The dp series is like web series. Every day a new episod is coming with lots of new things and leaving a suspence for the next one.. thank you striver for this wonderful effort..and "UNDERSTOOD"..
This series is so damn good that before watching this video, i thought of giving "min cost climbing stairs" question on leetcode a try because i felt i could do that. That is similar to this question only. And boom, i solved that question in all possible ways...recursion, memoization, tabulation and space optimised method. Thank you striver. You have kinda opened our brain and made our thinking broader. Thank you so much for the inspiration you are. Love you !!!
this is the only playlist on youtube where you will find only positive comments because this man really put his effort to make this playlist valuable. thank you so much striver.
After searching all over the world to grasp these DSA concepts, it wasn't until I watched this video that everything finally clicked. The way you explained it was beyond exceptional-thank you for creating such a clear and impactful resource!
We will take a for loop running from j=1 to k and instead of writing (ind-1) and (ind-2) we will going to simply write (ind-j) that will give me the solution. Time Complexity:-O(N*K) Space complexity:-O(N). Yes sir, it cannot be space optimised because we have to use K+1 variables . Nice video and yes you are absolutely right, I have learnt space optimisation in dp from you THANKS MAN 👍👍
You will get hired even if you can explain an interviewer till memoization thing but it will blow the mind of interviewer if you go till space optimisation to constant.... awesome video and striver is just god.
Learnt the concept, went straight to the editor, coded it in JAVA and submitted it...All test cases submitted 🔥👍 Striver's way of explaining recursion and DP is unbeatable and uncomparable
Understood, Honestly this is the very first time I am trying to learn any dp problem. And I am able to understand everything... Very clear explanation. Loved it.. Thanks a lot for all the efforts :)
Hey Striver! I tried solving this question myself and I was able to solve it, I must say this series is amazing. You started building my concepts in first 2 lectures itself. Gonna see your lecture now and look at your approach. Would surely complete whole dp series. Thank you for this series ❤
your explanation is so good that even before you write the pseudo code, i was able to write the recursion by starting from 0 to n-1 rather than the reverse approach that you have taken in the video. Great Work you are doing. Helps to improve the way a person can solve a problem.
Hey, striver i just want to say that concept in your videos are such a crystal clear and easy to understand that even i am studying for the first time i do not feel any difficulty to understand the solutions. so, thank you so much for providing such a video lecture free on youtube
Amazing video! Everything is clearly understandable. from day 1 of dp series I have watching your videos and Day by Day become big fan of your teaching way. God bless you sir. Came to know from your insta story that u have fever 🤒 Get well soon sir.
The code for k jumps: #Memoization def k_jumps(n,dp,arr,k,result): if n == 0: return 0 if dp[n] != -1: return dp[n] for i in range(1,k+1): if (n > (i - 1)): step = k_jumps(n-i,dp,arr,k,result) + abs(arr[n] - arr[n-i]) else: step = 9999999 result = min(result,step) dp[n] = result return dp[n] arr = [30,10,60,10,60,50] n = len(arr) k = 3 result = 999999 dp = [-1 for i in range(n)] print(k_jumps(n-1,dp,arr,k,result)) #Tabulation def k_jumps(n,arr,k): dp = [-1 for i in range(n)] if n == 0: return 0 dp[0] = 0 for index in range(1,n): result = 999999 for i in range(1,k+1): if (index > (i - 1)): step = dp[index-i] + abs(arr[index] - arr[index-i]) else: step = 9999999 result = min(result,step) dp[index] = result print(dp) return dp[n-1] arr = [60,30,10,60,10,60,50] n = len(arr) k = 3 result = 999999 print(k_jumps(n,arr,k))
"UNDERSTOOD"..... Bhaiya pls upload practice problem link other than coding ninja link. Leetcode interface is best for easy understanding. I hope you understand , Thanks
00:03 Find the minimum total energy used by the frog to reach the first stair 02:13 The minimum energy required to jump from one point to another is 20. 06:11 Recursion is a way to find the path with minimum energy. 08:30 The costing for this guy will be zero no matter what you do. 12:51 Applying memoization to optimize recursion 14:47 Return the minimum of the left and right subpoints 18:51 Recursion helps solve multiple sub problems by reusing previously solved solutions. 20:50 Convert a recursion solution to dynamic programming 24:48 Memoization and tabulation are two approaches to solve problems recursively and iteratively respectively. 26:43 The first step in tabulation is to initialize the dp array. 30:55 The space complexity can be optimized by using the index minus 1 and index minus 2 technique 32:49 Use variables instead of an array to store previous values. 36:49 The problem requires k jumps as a follow-up to i plus 1 and i plus 2. 38:38 The ultimate resource for placements
this is the only playlist on youtube where you will find only positive comments because this man really put his effort to make this playlist valuable. thank you so much striver. ❤❤From Gujarat !! Jordar (Gujarati)!
