Saturday, October 15, 2016

Patching Array

Given a sorted positive integer array nums and an integer n, add/patch elements to the array such that any number in range [1, n] inclusive can be formed by the sum of some elements in the array. Return the minimum number of patches required.
Example 1:
nums = [1, 3]n = 6
Return 1.
Combinations of nums are [1], [3], [1,3], which form possible sums of: 1, 3, 4.
Now if we add/patch 2 to nums, the combinations are: [1], [2], [3], [1,3], [2,3], [1,2,3].
Possible sums are 1, 2, 3, 4, 5, 6, which now covers the range [1, 6].
So we only need 1 patch.
Example 2:
nums = [1, 5, 10]n = 20
Return 2.
The two patches can be [2, 4].
Example 3:
nums = [1, 2, 2]n = 5
Return 0.

This is a very interesting problem, it's more like a brain storm question rather than an algorithm question. We define a variable missing, initialized to 1. This variable means the array can represent all sums from 0 to missing, exclusive, i.e., [0, missing). Now if the current number in array is smaller than missing, we can increase the range to [0, missing + num). However, if the current number is larger than missing, we know we have a gap to represent at least the number missing. So we need to add a number in the range of 0 to missing. In order to get the maximum range, we add the number missing. Now we can represent numbers from [0, missing + missing). So we iteratively process until missing is larger than n.

public int minPatches(int[] nums, int n) {
        if (nums == null)
            return 0;
        long patches = 0;
        long miss = 1;
        int pos = 0;
        while (miss <= n) {
            if (pos < nums.length && nums[pos] <= miss) 
                miss += nums[pos++];
            else {
                miss += miss;
                patches++;
            }
        }
        return (int)patches;
    }


No comments:

Post a Comment