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 ]
class Solution {
    public int[] getModifiedArray(int length, int[][] updates) {
        int[] ans = new int[length];
        
        for (int[] update : updates) {
            int start = update[0];
            int end = update[1];
            int inc = update[2];
            
            ans[start] += inc;
            if (end + 1 < length) {
                ans[end + 1] += -inc;
            }
        }
        for (int i = 1; i < length; i++) {
            ans[i] += ans[i -1];
        }
        
        return ans;
    }
}
Analysis:
Time complexity: O(n + k), where n is the length, and k is the number of update operations.
No comments:
Post a Comment