LeetCode: Range Addition

/**
Range Addition
Assume you have an array of length n initialized with all 0's and are given k update operations.
Each operation is represented as a triplet: [startIndex, endIndex, inc] which increments each element of subarray A[startIndex ... endIndex] (startIndex and endIndex inclusive) with inc.
Return the modified array after all k operations were executed.
Example:
Given:
length = 5,
updates = [
    [1,  3,  2],
    [2,  4,  3],
    [0,  2, -2]
]
Output:
[-2, 0, 3, 5, 3]
Explanation:
Initial state: [ 0, 0, 0, 0, 0 ]
After applying operation [1, 3, 2]: [ 0, 2, 2, 2, 0 ]
After applying operation [2, 4, 3]: [ 0, 2, 5, 5, 3 ]
After applying operation [0, 2, -2]: [-2, 0, 3, 5, 3 ]
Hint:
Thinking of using advanced data structures?
You are thinking it too complicated.
For each update operation, do you really need to update all elements between i and j?
Update only the first and end element is sufficient.
The optimal time complexity is O(k + n) and uses O(1) extra space.
*/

public class RangeAddition{
/**
-create a result array
-for each update, increase item at start index, decrease item at (end+1) index
-iterate through the result array, accumulate the sum and propagate the sum towards end of the result array
- as we have decreased the item at (end+1) index, the accumulated sum would not propagate the effect of the update from (end+1) index onwards
*/
public int[] getModifiedArray(int length, int[][] updates) {
int[] res = new int[length];
         for(int[] update : updates) {
            int value = update[2];
            int start = update[0];
            int end = update[1];
res[start] += value;
if(end < length - 1)
                res[end + 1] -= value;

        }
int sum = 0;
        for(int i = 0; i < length; i++) {
            sum += res[i];
            res[i] = sum;
        }

        return res;
    }
}