[LintCode] Minimum Size Subarray Sum

Problem

Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return -1 instead.

Example

Given the array [2,3,1,2,4,3] and s = 7, the subarray [4,3] has the minimal length under the problem constraint.

Challenge

If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(nlogn).

Note

做一个窗口minsize，满足sum >= s的左界到右界的距离最小值为所求。while循环的约束条件要注意，不要遗漏：right不能超过nums[]的长度，但可以等于，因为存在nums[]所有元素之和为s的极端情况。
在sum < s时，sum加上右指针的元素，同时right指针后移，但是注意当right = nums.length时，sum不能加（元素不存在！），所以sum += nums[right]要加限制条件。
在sum >= s时，先更新窗口为当前循环后的最小值，减去最左元素，left指针后移。

Solution

public class Solution {
    public int minimumSize(int[] nums, int s) {
        if (nums == null) return -1;
        int left = 0, right = 0, sum = 0, minsize = Integer.MAX_VALUE;
        while (right <= nums.length && left <= right) {
            if (sum < s) {
                if (right < nums.length) {
                    sum += nums[right];
                }
                right++;
            }
            else {
                minsize = Math.min(minsize, right - left);
                sum -= nums[left];
                left++;
            }
        }
        return minsize <= nums.length ? minsize : -1;
    }
}

Binary Search:

http://www.jyuan92.com/blog/leetcode-min...

public class Solution {
    public int minimumSize(int[] nums, int s) {
        if (nums == null || nums.length == 0) return -1;
        int[] sum = new int[nums.length+1];
        int minsize = Integer.MAX_VALUE;
        for (int i = 0; i < nums.length; i++) {
            sum[i+1] = sum[i] + nums[i];
            if (sum[i+1] >= s) {
                int left = helper(sum, sum[i+1]-s, 0, i+1);
                minsize = Math.min(minsize, i+1-left);
            }
        }
        return minsize > nums.length ? -1 : minsize;
    }
    public int helper(int[] A, int target, int start, int end) {
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (A[mid] < target) start = mid; 
            else end = mid; //A[mid] = target is in this branch
        }
        if (A[end] <= target) return end;//Make sure return the same brunch
                                         //---->end, when A[end] = target
        else return start;
    }
}