Q. Given an integer array nums, return the length of the longest strictly increasing subsequence.
A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7] is a subsequence of the array [0,3,1,6,2,2,7].
Example 1:
Input: nums = [10,9,2,5,3,7,101,18]
Output: 4Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
Answer:
My recursion code is working correct but TopDown DP code is not working. Even I am just storing the correct answers in dp vector and use them again.
Recursion code:
int lengthOfLIS(vector<int>& arr, int i=0, int prev= INT_MIN){
//........... base case............
if(i==arr.size()) return 0;
//........... recursive case...........
// take if it is grater than prev
int X = INT_MIN;
if(arr[i] > prev)
X = 1 + lengthOfLIS(arr, i+1, arr[i]);
// ignore
int Y = lengthOfLIS(arr, i+1, prev);
return max(X, Y);
}
TopDown DP code:-
int sol(vector<int> arr, vector<int>& dp, int i=0, int prev= INT_MIN){
//........... base case............
if(i==arr.size()) return dp[i]=0;
if(dp[i]!=-1) return dp[i];
//........... recursive case...........
// take if it is grater than prev
int X = INT_MIN;
if(arr[i] > prev)
X = 1 + sol(arr, dp, i+1, arr[i]);
// ignore
int Y = sol(arr, dp, i+1, prev);
return dp[i] = max(X, Y);
}
Basically you want to memoize using dp. I would suggest taking a 2d vector for memoization in this case.
So the final solution will be :
int lengthOfLIS(vector& nums){
int n=nums.size();
vector<vector<int>> dp(n,vector<int>(n+1,-1));
return sol(nums,0,n,-1,dp);
}
int sol(vector<int>& nums,int i,int n,int prev,vector<vector<int>>& dp){
if(i==n){
return 0;
}
if(dp[i][prev+1]!=-1){
return dp[i][prev+1];
}
int X=0;int Y=0;
if(prev==-1 or nums[i]>nums[prev]){
X=1+sol(nums,i+1,n,i,dp);
}
Y=sol(nums,i+1,n,prev,dp);
dp[i][prev+1]=max(X,Y);
return dp[i][prev+1];
}
};
The concept is the same as yours, I have just used a 2d vector dp.
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