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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)
- 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.
1 public class Solution { 2 public int[] getModifiedArray(int length, int[][] updates) { 3 int[] res = new int[length]; 4 for (int[] update : updates) { 5 res[update[0]] += update[2]; 6 if (update[1]+1 < length) res[update[1]+1] -= update[2]; 7 } 8 for (int i=1; i<length; i++) { 9 res[i] = res[i] + res[i-1]; 10 } 11 return res; 12 } 13 }
Leetcode: Range Addition
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