FB面经prepare: 3 Window Max Sum

Given a num array, find a window of size k, that has a maximum sum of the k entries.
follow-up: Find three non-overlap windows of size k, that together has a maximum sum of the 3k entries, time complexity O(n^2)

Follow Up:

这个其实可以优化到O(n)时间。
建从左端到每个下标的最大window数组，再建从右端到每个下标的最大window数组。
再从左往右走一遍所有的size k window，将其和与上一步建的两个数组一起加起来。遍历一遍取最大值即可。

package fbOnsite;

public class ThreeWindow {
    public static int find(int[] arr, int k) {
        int res = Integer.MIN_VALUE;
        int len = arr.length;
        int[] left = new int[len];
        
        int lsum = 0;
        for (int i=0; i<len; i++) {
            lsum += arr[i];
            if (i >= k) lsum -= arr[i-k];
            if (i >= k-1) left[i] = Math.max(lsum, i>=1? left[i-1] : 0);//find the window end at i with max sum
        }
        
        int[] right = new int[len];
        int rsum = 0;
        for (int j=len-1; j>=0; j--) {
            rsum += arr[j];
            if (j < len-k) rsum -= arr[j+k];
            if (j <= len-k) right[j] = Math.max(rsum, j<len-1? right[j+1] : 0);
        }
        
        int midSum = 0;
        for (int t=k; t<=len-k-1; t++) {
            midSum += arr[t];
            if (t >= k + k) midSum -= arr[t-k];
            if (t >= k + k - 1) { // for each size k window in the middle with sum as midSum
                //find midSum + left[t-k] + right[t+1] that gives the max
                res = Math.max(res, midSum + left[t-k] + right[t+1]);
            } 
        }
        return res;
    }

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int[] arr = new int[]{2,1,1,-1,3,6,-2,-4,1,2,-1,2,1,-1,2};
        int[] arr1 = new int[]{1,2,2,-1,1,2,1,3,5};
        int res = find(arr, 3);
        System.out.println(res);
    }

}