Maximize The Cut Segments | gfg potd today | POTD | GFG Problem of the Day | C++ |

Today's problem is really a good problem based on Maximize The Cut Segments ,stay with the video till the end definitely u will learn something from here and make sure to like , comment and share the video and subscribe the channel
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[@lucifercoxi8324]

Great

[@CodeThurst]

thnks

[@newglobal2056]

❤❤❤❤❤👍👍👍

[@lalitbisht8381]

here's the correct code
int solve(int n, int x, int y, int z, vector& dp){
if(n == 0) return 0;
if(n < 0) return INT_MIN;
if(dp[n]!=-1) return dp[n];
int one=INT_MIN, two=INT_MIN, three=INT_MIN;
one=solve(n-x, x,y,z,dp);
two=solve(n-y, x,y,z,dp);
three=solve(n-z, x,y,z,dp);
return dp[n] = 1+max(one,max(two,three));
}
int maximizeTheCuts(int n, int x, int y, int z)
{
vector dp(n+1,-1);
int ans=solve(n,x,y,z ,dp);
if(ans

[@tanurajput6759]

why we are adding 1 in return of 1+max(one,max(two,three)) ???

[@CodeThurst]

if we move with any of the option we are making a cut
so thts why

[@newglobal7271]

17/08/24-->
EASY PEASY with 1 Lakh 400 test case 😂😂🤣🤣 -->
class Solution {
public:
vector productExceptSelf(vector& nums) {
vectorresult(nums.size());
long long int temp=1;
int isZero=0;
for(auto&ele:nums){
if(ele==0){
isZero++;
continue;
}
temp=(temp*ele);
}
for(int i=0;i1)result[i]=0;
else
result[i]=(temp)/nums[i];
}
return result;
}
};

[@CodeThurst]

yess bro easy 🥱🥱

[@subrotomukherjee8571]

Why you have added 1 with max val?

[@CodeThurst]

if we move with any of the option we are making a cut
so thts why