Leetcode: Range Addition

时间:2023-03-10 07:27:50
Leetcode: 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:

Time Complexity: O(N+K), Space: O(1)

  1. For each update operation, do you really need to update all elements between i and j?
  2. Update only the first and end element is sufficient.
  3. The optimal time complexity is O(k + n) and uses O(1) extra space.
 public class Solution {
public int[] getModifiedArray(int length, int[][] updates) {
int[] res = new int[length];
for (int[] update : updates) {
res[update[0]] += update[2];
if (update[1]+1 < length) res[update[1]+1] -= update[2];
}
for (int i=1; i<length; i++) {
res[i] = res[i] + res[i-1];
}
return res;
}
}