Approach: Since the frog can make k jumps, we can use k recursive calls and find the minimum of them. used a variable ans to kept it as positive infinity to update the answer . base case is same as the frog jumps qn. but since frog can make k jumps, used a for loop to find the the answer for every jump (1st , 2nd ,.... till kth). also a check of i-w>=0 is made so that there is no infinite recursion. def frogJump(n,heights,k) -> int: i=n-1 ans=float('inf') def helper(i,ans): if i==0: return 0 for w in range(1,k+1): if i-w>=0: find=helper(i-w,ans)+abs(heights[i-w]-heights[i]) ans=min(find,ans) return ans return helper(i,ans) def frogjump_tabulation(n,heights,k): dp=[0]*(n+1) dp[0]=0 for i in range(1,n): ans=float('inf') for j in range(1,k+1): if i-j>=0: find=dp[i-j]+abs(heights[i-j]-heights[i]) ans=min(ans,find) dp[i]=ans return dp[n-1]
your intention is to make education easier...yes!! striver you did it....We learn from you.We will make you proud a Day...thankyou striver for this wonderful series..this is first comment on my youtube..try to ping my comment if u have time ❤
"understood" i came here with a doubt that i am from java if i will be able to understand or not but what happened i understood as fine as i can and i can write the same in c++ with little bit of help oviously but thanks a lot !
big thank you bhai for your hard work and make questions easy for us , again thank you with saying "we can't be lazy by seeing your high level of energy to teach us", "UNDERSTAND" 😍🥰🤩😇😘❤💘
As I am declaring the dp array globally it is giving incorrect answer. What could be the reason?? vector dp(100000,INT_MAX); int solve(int ind, vector& height){ if(ind==0) return 0; if(dp[ind]!=INT_MAX) return dp[ind]; int jumpTwo = INT_MAX; int jumpOne= solve(ind-1, height)+ abs(height[ind]-height[ind-1]); if(ind>1) jumpTwo = solve(ind-2, height)+ abs(height[ind]-height[ind-2]);
return dp[ind]=min(jumpOne, jumpTwo); } int frogJump(int n, vector &heights) { // Write your code here. return solve(n-1,heights);
I need your support, and you can do that by giving me a like, and commenting "understood" if I was able to explain you.
Please Upload the next One video
Understood 🥺🥺
Can't we start from 0 and using recursion ?
koi compiler theek krao yrr python vla coding ninjas ka
We are with u.
2 years later. Still the ultimate and unbeatable learning resource in the market.
But i thought adita verma was best for dp
@@salmanpandey8150yrr usne to mujko kisi pe bhi recursion karna bata diya bina soche uska base case nikap deta hun
The dp series is like web series. Every day a new episod is coming with lots of new things and leaving a suspence for the next one.. thank you striver for this wonderful effort..and "UNDERSTOOD"..
only if we could get a trailer for the next day's video😂
This series is so damn good that before watching this video, i thought of giving "min cost climbing stairs" question on leetcode a try because i felt i could do that. That is similar to this question only. And boom, i solved that question in all possible ways...recursion, memoization, tabulation and space optimised method. Thank you striver. You have kinda opened our brain and made our thinking broader. Thank you so much for the inspiration you are. Love you !!!
this is the only playlist on youtube where you will find only positive comments because this man really put his effort to make this playlist valuable.
thank you so much striver.
i can see 2, striver and love babbar
@@manikantadevadiga7146 Love Babbar copies from striver. I have seen many videos taking the same sample code and examples from striver.
