Sharing an answer code of mine about MinMaxDivision problem of Codility lesson 14.

# Lesson 14: MinMaxDivision

You are given integers K, M and a non-empty zero-indexed array A consisting of N integers. Every element of the array is not greater than M.

You should divide this array into K blocks of consecutive elements. The size of the block is any integer between 0 and N. Every element of the array should belong to some block.

The sum of the block from X to Y equals A[X] + A[X + 1] + … + A[Y]. The sum of empty block equals 0.

The large sum is the maximal sum of any block.

For example, you are given integers K = 3, M = 5 and array A such that:

  A = 2
A = 1
A = 5
A = 1
A = 2
A = 2
A = 2


The array can be divided, for example, into the following blocks:

• [2, 1, 5, 1, 2, 2, 2], [], [] with a large sum of 15;
• , [1, 5, 1, 2], [2, 2] with a large sum of 9;
• [2, 1, 5], [], [1, 2, 2, 2] with a large sum of 8;
• [2, 1], [5, 1], [2, 2, 2] with a large sum of 6.

The goal is to minimize the large sum. In the above example, 6 is the minimal large sum.

Write a function:

def solution(K, M, A)


that, given integers K, M and a non-empty zero-indexed array A consisting of N integers, returns the minimal large sum.

For example, given K = 3, M = 5 and array A such that:

  A = 2
A = 1
A = 5
A = 1
A = 2
A = 2
A = 2


the function should return 6, as explained above.

Assume that:

• N and K are integers within the range [1..100,000];
• M is an integer within the range [0..10,000];
• each element of array A is an integer within the range [0..M].

Complexity:

• expected worst-case time complexity is O(N*log(N+M));
• expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

# Answer Code in Python 3

• Time complexity: $O(N\log(N+M))$
• Space complexity: $O(1)$
# In Python 3.6
# Time complexity: O(N*log(N+M))
# Space complexity: O(1)
import math
def check(K, A, mid):
n = len(A)
ls = 0
sum = 0
block = 0
for i in range(n):
if sum+A[i]>mid:
ls = max(ls, sum)
block += 1
sum = A[i]
elif sum+A[i]==mid:
sum = sum + A[i]
ls = max(ls, sum)
block += 1
sum = 0
else:
sum+=A[i]
if i == n-1 and sum != 0:
ls = max(ls, sum)
block += 1

return block, ls

def solution(K, M, A):
n = len(A)
beg = 1
end = M*n
min_ = sum(A)
while beg <= end:
mid = (beg+end)//2
block, ls = check(K, A, mid)
#print (mid, block, ls)
if block <= K:
end = mid - 1
min_ = min(min_, ls)
else:
beg = mid + 1

return min_


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