我正在解决 Codility 问题作为练习，但无法回答其中一个问题。我在互联网上找到了答案，但我不明白这个算法是如何工作的。有人可以引导我逐步完成它吗？
这是问题：
/*
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:
class Solution { public int solution(int K, int M, int[] 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[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).
Elements of input arrays can be modified.
*/
这是我通过对我不理解的部分的评论找到的解决方案：
public static int solution(int K, int M, int[] A) {
int lower = max(A); // why lower is max?
int upper = sum(A); // why upper is sum?
while (true) {
int mid = (lower + upper) / 2;
int blocks = calculateBlockCount(A, mid); // don't I have specified number of blocks? What blocks do? Don't get that.
if (blocks < K) {
upper = mid - 1;
} else if (blocks > K) {
lower = mid + 1;
} else {
return upper;
}
}
}
private static int calculateBlockCount(int[] array, int maxSum) {
int count = 0;
int sum = array[0];
for (int i = 1; i < array.length; i++) {
if (sum + array[i] > maxSum) {
count++;
sum = array[i];
} else {
sum += array[i];
}
}
return count;
}
// returns sum of all elements in an array
private static int sum(int[] input) {
int sum = 0;
for (int n : input) {
sum += n;
}
return sum;
}
// returns max value in an array
private static int max(int[] input) {
int max = -1;
for (int n : input) {
if (n > max) {
max = n;
}
}
return max;
}