@@manikantadevadiga7146 Aditya Verma ka bhi dekh lo.
After searching all over the world to grasp these DSA concepts, it wasn't until I watched this video that everything finally clicked. The way you explained it was beyond exceptional-thank you for creating such a clear and impactful resource!
I recently started this dp series and i want to complete this series in 20 days this is my day 1
Thank you for this awesome series
We will take a for loop running from j=1 to k and instead of writing (ind-1) and (ind-2) we will going to simply write (ind-j) that will give me the solution.
Time Complexity:-O(N*K)
Space complexity:-O(N).
Yes sir, it cannot be space optimised because we have to use K+1 variables . Nice video and yes you are absolutely right, I have learnt space optimisation in dp from you
THANKS MAN 👍👍
baniya buddhi mat chalao jyada striver bhai ko appreciate kaaro khud toh kuch kar nahi raahe
@@cipherCrafters14 XD
Amazing video! Do watch till the end because in the space optimization part, the actual value of watching the past two videos comes.
You will get hired even if you can explain an interviewer till memoization thing but it will blow the mind of interviewer if you go till space optimisation to constant.... awesome video and striver is just god.
Learnt the concept, went straight to the editor, coded it in JAVA and submitted it...All test cases submitted 🔥👍
Striver's way of explaining recursion and DP is unbeatable and uncomparable
Understood, Honestly this is the very first time I am trying to learn any dp problem. And I am able to understand everything... Very clear explanation. Loved it.. Thanks a lot for all the efforts :)
Hey Striver!
I tried solving this question myself and I was able to solve it, I must say this series is amazing. You started building my concepts in first 2 lectures itself. Gonna see your lecture now and look at your approach. Would surely complete whole dp series.
Thank you for this series ❤
Thanks so much Striver. We really appreciate your effort. You are a rare gem.
Best video I have ever seen on DP. Thank you Striver for helping out fellow developers.
your explanation is so good that even before you write the pseudo code, i was able to write the recursion by starting from 0 to n-1 rather than the reverse approach that you have taken in the video. Great Work you are doing. Helps to improve the way a person can solve a problem.
Recursion code at 24:33
Memoization code at 25:25
Tabulation code at 30:54
Space optimization at 36:22
Thanks dude
Would recommend to watch completely without skipping though
The insights given by him are great
@@yournemesis8232 Yes, you are right. This is just for jumping to the desired explanation without wasting time - when you are revising
follow ups solution:
#include
int frogJump(int n, vector &heights, int k){
vector dp(n, INT_MAX);
dp[0] = 0;
for(int i=1; i
this is what i thought in the very first second.
@@letslearn1055 me too
@@letslearn1055 same here when he said k jumps then i thought of a nested for loop till k
mee too
Code Of Frog Jump
#include
using namespace std;
//Top-Down Approach
int dp1(int ind,vector&heights,vector&dp)
{
if(ind == 0) return 0;
if(dp[ind] != -1) return dp[ind];
int left = dp1(ind-1,heights,dp) + abs(heights[ind]-heights[ind-1]);
int right = INT_MAX;
if(ind > 1) right = dp1(ind-2,heights,dp) + abs(heights[ind]-heights[ind-2]);
return dp[ind] = min(left,right);
}
//Tabulation Method - Bottom Up Approach
/*
vectordp(n,0);
dp[0]=0;
for(int i=1;i 1) secondstep = dp[i-2] + abs(heights[i] - heights[i-2]);
dp[i] = min(firststep,secondstep);
}
return dp[n-1];
*/
//Space Optimization
/*
int prev=0;
int prev2=0;
for(int i=1;i1) secondstep = prev2+abs(heights[i]-heights[i-2]);
int curi = min(firststep,secondstep);
prev2 = prev;
prev = curi;
}
return prev;
*/
int frogJump(int n, vector &heights)
{
vectordp(n+1,-1);
return dp1(n-1,heights,dp);
}
Hey, striver
i just want to say that concept in your videos are such a crystal clear and easy to understand that even i am studying for the first time i do not feel any difficulty to understand the solutions.
