feat : lis

Author: ekgns33

longest-increasing-subsequence/ekgns33.java

@@ -0,0 +1,60 @@
+/**
+ input : array of integers
+ output : length of the longest strictly increasing subsequence;
+ example
+ 0 1 0 3 2 3 > 0 1 2 3  >> length : 4
+ 4 3 2 1 > 1 >> length : 1
+ 1 1 1 1 > 1 >> length : 1
+
+ constraints :
+ 1) empty array allowed?
+ no, at least one
+
+ solution 1) brute force
+ nested for loop
+ iterate through the array from index i = 0 to n-1 when n is the lenght of input
+ iterate throught the array from index j = i + 1 to n
+ update current Maximum value
+ if bigger than prev max value count ++
+ tc : O(n^2);
+ sc : O(1)
+
+ solution 2) binary search + dp
+ iterate through the array
+ search from saved values.
+ if current value is bigger than everything add
+ else change value
+
+ let val[i] save the smallest value of length i subsequence
+ tc : O(nlogn)
+ sc : O(n)
+
+ */
+
+class Solution {
+  public int lengthOfLIS(int[] nums) {
+    List<Integer> vals = new ArrayList<>();
+    for(int num : nums) {
+      if(vals.isEmpty() || vals.getLast() < num) {
+        vals.add(num);
+      } else {
+        vals.set(binarySearch(vals, num), num);
+      }
+    }
+    return vals.size();
+  }
+  int binarySearch(List<Integer> vals, int num) {
+    int l = 0, r = vals.size()-1;
+    while(l <= r) {
+      int mid = (r - l) / 2+ l;
+      if(vals.get(mid) == num) {
+        return mid;
+      } else if (vals.get(mid) > num) {
+        r = mid - 1;
+      } else {
+        l = mid + 1;
+      }
+    }
+    return l;
+  }
+}