[LeetCode] 689. Maximum Sum of 3 Non-Overlapping Subarrays

In a given array nums of positive integers, find three non-overlapping subarrays with maximum sum.
Each subarray will be of size k, and we want to maximize the sum of all 3*k entries.
Return the result as a list of indices representing the starting position of each interval (0-indexed). If there are multiple answers, return the lexicographically smallest one.
Example:
Input: [1,2,1,2,6,7,5,1], 2
Output: [0, 3, 5]
Explanation: Subarrays [1, 2], [2, 6], [7, 5] correspond to the starting indices [0, 3, 5].
We could have also taken [2, 1], but an answer of [1, 3, 5] would be lexicographically larger.
Note:

  • nums.length will be between 1 and 20000.
  • nums[i] will be between 1 and 65535.
  • k will be between 1 and floor(nums.length / 3).


    • Thought process:
    Dynamic programming.
    1. Create an array of sums of windows of size k. This array's length will be nums.length - k + 1, because an array can have this many choices of windows.
    2. Iterate through the windows and find the index of the window with the max sum to the left and to the right.
    3. Iterate through the windows again and get and three windows with max sum.

    Solution:
     1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
    class Solution {
    public int[] maxSumOfThreeSubarrays(int[] nums, int k) {
        int[] win = new int[nums.length - k + 1];
        int sum = 0;
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            if (i >= k) {
                sum -= nums[i - k];
            }
            if (i >= k - 1) {
                win[i - (k - 1)] = sum;
            }
        }
        
        int j = 0;
        int[] left = new int[win.length];
        for (int i = 0; i < win.length; i++) {
            if (win[i] > win[j]) {
                j = i;
            }
            left[i] = j;
        }
        
        j = win.length - 1;
        int[] right = new int[win.length];
        for (int i = win.length - 1; i >= 0; i--) {
            if (win[i] >= win[j]) {
                j = i;
            }
            right[i] = j;
        }
        
        int[] max = new int[]{ -1, -1, -1 };
        for (int b = k; b < win.length - k; b++) {
            int a = left[b - k];
            int c = right[b + k];
            if (max[0] == -1 || win[a] + win[b] + win[c] > win[max[0]] + win[max[1]] + win[max[2]]) {
                max[0] = a;
                max[1] = b;
                max[2] = c;
            }
        }
        return max;
    }
}

    Time complexity: O(n).
    Location: Philadelphia, PA, USA

    Comments

    1. howard25330November 7, 2017 at 7:08 PM

      your right function wrong dude.
      if (win[i] > win[j]) {
      j = i;
      }
      should be
      if (win[i] >= win[j]) {
      j = i;
      }
      since the answer is
      If there are multiple answers, return the lexicographically smallest one.

      ReplyDelete
    2. UnknownJuly 19, 2020 at 9:14 AM

      what if they asked for m subarray in general?

      ReplyDelete

    Post a Comment

    Popular posts from this blog

    [LeetCode] 269. Alien Dictionary

    There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of  non-empty  words from the dictionary, where  words are sorted lexicographically by the rules of this new language . Derive the order of letters in this language. Example 1: Given the following words in dictionary, [ "wrt", "wrf", "er", "ett", "rftt" ] The correct order is:  "wertf" . Example 2: Given the following words in dictionary, [ "z", "x" ] The correct order is:  "zx" . Example 3: Given the following words in dictionary, [ "z", "x", "z" ] The order is invalid, so return  "" . Note: You may assume all letters are in lowercase. You may assume that if a is a prefix of b, then a must appear before b in the given dictionary. If the order is invalid, return an empty string
    Read more

    [HackerRank] Roads and Libraries

    Image
    The Ruler of HackerLand believes that every citizen of the country should have access to a library. Unfortunately, HackerLand was hit by a tornado that destroyed all of its libraries and obstructed its roads! As you are the greatest programmer of HackerLand, the ruler wants your help to repair the roads and build some new libraries efficiently. HackerLand has   cities numbered from   to  . The cities are connected by   bidirectional roads. A citizen has access to a library if: Their city contains a library. They can travel by road from their city to a city containing a library. The following figure is a sample map of HackerLand where the dotted lines denote obstructed roads: The cost of repairing any road is   dollars, and the cost to build a library in any city is   dollars. You are given   queries, where each query consists of a map of HackerLand and value of   and  . For each query, find the minimum cost of making libraries accessible to all the citizens and prin
    Read more

    [LeetCode] 631. Design Excel Sum Formula

    Your task is to design the basic function of Excel and implement the function of sum formula. Specifically, you need to implement the following functions: Excel(int H, char W):  This is the constructor. The inputs represents the height and width of the Excel form.  H  is a positive integer, range from 1 to 26. It represents the height.  W  is a character range from 'A' to 'Z'. It represents that the width is the number of characters from 'A' to  W . The Excel form content is represented by a height * width 2D integer array  C , it should be initialized to zero. You should assume that the first row of  C  starts from 1, and the first column of  C  starts from 'A'. void Set(int row, char column, int val):  Change the value at  C(row, column)  to be val. int Get(int row, char column):  Return the value at  C(row, column) . int Sum(int row, char column, List of Strings : numbers):  This function calculate and set the value at  C(row, column) , wh
    Read more