+ 3

MinMaxDivision(Divide array A into K blocks and minimize the largest sum of any block.)

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[0] = 2 A[1] = 1 A[2] = 5 A[3] = 1 A[4] = 2 A[5] = 2 A[6] = 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; [2], [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: int solution(int K, int M, int A[], int N); 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[0] = 2 A[1] = 1 A[2] = 5 A[3] = 1 A[4] = 2 A[5] = 2 A[6] = 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).

9th Jan 2018, 2:07 PM
King David
3 Antworten
9th Jan 2018, 2:21 PM
King David
+ 3
Is this your homework?
7th Dec 2017, 10:29 AM
blackcat1111
blackcat1111 - avatar
0
No
19th Dec 2017, 5:51 PM
King David