HackerRank "Angry Children 2"

Fun one! A combination of Greedy and DP. The solution sparkled in my mind - I almost lost it..

Greedy: we sort the input numbers and always pick k continuous numbers - can be proved using contradiction
DP: Visualize it in your mind and you will get it : ) Just like a 2D geometry drawing.

# Get Input
n = int(input())
k = int(input())
arr = []
for _ in range(n):
v = int(input())
arr.append(v)

# Sort - kinda Greedy: we only pick k continuous nums
arr.sort()

# Step1: calc D of first k nums
d = 0
areaUp = 0
areaDw = 0
for i in range(1, k):
areaUp += i * (arr[i] - arr[i - 1])
areaDw += arr[i] - arr[0]
d += areaUp

ret = d
# Step2: go over rest numbers
for i in range(k, n):
dd = d
# removing areaDw
dd -= areaDw
areaDw -= (k - 1) * (arr[i - k + 1] - arr[i - k])
areaDw += arr[i] - arr[i - k + 1]
# adding new areaUp
areaUp = areaUp - (arr[i - 1] - arr[i - k]) + (k - 1) * (arr[i] - arr[i - 1])
dd += areaUp
d = dd
ret = min(ret, d)

print(ret)