so, thank you so much for providing such a video lecture free on youtube
UNDERSTOOD , Now I am loving dp its become easy to think from recursion to tabulation step by step, thank you so much for the hardwork!
Amazing video! Everything is clearly understandable. from day 1 of dp series I have watching your videos and Day by Day become big fan of your teaching way. God bless you sir. Came to know from your insta story that u have fever 🤒 Get well soon sir.
Yes his teaching is absolutely marvelous and I become big fan of him he keep rocking now I am happy to be cse student 😇😇
Understood!! After seeing the recursion, I was able to convert it on my own into a dp. Thank you STRIVER. ❤
You need to learn Recursion bro
The code for k jumps:
#Memoization
def k_jumps(n,dp,arr,k,result):
if n == 0:
return 0
if dp[n] != -1:
return dp[n]
for i in range(1,k+1):
if (n > (i - 1)):
step = k_jumps(n-i,dp,arr,k,result) + abs(arr[n] - arr[n-i])
else:
step = 9999999
result = min(result,step)
dp[n] = result
return dp[n]
arr = [30,10,60,10,60,50]
n = len(arr)
k = 3
result = 999999
dp = [-1 for i in range(n)]
print(k_jumps(n-1,dp,arr,k,result))
#Tabulation
def k_jumps(n,arr,k):
dp = [-1 for i in range(n)]
if n == 0:
return 0
dp[0] = 0
for index in range(1,n):
result = 999999
for i in range(1,k+1):
if (index > (i - 1)):
step = dp[index-i] + abs(arr[index] - arr[index-i])
else:
step = 9999999
result = min(result,step)
dp[index] = result
print(dp)
return dp[n-1]
arr = [60,30,10,60,10,60,50]
n = len(arr)
k = 3
result = 999999
print(k_jumps(n,arr,k))
understood and I was always afraid of dp and now it looks easy.... Thank you!
Understood! Your series is making me love DP. Thanks a ton! Excited for the next one.
At 25:23, can't you initialize the dp array for only n elements instead of n+1? Like you did for tabulation?
"UNDERSTOOD".....
Bhaiya pls upload practice problem link other than coding ninja link. Leetcode interface is best for easy understanding.
I hope you understand , Thanks
Its not there on leetcode(Same problem)
Sadly leetcode is missing these kind of problems. No platform can match leetcode interface
A similar problem on Leetcode is Min cost Climbing stairs 746
@@zaidshaikh2536 bro i didnt ask for a similar problem on leetcode, I asked Striver bro to give the leetcode link and not the coding ninja link.
00:03 Find the minimum total energy used by the frog to reach the first stair
02:13 The minimum energy required to jump from one point to another is 20.
06:11 Recursion is a way to find the path with minimum energy.
08:30 The costing for this guy will be zero no matter what you do.
12:51 Applying memoization to optimize recursion
14:47 Return the minimum of the left and right subpoints
18:51 Recursion helps solve multiple sub problems by reusing previously solved solutions.
20:50 Convert a recursion solution to dynamic programming
24:48 Memoization and tabulation are two approaches to solve problems recursively and iteratively respectively.
26:43 The first step in tabulation is to initialize the dp array.
30:55 The space complexity can be optimized by using the index minus 1 and index minus 2 technique
32:49 Use variables instead of an array to store previous values.
36:49 The problem requires k jumps as a follow-up to i plus 1 and i plus 2.
