Maximum contiguous sub-array (With most number of elements)

Question

Given an array of natural numbers and an another natural T, how to find the contiguous subarray with sum less than or equal to T but the number of element in this subarray is maximized?

For example, if the given array is:

{3, 1, 2, 1, 1} and T = 5 . Then the maximum contigous subarray is {1, 2, 1, 1} because it will contain 5 elements and the sum is equal to 5.

Another example: {10,1,1,1,1,3,6,7} with T = 8 . Then the maximum contigous subarray is ${1,1,1,1,3}$

I can do it with O(n^2) operation. However I am looking for a linear time solution for this problem. Any ideas?

Answer 1

It ought to be possible to do this with O(n). I've not tested this, but it looks OK:

int start = 0, end = 0;
int beststart = 0, bestend = 0;
int sum = array[0];

while (end + 1 < arraysize) {
  if (array[end + 1] + sum <= T)
    sum += array[end++];
  else
    sum -= array[start++];
  if ((end - start) > (bestend - beststart)) {
    beststart = start;
    bestend = end;
  }
}

So, basically, it moves a sliding window along the array and records the point at which end - start is the greatest.

Answer 2

它似乎是最大子数组问题的上限版本： http : //en.wikipedia.org/wiki/Maximum_subarray_problem我想您可以从现有算法中找到灵感。