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The constraints given say that any element in the array will be at least -9 and be at most 9. The smallest hourglass sum will be -9*7 = -63. So in this case, you just need to initialize max sum to be any value <= -63.
jj=(jj<3)?jj+1:0;// set 0 if jj(collumn pointer) set in one position which allows calculate an hourglassif(sum>maxsum)maxsum=sum;if(jj!=0)i--;// if jj is in a position where a hourglass can be calculated, the line pointer(i) continue in the same line
His algorithm is getting the middle element of the hourglass on the first pass through and then incrementing the top and bottom rows of the hourglass. He increments count to track the amount of loops needed to read a hourglass and increments x to correspond to the position in the hourglass; 0,1,2 respectively.
res.reserve(18) is like declaring an array of size 18 though am not sure why size 18 was choosen ,I think 16 would have done the job.
-the two for loops with contraints <4 is for the purpose of forming the hour glass (had it been <6 it would have resulted in indexing issues)
-arr[i][j]+arr[i][j+1]+arr[i][j+2]+ arr[i+1][j+1]+ arr[i+2][j]+arr[i+2][j+1]+arr[i+2][j+2] these are the indexes which form an hour glass, the sum of every hour glass is calculated and push to the res
-cout<<*max_element(res.begin(),res.end()) the maximum value in res is choosen as the output
I used a similar approach but dint keep on storing all of the values of hour glass instead just kept on swaping the values incase the value of new hour glass was greater than the previous one.
Is there any specific reason why you ran the j loop instead of the i loop first while finding the sum of the hourglass or is it just a matter of choice.
2D Array - DS
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how to do this task in C???
sorry, i am not familiar with c language.
why have u assigned maxsum=-1000 ? can u elobarate
the inputs also given in negitive....
The constraints given say that any element in the array will be at least -9 and be at most 9. The smallest hourglass sum will be -9*7 = -63. So in this case, you just need to initialize max sum to be any value <= -63.
why did you use j=jj
what the below statement does in the code?
if(jj != 0) i--;
To run loop for same value of i with different values for j untill j maximum limit is reached
why not take -1
could anyone identify the mistake
in for loop try i<4 instead of i<3
Whats the reason behind i<4..Not able to get the reason.Can you explain please
there are 4 hourglasses vertically and 4 hourglasses horizontally
I didnt understand this part. Can somebody help me out in this?
His algorithm is getting the middle element of the hourglass on the first pass through and then incrementing the top and bottom rows of the hourglass. He increments count to track the amount of loops needed to read a hourglass and increments x to correspond to the position in the hourglass; 0,1,2 respectively.
This code passes all the test cases
can anyone explain this part
res.reserve(18); for(unsigned int j=0; j<4;++j){ for(unsigned int i=0; i<4;++i){ res.push_back(arr[i][j]+arr[i][j+1]+arr[i][j+2]+ arr[i+1][j+1]+ arr[i+2][j]+arr[i+2][j+1]+arr[i+2][j+2]); } } cout<<*max_element(res.begin(),res.end())<
-the two for loops with contraints <4 is for the purpose of forming the hour glass (had it been <6 it would have resulted in indexing issues)
-arr[i][j]+arr[i][j+1]+arr[i][j+2]+ arr[i+1][j+1]+ arr[i+2][j]+arr[i+2][j+1]+arr[i+2][j+2] these are the indexes which form an hour glass, the sum of every hour glass is calculated and push to the res
-cout<<*max_element(res.begin(),res.end()) the maximum value in res is choosen as the output
I used a similar approach but dint keep on storing all of the values of hour glass instead just kept on swaping the values incase the value of new hour glass was greater than the previous one.
Hi, you're including a bunch of extraneous libraries. You are just using vector, iostream, and algorithm.
The rest could be deleted :) Cheers
I didnot include any extraneous libraries, I did my code in c++,kindly help me...thank you for your time and help my code:
include
using namespace std; void sum(int[][6]); int main() { int a[6][6]; int i,j; cout<<"enter val\n"; for(i=0;i<6;i++) { for(j=0;j<6;j++) { cin>>a[i][j]; } } sum(a); return 0; } void sum(int a[][6]) { int l=0,m=0; int i,j,max; int sum[4][4]; for(i=0,j=0;i<=3,j<=3;j++) { if(j==3) { sum[l][m]=a[i][j]+a[i][j+1]+a[i][j+2]+a[i+1][j+1]+a[i+2][j]+a[i+2][j+1]+a[i+2][j+2]; i++; j=-1; } else { sum[l][m]=a[i][j]+a[i][j+1]+a[i][j+2]+a[i+1][j+1]+a[i+2][j]+a[i+2][j+1]+a[i+2][j+2]; } m++; if(m==4) { l++; m=0; } } max=sum[0][0]; for(i=0;i<4;i++) { for(j=0;j<4;j++) { if(sum[i][j]>=max) { max=sum[i][j]; } } } cout<<"max"<
hi, i think you're more likely to get help if your code is more readable. Wrap your code with three back-ticks and that should help improve the
include
using namespace std; void sum(int[][6]); int main() { int a[6][6]; int i,j; cout<<"enter val\n"; for(i=0;i<6;i++) { for(j=0;j<6;j++) { cin>>a[i][j]; } } sum(a); return 0; } void sum(int a[][6]) { int l=0,m=0; int i,j,max; int sum[4][4]; for(i=0,j=0;i<=3,j<=3;j++) { if(j==3) { sum[l][m]=a[i][j]+a[i][j+1]+a[i][j+2]+a[i+1][j+1]+a[i+2][j]+a[i+2][j+1]+a[i+2][j+2]; i++; j=-1; } else { sum[l][m]=a[i][j]+a[i][j+1]+a[i][j+2]+a[i+1][j+1]+a[i+2][j]+a[i+2][j+1]+a[i+2][j+2]; } m++; if(m==4) { l++; m=0; } } max=sum[0][0]; for(i=0;i<4;i++) { for(j=0;j<4;j++) { if(sum[i][j]>=max) { max=sum[i][j]; } } } cout<<"max"<
how to add back ticks? i am unable to add the pdf file as well plzz help.
:O Are you using all those includes?
Simple and efficient and also easy to understand . Thank you sir.
Is there any specific reason why you ran the j loop instead of the i loop first while finding the sum of the hourglass or is it just a matter of choice.
Can you go over this a little bit?What are we trying to solve and how?^
I think python solution is simple and consume less memory space
print(max([sum(arr[i-1][j-1:j+2] + [arr[i][j]] + arr[i+1][j-1:j+2]) for j in range(1, 5) for i in range(1, 5)]))
include
int main() { int a[6][6],hg[3][3]; int i,j,x,y,sum=0,max=-63; for(i=0;i<6;i++) { for(j=0;j<6;j++) { scanf("%d",&a[i][j]); } }
}
Holy for-loops, Batman