38:38 The ultimate resource for placements
this is the only playlist on youtube where you will find only positive comments because this man really put his effort to make this playlist valuable.
thank you so much striver.
❤❤From Gujarat !! Jordar (Gujarati)!
Can we write base case for n=1, then can write Math.abs(heights[1]-heights[0])?
Yes
If possible pls provide leetcode problem link also of the particular problem since most of us are following leetcode
Similar problem on Leetcode is 746
Count all possible palindrome in string with rearranging characters - Infosys 🙄
This can be solve using combinatrix also , right??😅
Understood, Thanks a lot bhaiya ❤
I did not see this series feeling that DP is soo tough but after listening to your explanation, I've changed my mind. All credits to you Striver !
Understood, Now I'm loving Dp Thank you Striver ❤
Approach:
Since the frog can make k jumps, we can use k recursive calls and find the minimum of them. used a variable ans to kept it as positive infinity to update the answer . base case is same as the frog jumps qn. but since frog can make k jumps, used a for loop to find the
the answer for every jump (1st , 2nd ,.... till kth). also a check of i-w>=0 is made so that there is no infinite recursion.
def frogJump(n,heights,k) -> int:
i=n-1
ans=float('inf')
def helper(i,ans):
if i==0:
return 0
for w in range(1,k+1):
if i-w>=0:
find=helper(i-w,ans)+abs(heights[i-w]-heights[i])
ans=min(find,ans)
return ans
return helper(i,ans)
def frogjump_tabulation(n,heights,k):
dp=[0]*(n+1)
dp[0]=0
for i in range(1,n):
ans=float('inf')
for j in range(1,k+1):
if i-j>=0:
find=dp[i-j]+abs(heights[i-j]-heights[i])
ans=min(ans,find)
dp[i]=ans
return dp[n-1]
semester 1 is about to start,and i am on this lecture,super confident>3000
bhai itna jaldi
understood this very well. Thanks to striver
JavaScript/ typescript all 4 ways
// simple recursion
function f(ind: number, heights: number[]): number {
if (ind == heights.length - 1) return 0;
let left = f(ind - 1, heights) + Math.abs(heights[ind] - heights[ind - 1]);
let right = Infinity;
if (ind > 1) right = f(ind - 2, heights) + Math.abs(heights[ind] - heights[ind - 2]);
return Math.min(left, right);
}
// recusion with memoization
function memoF(ind: number, heights: number[], dp: number[]): number {
if (ind == heights.length - 1) return 0;
if (dp[ind] != -1) return dp[ind];
let left = memoF(ind - 1, heights, dp) + Math.abs(heights[ind] - heights[ind - 1]);
let right = Infinity;
if (ind > 1) right = memoF(ind - 2, heights, dp) + Math.abs(heights[ind] - heights[ind - 2]);
return dp[ind] = Math.min(left, right);
}
// tabulation
function TabulationFrogJump(n: number, heights: number[]): number {
let dp = new Array(n + 1).fill(0);
dp[0] = 0;
dp[1] = Math.abs(heights[1] - heights[0]);
for (let i = 2; i 1) right = dp[i - 2] + Math.abs(heights[i] - heights[i - 2]);
dp[i] = Math.min(left, right);
}
return dp[n];
}
// space optimised tabulation
function SpaceOptimisedTabulationFrogJump(n: number, heights: number[]): number {
let prev1 = 0;
let prev2 = Math.abs(heights[1] - heights[0]);
for (let i = 2; i
Understood!! I was always scared of Dynamic Programming but you made it so easy 🔥
I managed to do the space optimization on my own! That's just how good your explanation is!
I am nothing but greatful. Wanna finish the whole playlist. Thanks a lot
Understood!
