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.
reference: https://leetcode.com/discuss/82822/solution-explanation
Let miss
be the smallest sum in [1,n]
that we might be missing. Meaning we already know we can build all sums in [1,miss)
. Then if we have a number num <= miss
in the given array, we can add it to those smaller sums to build all sums in [1,miss+num)
. If we don't, then we must add such a number to the array, and it's best to add miss
itself, to maximize the reach.
1 public class Solution { 2 public int minPatches(int[] nums, int n) { 3 long missed=1; 4 int added=0, i=0; 5 while (missed <= n) { 6 if (i<nums.length && nums[i]<=missed) { 7 missed = missed + nums[i]; 8 } 9 else { 10 added++; 11 missed = missed + missed; 12 i--; 13 } 14 i++; 15 } 16 return added; 17 } 18 }