Range Addition

ssume 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.

Tips:

注意看hint，每次更新时只更新 start 和 end + 1 位，start加上value代表从这里开始变， end + 1位减去value代表到这里变回来。iterate一遍之后算前向和，存入答案数组即为答案。这个解法真妙！

tricky one. For each update, increment the start index by inc, decrement the end index + 1 by inc. Then do a moving sum at last.

Just store every start index for each value and at end index plus one minus it

for example it will look like:

[1 , 3 , 2] , [2, 3, 3] (length = 5)

res[ 0, 2, ,0, 0 -2 ]

res[ 0 ,2, 3, 0, -5]

sum 0, 2, 5, 5, 0

res[0, 2, 5, 5, 0]

Code:

public class Solution {
    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;
    }
}