Also, I could do tabulation and space optimization on my own after watching the explanation! A very big Thank you🥺
your intention is to make education easier...yes!! striver you did it....We learn from you.We will make you proud a Day...thankyou striver for this wonderful series..this is first comment on my youtube..try to ping my comment if u have time ❤
understood. Thanks for DP web series with addiction
I also started understanding dsa , you are magical man of programming world✌
UNDERSTOOD...........😎😎😎😎😎😎
Thank You Much for this wonderful video.............🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
Amazing explanation. Perfect solution to explain in an interview..
This question became a piece of cake for the one's following previous vedios.
Thankyou Striver Bhaiya.
US striver! TUF always takes us forward...No one can beat striver in explaining the problem
Understood ❤
K jumps code:
int memomization(int index,vector &h)
{
vector dp(index+1,0);
int fs=0;int ss=INT_MAX;
for(int i=1;i=0)
ss=dp[i-k-1]+abs(h[i-k-1]-h[i]);
smallest=min(smallest,min(fs,ss));
dp[i]=smallest;
}
}
}
return dp[index-1];
}
Bro the way u explained is just amazing
Thank you vai. Its really awesome. And love from Bangladesh
Very helpful and minute problems were explained nicely. Nothing best than this.
Code for K jumps :
int f(int idx, int k, vector &heights, vector &dp)
{
if (idx == 0)
{
return 0;
}
if (dp[idx] != -1)
{
return dp[idx];
}
int minCost = INT_MAX;
for (int i = 1; i = 0)
{
int jumpCost = f(idx - i, k, heights, dp) + abs(heights[idx] - heights[idx - i]);
minCost = min(minCost, jumpCost);
}
}
return dp[idx] = minCost;
}
int frogJump(int n, int k, vector &heights)
{
vector dp(n + 1, -1);
return f(n - 1, k, heights, dp);
}
Frog jump with k jumps:
#include
using namespace std;
int min_dp(int n, vector &heights, int ind, vector &dp,int k) {
if (ind == 0)
return dp[0]=0;
if (ind == 1)
return dp[1] = abs(heights[ind] - heights[ind - 1]);
if (dp[ind] == -1) {
vector temp(k);
for(int i=1 ; i> n>>k;
vector heights(n);
for (int i = 0; i < n; i++)
{
cin >> heights[i];
}
int ind = n-1;
vector dp(n + 1, -1);
cout> t;
while (t--)
{
solve();
}
return 0;
}
understood every bit of content ...thank u for this amazing lecture
Regular Recursion using java:
public static int frogJump(int n, int heights[]) {
return dpRecursion(n - 1, heights);
}
private static int dpRecursion(int n, int[] heights) {
if (n == 0) return 0;
int left = dpRecursion(n - 1, heights) + Math.abs(heights[n] - heights[n - 1]);
int right = n > 1 ? dpRecursion(n - 2, heights) + Math.abs(heights[n] - heights[n - 2]) : Integer.MAX_VALUE;
return Math.min(left, right);
}
US😎 Before the lecture, i was having a lot of fear about tabulation. Bt now its totally clear.😇 Thanks Raj Bhaiyya
this space optimization technique is way too classic, thanks bhaiya
#include
int frogJump(int n, vector &heights)
{
if(n == 0) return n;
int left = frogJump(n-1,heights) + abs(heights[n] - heights[n-1]);
int right;
if(n > 1)
{
right = frogJump(n-2,heights) + abs(heights[n] - heights[n-2]);
}
return min(left,right);
}
whats wrong?
declare right = MAX_INT.
UNDERSTOOD 🥰
Take the base case of index == 1 as return abs(heights[1]-heights[0]), makes the code a bit simpler without if else conditions.
