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Leetcode[209][M]Minimum Size Subarray Sum

18 Nov 2015

###Task1 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 0 instead.

For 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.

click to show more practice.

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

###Python ####O(n)

class Solution(object):
    def minSubArrayLen(self, s, nums):
        """
        :type s: int
        :type nums: List[int]
        :rtype: int
        """
        if not nums or len(nums) == 0:
            return 0
        if sum(nums) < s:
            return 0
        ret = len(nums)
        start = 0
        for i in range(1, len(nums) + 1):
            while sum(nums[start:i]) >= s:
                ret = min(ret, i - start)
                start += 1
        return ret

###Java

public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        int sum = 0;
        int count = Integer.MAX_VALUE;
        int start = 0;
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            if (sum >= s) {
                while (sum - nums[start] >= s) {
                    sum -= nums[start];
                    start++;
                }
                int curCount = i - start + 1;
                count = Math.min(count, curCount);
            }
        }
        
        return sum >= s ? count : 0;
    }
}

####O(nlog)

public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        if (null == nums || nums.length == 0) {
            return 0;
        }
        int min = nums.length + 1;
        int[] sum = new int[nums.length + 1];
        for (int i = 0; i < nums.length; i++) {
            sum[i + 1] = sum[i] + nums[i];
            if (sum[i + 1] >= s) {
                int left = binarySearch(sum, sum[i + 1] - s, 0, i + 1);
                min = Math.min(min, i + 1 - left);
            }
        }
        return min == nums.length + 1 ? 0 : min;
    }
     
    private int binarySearch(int[] nums, int target, int start, int end) {
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] >= target) {
                end = mid;
            } else {
                start = mid;
            }
        }
        if (nums[end] <= target) {
            return end;
        } else {
            return start;
        }
    }

}

###Points

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