Number of Longest Increasing Subsequence in C++

Suppose we have one unsorted array of integers. we have to find the number of longest increasing subsequence, so if the input is like [1, 3, 5, 4, 7], then the output will be 2, as increasing subsequence are [1,3,5,7] and [1, 3, 4, 7]

To solve this, we will follow these steps −

n := size of the num array, create two arrays len and cnt of size n, and fill them with value 1.
lis := 1
for i in range 1 to n
- for j in range 0 to i – 1
  - if nums[i] > nums[j], then
    - if len[j] + 1 > len[i], then len[i] := len[j] + 1, and cnt[i] := cnt[j]
    - otherwise when len[j] + 1 = len[j], then cnt[i] := cnt[i] + cnt[j]
  - lis := max of lis and len[j]
ans := 0
for i in range 0 to n – 1
- if len[i] = lis, then ans := ans + cnt[j]
return ans

Example(C++)

Let us see the following implementation to get better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   int findNumberOfLIS(vector<int>& nums) {
      int n = nums.size();
      vector <int> len(n, 1), cnt(n, 1);
      int lis = 1;
      for(int i = 1; i < n; i++){
         for(int j = 0; j < i; j++){
            if(nums[i] > nums[j]){
               if(len[j] + 1 > len[i]){
                  len[i] = len[j] + 1;
                  cnt[i] = cnt[j];
               }
               else if(len[j] + 1 == len[i]){
                  cnt[i] += cnt[j];
               }
            }
            lis = max(lis, len[i]);
         }
      }
      int ans = 0;
      for(int i = 0; i < n; i++){
         if(len[i] == lis)ans += cnt[i];
      }
      return ans;
   }
};
main(){
   Solution ob;
   vector<int> v = {1,3,5,4,7};
   cout << (ob.findNumberOfLIS(v));
}