Does there exist a Top Down Dynamic Programming solution for Longest Increasing Subsequence?
Recursive approach to solve LIS length in java would be like below -
public int LIS(int[] arr) {
return LISLength(arr, Integer.MIN_VALUE, 0);
}
public int LISLength(int[] arr, int prev, int current) {
if (current == arr.length) {
return 0;
}
int include = 0;
if (arr[current] > prev) {
include = 1 + LISLength(arr, arr[current], current + 1);
}
int exclude = LISLength(arr, prev, current + 1);
return Math.max(include, exclude);
}
But it would work with O(2^n) time complexity so we need to use memoization technique to reduce the complexity with below approach -
public int LIS(int[] arr) {
int memoTable[][] = new int[arr.length + 1][arr.length];
for (int[] l : memoTable) {
Arrays.fill(l, -1);
}
return LISLength(arr, -1, 0, memoTable);
}
public int LISLength(int[] arr, int prev, int current, int[][] memoTable) {
if (current == arr.length) {
return 0;
}
if (memoTable[prev + 1][current] >= 0) {
return memoTable[prev + 1][current];
}
int include = 0;
if (prev < 0 || arr[current] > arr[prev]) {
include = 1 + LISLength(arr, current, current + 1, memoTable);
}
int exclude = LISLength(arr, prev, current + 1, memoTable);
memoTable[prev + 1][current] = Math.max(include, exclude);
return memoTable[prev + 1][current];
}
So O(n^2) would be optimized time complexity with memoization technique.
Sure. Define:
F(n) = longest increasing subsequence of sequence 1..n , and the sequence must ends with elementn
Then we get that recursion function (Top down):
F(n) = max(len(F(i)) + 1) which 0 <= i < n and array[i] < array[n]
So the answer is:
Longest increasing subsequence of F(1..n)
With memoization, we come to this code(That's Python, it's better than pseudo-code):
d = {}
array = [1, 5, 2, 3, 4, 7, 2]
def lis(n):
if d.get(n) is not None:
return d[n]
length = 1
ret = [array[n]]
for i in range(n):
if array[n] > array[i] and len(lis(i)) + 1 > length:
length = len(lis(i)) + 1
ret = lis(i) + [array[n]]
d[n] = ret
return ret
def get_ans():
max_length = 0
ans = []
for i in range(len(array)):
if max_length < len(lis(i)):
ans = lis(i)
max_length = len(lis(i))
return ans
print get_ans() # [1, 2, 3, 4, 7]