Maximum Subarray IV

Given an integer arrays, find a contiguous subarray which has the largest sum and length should be greater or equal to given length k.
Return the largest sum, return 0 if there are fewer than k elements in the array.

 Notice

Ensure that the result is an integer type.

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Example

Given the array [-2,2,-3,4,-1,2,1,-5,3] and k = 5, the contiguous subarray [2,-3,4,-1,2,1] has the largest sum = 5.

java

public class Solution {    /*     * @param nums: an array of integer     * @param k: an integer     * @return: the largest sum     */    public int maxSubarray4(int[] nums, int k) {        // write your code here        if (nums == null || nums.length == 0 || k <= 0 || nums.length < k) {            return 0;        }        int[] preSum = new int[nums.length + 1];        preSum[0] = 0;        int max = Integer.MIN_VALUE;        int min = 0;        for (int i = 1; i <= nums.length; i++) {            preSum[i] = preSum[i - 1] + nums[i - 1];            if (i >= k && max < preSum[i] - min) {                max = preSum[i] - min;            }            if (i >= k) {                min = Math.min(min, preSum[i - k + 1]);            }        }        return max;    }}

python

class Solution:    """    @param: nums: an array of integer    @param: k: an integer    @return: the largest sum    """    def maxSubarray4(self, nums, k):        # write your code here        if nums is None or len(nums) == 0 or k <= 0 or len(nums) < k:            return 0        preSum, maxVal, minVal = [0] * (len(nums) + 1), float('-inf') ,0        for i in range(1, len(nums) + 1):            preSum[i] = preSum[i - 1] + nums[i - 1]            if i >= k and maxVal < preSum[i] - minVal:                maxVal = preSum[i] - minVal            if i >= k:                minVal = min(minVal, preSum[i - k + 1])        return maxVal