LeetCode Patching Array
Tags: Algorithm, LeetCodeGiven 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
.
Java Solution
The following algorithm is from https://discuss.leetcode.com/topic/35494/solution-explanation
We define miss
to be the smallest number that will miss, which means, given an input {1…m}, we can build all sums in [0, miss). For the next number x
in the array, if x <= miss
, then we can build sums in the range [0, miss + x)
, thus there is no need to add a number.
If the array is empty, or the next number x
in the array is larger than miss
, we have to add a number to get miss
. The best strategy is to add miss
itself. Then our next range is [0, miss + miss)
.
I also used an example below to describe how this method works.
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public class Solution { public int minPatches(int[] nums, int n) { // [1,2,5,9] //set miss = 1 // 1 ==> {1} [1, miss <= miss + 1 == 2) // since 2 <= miss, {1} + 2 ==> {1, 2, 3} === [1, miss + 2 == 4) // since 5 is larger than 4, we have to add 4 // with 4, we can build {1, 2, 3} + 4 ==> {1, 2, 3, 4, 5, 6, 7} == [1, 4 + 4 == 8) // 9 > 8, so we have to add 8: {1, 2, 3, ... 7} + {8} ==> {1, .... 8 + 7} == [1, 8 + 8 == 16) //{1}, 1 >> 0 //{1, 2}, 3 >> 0 //{1, 2}, 4 >> 1 long miss = 1; int cnt = 0; int i = 0; while(miss <= n){ if(i < nums.length && nums[i] <= miss) { miss += nums[i]; i++; } else { miss += miss; // add miss to the array cnt++; } } return cnt; } } |