index == 0 condition stays as it is so basically two base cases
I think it is the best solution to previous videos
time complexity and space complexity of memoization, 22:20
Solution for Frog Jump:
#include
#include
#include
#include
using namespace std;
// recursion
// Time Complexity: O(2^N)
// Space Complexity: O(N) - stack space
int helper(vector& nums, int index) {
if(index == 0) return 0;
int step1 = abs(nums[index] - nums[index-1]) + helper(nums, index-1);
int step2 = INT_MAX;
if(index - 2 >= 0) {
step2 = abs(nums[index] - nums[index-2]) + helper(nums, index-2);
}
return min(step1, step2);
}
int forgJump(vector& nums) {
int size = nums.size();
return helper(nums, size-1);
}
// memoization
// Time Complexity: O(N)
// Space Complexity: O(N) + O(N) - stack space
int helper(vector& nums, int index, vector& dp) {
if(index == 0) return 0;
if(dp[index] != -1) return dp[index];
int step1 = abs(nums[index] - nums[index-1]) + helper(nums, index-1, dp);
int step2 = INT_MAX;
if(index - 2 >= 0) {
step2 = abs(nums[index] - nums[index-2]) + helper(nums, index-2, dp);
}
return dp[index] = min(step1, step2);
}
int forgJump(vector& nums) {
int size = nums.size();
vector dp(size, -1);
return helper(nums, size-1, dp);
}
// Tabulation
// Time Complexity: O(N)
// Space Complexity: O(N)
int forgJump(vector& nums) {
int size = nums.size();
vector dp(size, 0);
dp[0] = 0;
for(int i=1;i= 0) {
step2 = abs(nums[i] - nums[i-2]) + dp[i-2];
}
dp[i] = min(step1, step2);
}
return dp[size-1];
}
// Space Optimized
// Time Complexity: O(N)
// Space Complexity: O(1)
int forgJump(vector& nums) {
int size = nums.size();
int prev = 0, secondPrev = 0;
for(int i=1;i= 0) {
step2 = abs(nums[i] - nums[i-2]) + secondPrev;
}
int curr = min(step1, step2);
secondPrev = prev;
prev = curr;
}
return prev;
}
Thanks a lot!!!! I was able to solve the video without seeing the video all because of you!!!!
Understood ! much appreciations to your efforts .
understood !
I am actually watching this to give a google interview sometime later but god damn, this is actually fun to learn lol
Understood each step in detail 💯
"UNDERSTOOD" Great series Striver
thanku striver bhaiya
Understood! Thanks for explaining beautifully!
4/57 done!!! understood!
Understood. Amazing explanation. Thanks a lot for making these videos.
Understood! hauuwa bana rakha tha dp ka 1 saal se man mein
thanks 900
Understood. Great Playlist !!!!
"understood"
i came here with a doubt that i am from java if i will be able to understand or not but what happened i understood as fine as i can and i can write the same in c++ with little bit of help oviously but thanks a lot !
Best explanations indeed, love you man.
big thank you bhai for your hard work and make questions easy for us , again thank you with saying "we can't be lazy by seeing your high level of energy to teach us", "UNDERSTAND" 😍🥰🤩😇😘❤💘
understood;
possibly wont have to watch this topic again :)
As I am declaring the dp array globally it is giving incorrect answer. What could be the reason??
vector dp(100000,INT_MAX);
int solve(int ind, vector& height){
if(ind==0) return 0;
if(dp[ind]!=INT_MAX) return dp[ind];
int jumpTwo = INT_MAX;
int jumpOne= solve(ind-1, height)+ abs(height[ind]-height[ind-1]);
if(ind>1)
jumpTwo = solve(ind-2, height)+ abs(height[ind]-height[ind-2]);
return dp[ind]=min(jumpOne, jumpTwo);
}
int frogJump(int n, vector &heights)
{
// Write your code here.
return solve(n-1,heights);
}
Instead of if clause sir you could have inserted the val of dp[1] and started loop from i = 2
Understood!! Thanks a lot for creating this series.
UNDERSTOOD!!!
Thank you sir
Understood
Clearly understood💯. Thank you so much for the series
Understood
NICE SUPER EXCELLENT MOTIVATED
Understand
Explanation is very high level❤❤❤
Understood Striver Sir !!
very clear explanation
Understood...Completed 3/56
understood
Thanks baiya very wonderfull video. I understood the video. Please make more and more videos. God will bless you. HK.