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.
Credits:
Special thanks to @vinod23 for adding this problem and creating all test cases.
1 public class Solution { 2 public int[] getModifiedArray(int length, int[][] updates) { 3 int[] temp = new int[length]; 4 for (int[] update : updates) { 5 int start = update[0], end = update[1] + 1, inc = update[2]; 6 7 temp[start] += inc; 8 if (end < length) 9 temp[end] -= inc; 10 } 11 12 int[] result = new int[length]; 13 int sum = 0; 14 for (int i = 0; i < length; i++) { 15 result[i] = sum + temp[i]; 16 sum = result[i]; 17 } 18 19 return result; 20 } 21 }
时间: 2024-10-26 16:10:24