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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]
- if nums[i] > nums[j], then
- for j in range 0 to i – 1
- 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 −
#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));
}Input
[1,3,5,4,7]
Output
2
Updated on: 29-Apr-2020
280 Views
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