Create Number of Longest Increasing Subsequence.java

Author: py3-coder

DP LIS Pattern/Number of Longest Increasing Subsequence.java

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+
+/*
+673. Number of Longest Increasing Subsequence
+Medium
+
+Given an integer array nums, return the number of longest increasing subsequences.
+Notice that the sequence has to be strictly increasing.
+
+Example 1:
+Input: nums = [1,3,5,4,7]
+Output: 2
+Explanation: The two longest increasing subsequences are [1, 3, 4, 7] and [1, 3, 5, 7].
+
+Example 2:
+Input: nums = [2,2,2,2,2]
+Output: 5
+Explanation: The length of the longest increasing subsequence is 1, and there are 5 increasing subsequences of length 1, so output 5.
+
+Constraints:
+1 <= nums.length <= 2000
+-106 <= nums[i] <= 106
+*/
+
+class Solution {
+    public int findNumberOfLIS(int[] nums) {
+        // LIS Pure::
+        //edge case :: 
+        if(nums.length==1){
+            return 1;
+        }
+        return solveLIS(nums);
+        
+    }
+    public static int solveLIS(int[] arr){
+        int n=arr.length;
+        HashMap<Integer,Integer> map =new HashMap<>();
+        int len[] = new int[n];
+        int cnt[] = new int[n];
+        Arrays.fill(len,1);
+        Arrays.fill(cnt,1);
+
+        int maxi =0;
+        for(int i=1;i<n;i++){
+            for(int j=0;j<i;j++){
+                if(arr[i]>arr[j]){
+                    if(1+len[j]>len[i]){
+                        len[i] =1+len[j];
+                        cnt[i]=cnt[j];
+                    }else if(1+len[j]==len[i]){
+                        cnt[i]+=cnt[j];
+                    }
+                }
+            }
+            maxi =Math.max(maxi,len[i]);
+        }
+        int num=0;
+        for(int i=0;i<n;i++){
+            if(maxi==len[i]){
+                num +=cnt[i];
+            }
+        }
+        return num;
+    }
+}