Diagonal Difference | Solving Hackerrank with python | Ep5

Here we are with the fifth episode of Solving #Hackerrank with Python series where we will explain and solve #Diagonal_Difference using #python.

timestamps
[00:00] Intro
[00:33] What is Diagonal Difference?
[00:50] Solving Diagonal Difference with math.
[04:00] Let's Write Code.

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[roopkummar]

Why you stop solving hacker rank problems, you explained in how to approach a problem without directly to the solution great way please explain all the questions it will help to crack

[Redd]

What would you suggest to get better with arrays like that used in counter2? I had a hard time making that connection with the array length. I would have thought the array length would be 9 and not 3.

[tanercoder]

Thanks for your videos! Here's my function:

def diagonalDifference(arr):
# Write your code here
left_diagonal = []
right_diagonal = []
#print(len(arr))
for i in range(len(arr)):

#print(left_diagonal)
for i in range(len(arr)-1, -1, -1):

#print(right_diagonal)
ldiag_sum=sum(left_diagonal)
rdiag_sum=sum(right_diagonal)
diff = ldiag_sum-rdiag_sum
return abs(diff)

[qc]

damm ... you have fantastic explanation skills. which is rare for programmers. good job man :) keep uploading hackerrank videos man

[modaryaghi]

I love the challenge in this problem. thanx

[chanhuangsuan]

Found a much simpler yet less intuitive solution via stack overflow:
difference = sum(row[i] - row[-i-1] for i, row in enumerate(arr))
return abs(difference)

Can't seem to understand it yet, but it works.

[electricimpulsetoprogramming]

Listen to me, you are a fucking amazing person the most bad ass that I've ever seen, i tried to figure out by my own and i couldn't solve, then i went to discussions and saw the answer, but i didn't understand why it was the answer and you basically explained very well so i wish you to be happy with health and money!!!

[royerguerrerop]

def diagonalDifference(arr):
# pd => Primary Diagonal && sd => Secondary Diagonal
pd = 0
sd = 0
j = 0
k = len(arr) - 1
for i in range(len(arr)):
pd += arr[i][j]
sd += arr[i][k]
k -= 1
j += 1

return abs(pd - sd)


Hi, here are other solution for this challenge, in this algorithm I use one only for loop
Sorry my English is not my native language

[DaniLo]

I didnt understand (n-i-1) but you clarify. Thank you

[riashrivastava]

Hi !! I trie doing the same coding today and 9/11 test cases failed Can I somehow send u the code so that you could have a look and tell me where am I wrong as I cant seem to debug it :(

[sefabockun]

I did like that :

first_diag = arr[0][0] + arr[1][1] + arr[2][2]
sec_diag = arr[0][2] + arr[1][1] + arr[2][0]
return abs(first_diag - sec_diag)

[technoinfoworldwide]

lets try this asked question in company, help me to solve this by making video:


From an N * N sized matrix, compute the smallest elements from the
two diagonals, and output the multiplication of these two smallest
numbers

[benjamindiaz]

my second counter isn't working when following along