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.

Solution

public class Solution {
    public int[] getModifiedArray(int length, int[][] updates) {
        if(length == 0) return new int[]{};
        int ans[] = new int[length];
        for(int i = 0; i < updates.length; i++) {
            int inc = updates[i][2];
            ans[updates[i][0]] += inc;
            if(updates[i][1] + 1 < length) ans[updates[i][1] + 1] -= inc;
        }

        for(int i = 1; i < length; i++) {
            ans[i] += ans[i-1];
        }

        return ans;
    }
}