I have been following your channel since leetcode monthly challenge started but never fully explored your channel. I was just watching videos whenever I had doubts about solving some problems. Today I explored all the playlists and checked the content. It's wonderful how you are contributing to the community. In the future, your channel will be one of the best resources for Interview preparation and particularly learning DS and algorithms(it still is but need more impact and viewers). Now I am going to suggest all my juniors to learn from this channel instead of buying any course. Best of luck and thanks for sharing such great content.
Sir for this I think we can take left pointer and right pointer and traverse array completely making hashmap of element and count if we find complementary elements we can find minimum count
Hi Arjun, I think he has concatenated x, so that if user passes the same x multiple times, it will straight away hit memoization value. So successive calls of x will take very little time. If you take the solution for just one x value, it will not make sense. If this problem was a very small part of a large solution where this block is called multiple times with repeating x values, it will show the true usefulness of concatenating x.
bhaiya yeh jo 2 base cases hain unko ooper neeche karne se toh test cases ke galat answer aa rhe hain...par aisa kyon ho rha hai line number 8 ki condition pehle aur line number 6 ki baad mein likhne se wrong output aa rha hai..
@@techdose4u we can use sliding window technique such a minimum window that sums to target will be our ans. leetcode.com/problems/minimum-operations-to-reduce-x-to-zero/discuss/939891/Sliding-Window-C%2B%2B
public int minOperations(int[] nums, int x) { int[] dp = new int[nums.length]; int count = helper(x, nums, 0, nums.length-1, 0, dp); return count == Integer.MAX_VALUE?-1:count; }
int helper(int x, int[] arr, int i, int j, int count, int[] dp){ if(xj) return Integer.MAX_VALUE; if(dp[i]!=0) return dp[i]; return dp [i] = Math.min( helper(x - arr[i], arr, i+1, j, count+1, dp), helper (x - arr[j], arr, i, j-1, count+1, dp) ); } I tried using 2 player stone game technique, but got TLE.
Anyone wondering if it's still giving TLE, please watch the complete video.. (y) 21:35
I have been following your channel since leetcode monthly challenge started but never fully explored your channel. I was just watching videos whenever I had doubts about solving some problems. Today I explored all the playlists and checked the content. It's wonderful how you are contributing to the community. In the future, your channel will be one of the best resources for Interview preparation and particularly learning DS and algorithms(it still is but need more impact and viewers). Now I am going to suggest all my juniors to learn from this channel instead of buying any course. Best of luck and thanks for sharing such great content.
Thanks for your motivation 🙂
@@techdose4u can you tell me why it doesn't work
int solve(int l,int r,vector &nums,int x){
if(x
I was wondering about the solution as I was getting TLE and here you are, I am so lucky to have you over youtube.
😊
When i saw this problem first time i approached it using recursion which is intuitive for this problem.. then i came here.. thank for this video....
Nice. Welcome bro :)
one of the best tutorial for Dp
Not sure the x has a meaning in the calculation of the key for the memoization. If you have left and right indexes the x should always be the same.
next video of playlist : th-cam.com/video/7nzwrX4N_A0/w-d-xo.html
Sir for this I think we can take left pointer and right pointer and traverse array completely making hashmap of element and count if we find complementary elements we can find minimum count
Yes that's the optimal approach
is there a way to apply bottom up in this as memoization is giving tle
Nice Explanation ❤❤❤
Thanks
recursion is intiuative in this questions ,but the memoization part can be thought only if you have already seen solution
Can i use dp[left][right]=count in this manner to store minimum count?
no u need 3 values, x as well.
Nice explanation!
Thanks
This approach is giving TLE if we convert the approach to Java based. Any suggestion to resolve TLE?
Watch the complete video
Leetcode mein TLE a raha hain
It would be great if you can upload leetcode weekly contest solutions
I would love to, but I don't get time 😅
Ok
Can we use dp tabulatio method for this problem?
@TECH DOSE Can't this problem be solved by making two dimensional table as u have made in your previous videos of DP??
Is there any follow up?
can greedy approch be used in this question?
amazing sir!!
:)
Did not get as to why we need the count in making a unique key...could we form a unique state only using left+"*" +right (each are string) ??
Hi Arjun, I think he has concatenated x, so that if user passes the same x multiple times, it will straight away hit memoization value. So successive calls of x will take very little time.
If you take the solution for just one x value, it will not make sense. If this problem was a very small part of a large solution where this block is called multiple times with repeating x values, it will show the true usefulness of concatenating x.
bhaiya yeh jo 2 base cases hain unko ooper neeche karne se toh test cases ke galat answer aa rhe hain...par aisa kyon ho rha hai line number 8 ki condition pehle aur line number 6 ki baad mein likhne se wrong output aa rha hai..
it is better to use pair for memorization
I did exactly this but still getting heap out of memory exception.
Maybe out of bounds
this is giving tle error although explanation is awesome
it is similar to coin change problem 1
can i get this code in python
I submitted this question using greedy algorithm
Nice :)
@@techdose4u we can use sliding window technique such a minimum window that sums to target will be our ans.
leetcode.com/problems/minimum-operations-to-reduce-x-to-zero/discuss/939891/Sliding-Window-C%2B%2B
class Solution {
HashMap dp;
int min;
int solve(int [] nums,int l,int r,int x ,int count){
if(x==0)
{
return count;
}
if( l>r || x
You have commented on top MEMOIZATION->TLE ..?
public int minOperations(int[] nums, int x) {
int[] dp = new int[nums.length];
int count = helper(x, nums, 0, nums.length-1, 0, dp);
return count == Integer.MAX_VALUE?-1:count;
}
int helper(int x, int[] arr, int i, int j, int count, int[] dp){
if(xj) return Integer.MAX_VALUE;
if(dp[i]!=0) return dp[i];
return
dp [i] = Math.min( helper(x - arr[i], arr, i+1, j, count+1, dp),
helper (x - arr[j], arr, i, j-1, count+1, dp) );
}
I tried using 2 player stone game technique, but got TLE.
giving tle