3180 Maximum Total Reward Using Operations I || How to 🤔 in Interview || Sort+Recursion+Memoization

[@MiteshTank1]

Amazing!! Nice Explanation

[@Code_With_Confidence]

Thank you

[@reqzonegaming7157]

happy to see you after while.

[@Code_With_Confidence]

Thank you 😊

[@danusshkumar]

Super explanation ❤❤

[@Code_With_Confidence]

Thank you

[@sanasainath7758]

awesome explanation within minutes

[@Code_With_Confidence]

Thank you 😊

[@ShortDecoded1]

not understood why you take dp[2001][4001] instead of dp[2001][4000001] i know it will give tle but sum can go upto that ...

[@pradnyeshaglawe351]

He explained in the video, the max rewardValue is 2000 and it can be taken only if x is 1999. So, in worst case when x is 1999, we include rewardValue 2000. Which equals 2000+1999=3999

[@bhupenderjaglanvlogs486]

I had written the same code but mine is giving memory limit exced
int solve(int n,int x,vector r,vector &dp,int i)
{
if(i == n)
return 0;

if(dp[i][x] != -1)
return dp[i][x];

int take = 0;

if(r[i] > x)
take = r[i] + solve(n,x+r[i],r,dp,i+1);

int ntake = solve(n,x,r,dp,i+1);

return dp[i][x] = max(take,ntake);



}
int maxTotalReward(vector& r) {
sort(r.begin(),r.end());
int ans = 0;
int n = r.size();
vector dp(2001,vector (4001,-1));
ans = solve(n,0,r,dp,0);
return ans;

}

[@ShortDecoded1]

keep it up bhaiya

[@Code_With_Confidence]

Thank you 😊

[@RohitSingh-hc8yi]

Nice explanation

[@Code_With_Confidence]

Thank you 😊

[@RitikGautam-c2p]

Improve your dp explanation.

[@RohitRohit-j6h5q]

bhug

[@Code_With_Confidence]

Noted