370. Range Addition (Medium)

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:

1. Thinking of using advanced data structures? You are thinking it too complicated.
2. For each update operation, do you really need to update all elements between i and j?
3. Update only the first and end element is sufficient.
4. The optimal time complexity is O(k+n) and uses O(1) extra space.

Solution:

class Solution {
public:
    vector<int> getModifiedArray(int length, vector<vector<int>>& updates) {
        vector<int> res(length, 0);
        for (int i = 0; i < updates.size(); ++i) {
            res[updates[i][0]] += updates[i][2];
            if (updates[i][1]+1 < length) res[updates[i][1]+1] -= updates[i][2];
        }
        for (int i = 1; i < length; ++i) {
            res[i] += res[i-1];
        }
        return res;
    }
};