JavaAlgorithms/DynamicProgramming/LongestIncreasingSubsequence.java
2021-03-24 18:51:20 +02:00

98 lines
2.3 KiB
Java

package DynamicProgramming;
import java.util.Scanner;
/** @author Afrizal Fikri (https://github.com/icalF) */
public class LongestIncreasingSubsequence {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int arr[] = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
System.out.println(LIS(arr));
System.out.println(findLISLen(arr));
sc.close();
}
private static int upperBound(int[] ar, int l, int r, int key) {
while (l < r - 1) {
int m = (l + r) >>> 1;
if (ar[m] >= key) r = m;
else l = m;
}
return r;
}
private static int LIS(int[] array) {
int N = array.length;
if (N == 0) return 0;
int[] tail = new int[N];
// always points empty slot in tail
int length = 1;
tail[0] = array[0];
for (int i = 1; i < N; i++) {
// new smallest value
if (array[i] < tail[0]) tail[0] = array[i];
// array[i] extends largest subsequence
else if (array[i] > tail[length - 1]) tail[length++] = array[i];
// array[i] will become end candidate of an existing subsequence or
// Throw away larger elements in all LIS, to make room for upcoming grater elements than
// array[i]
// (and also, array[i] would have already appeared in one of LIS, identify the location and
// replace it)
else tail[upperBound(tail, -1, length - 1, array[i])] = array[i];
}
return length;
}
/** @author Alon Firestein (https://github.com/alonfirestein) */
// A function for finding the length of the LIS algorithm in O(nlogn) complexity.
public static int findLISLen(int a[]) {
int size = a.length;
int arr[] = new int[size];
arr[0] = a[0];
int lis = 1;
for (int i = 1; i < size; i++) {
int index = binarySearchBetween(arr, lis, a[i]);
arr[index] = a[i];
if (index > lis)
lis++;
}
return lis;
}
// O(logn)
private static int binarySearchBetween(int[] t, int end, int key) {
int left = 0;
int right = end;
if (key < t[0])
return 0;
if (key > t[end])
return end + 1;
while (left < right - 1) {
int middle = (left + right) / 2;
if (t[middle] < key)
left = middle;
else
right = middle;
}
return right;
}
}