |   0 well, they link back to this place so a non-ending recursive loop has been created :)
 » 12 months ago, # |   0 If array elements are not distinct, the worst case is when all elements of both arrays are equal and the sum we are looking for is a[0] + b[0]. In this case all n2 pairs of indices form the needed summation, thus the best approach is an O(n2) brute force, so an O(n) solution will not work.
•  » » 12 months ago, # ^ |   +1 If you need the indices of all pairs, you may have O(n^2) results, so a O(n) is not possible, but if you need just how many index pairs there are, it's still solveable in O(n), just count how many there equal there are of a and b, and result is equalA*equalB.