I'm having problems trying to find the elements that form the Longest Increasing Subsequence of a given list.
I have the algorithm to find the value of a given item of the list, and I understand the method it uses, I just don't know what to add and where to add it so that I have the numbers that compose the LIS
Here is what I'm doing now:
for (A[0] = N[0], i=lis=1; i<n; i++) {
int *l = lower_bound(A, A+lis, N[i]);
lis = max(lis, (l-A)+1);
*l = N[i];
}
A
is an array that stores the partial LIS, but at some point it changes because there may be a different solution. N
is the array of elements.
How can I get from here to finding the longest increasing subsequence of N
?
You can use two additional array to find the LIS. For example, if your source is put in an array A
1 8 4 12 6 6 1
and we have an array B to store the elements of A which are more likely to be elements of LIS. More precisely, B will be maintained as an LIS at position i . Plus an array idx to record the positions.
We begin from A[0] , place A[0] at B[0]. Since A[0] is appended at position 0 in B , idx[0] = 0.
[0] 1 2 3 4 5 6
A | 1 8 4 12 6 6 1
B | (1)
idx | 0
Then for position 1, since element in B is smaller than A[1], A[1] is appended to B. idx[1] records the position in B which is 1.
0 [1] 2 3 4 5 6
A | 1 8 4 12 6 6 1
B | 1 (8)
idx | 0 1
For position 2, A[2], or 4, is more likely to be an element of LIS compared to elements in B in order to maintain B as an LIS. So find the element in B which is the smallest one no less than 4 and replace, which is 8. idx[2] is set to the position where 8 is replaced in B . I think you can use your searching algorithm to find such an element.
0 1 [2] 3 4 5 6
A | 1 8 4 12 6 6 1
B | 1 (4)
idx | 0 1 1
So continue this manner, we gradually set up idx .
position 3
0 1 2 [3] 4 5 6
A | 1 8 4 12 6 6 1
B | 1 4 (12)
idx | 0 1 1 2
position 4
0 1 2 3 [4] 5 6
A | 1 8 4 12 6 6 1
B | 1 4 (6)
idx | 0 1 1 2 2
position 5
0 1 2 3 4 [5] 6
A | 1 8 4 12 6 6 1
B | 1 4 (6)
idx | 0 1 1 2 2 2
position 6
0 1 2 3 4 5 [6]
A | 1 8 4 12 6 6 1
B | (1) 4 6
idx | 0 1 1 2 2 2 0
We have idx recorded positions, now we scan idx backwards and will find out the LIS.
0 1 2 3 4 5 6
A | 1 8 4 12 6 6 1
idx | 0 1 1 2 2 (2) 0 | 6
0 1 2 3 4 5 6
A | 1 8 4 12 6 6 1
idx | 0 1 (1) 2 2 2 0 | 4 6
0 1 2 3 4 5 6
A | 1 8 4 12 6 6 1
idx | (0) 1 1 2 2 2 0 | 1 4 6
Hence, the output LIS is {1, 4, 6}
The code and A = {1, 8, 4, 12, 6, 6, 1} as source
#include <stdio.h>
#include <stdlib.h>
#define INT_INF 10000
int search_replace(int *lis, int left, int right, int key) {
int mid;
for (mid = (left+right)/2; left <= right; mid = (left+right)/2) {
if (lis[mid] > key) {
right = mid - 1;
} else if (lis[mid] == key) {
return mid;
} else if (mid+1 <= right && lis[mid+1] >= key) {
lis[mid+1] = key;
return mid+1;
} else {
left = mid + 1;
}
}
if (mid == left) {
lis[mid] = key;
return mid;
}
lis[mid+1] = key;
return mid+1;
}
int main(void) {
int i, tmp, size = 7, lis_length = -1;
int *answer;
int A[7] = {1,8,4,12,6,6,1},
LIS[7],
index[7] = {0};
LIS[0] = A[0];
for (i = 1; i < size; ++i) {
LIS[i] = INT_INF;
}
for (i = 1; i < size; ++i) {
index[i] = search_replace(LIS, 0, i, A[i]);
if (lis_length < index[i]) {
lis_length = index[i];
}
}
answer = (int*) malloc((lis_length+1) * sizeof(int));
for (i = size-1, tmp = lis_length; i >= 0; --i) {
if (index[i] == tmp) {
answer[tmp] = A[i];
--tmp;
}
}
printf("LIS: ");
for (i = 0; i < lis_length+1; ++i) {
printf("%d ", answer[i]);
}
printf("\n");
return 0;
}
And the output of the code
LIS: 1 4 6
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