LeetCode 0330. Patching Array Solution in Java, Python, C++, JavaScript, Go & Rust | Explanation + Code

Description

Given a sorted integer array nums and an integer n, add/patch elements to the array such that any number in the 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:

Input: nums = [1,3], n = 6
Output: 1
Explanation:
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:

Input: nums = [1,5,10], n = 20
Output: 2
Explanation: The two patches can be [2, 4].

Example 3:

Input: nums = [1,2,2], n = 5
Output: 0

 

Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 104
• nums is sorted in ascending order.
• 1 <= n <= 231 - 1

Solutions

Solution 1: Greedy

Let's assume that the number x is the smallest positive integer that cannot be represented. Then all the numbers in [1,..x-1] can be represented. In order to represent the number x, we need to add a number that is less than or equal to x:

• If the added number equals x, since all numbers in [1,..x-1] can be represented, after adding x, all numbers in the range [1,..2x-1] can be represented, and the smallest positive integer that cannot be represented becomes 2x.
• If the added number is less than x, let's assume it's x', since all numbers in [1,..x-1] can be represented, after adding x', all numbers in the range [1,..x+x'-1] can be represented, and the smallest positive integer that cannot be represented becomes x+x' \lt 2x.

Therefore, we should greedily add the number x to cover a larger range.

We use a variable x to record the current smallest positive integer that cannot be represented, initialized to 1. At this time, [1,..x-1] is empty, indicating that no number can be covered; we use a variable i to record the current index of the array being traversed.

We perform the following operations in a loop:

• If i is within the range of the array and nums[i] \le x, it means that the current number can be covered, so we add the value of nums[i] to x, and increment i by 1.
• Otherwise, it means that x is not covered, so we need to supplement a number x in the array, and then update x to 2x.
• Repeat the above operations until the value of x is greater than n.

The final answer is the number of supplemented numbers.

The time complexity is O(m + log n), where m is the length of the array nums. The space complexity is O(1).

PythonJavaC++GoTypeScript

class Solution: def minPatches(self, nums: List[int], n: int) -> int: x = 1 ans = i = 0 while x <= n: if i < len(nums) and nums[i] <= x: x += nums[i] i += 1 else: ans += 1 x <<= 1 return ans(code-box)

class Solution { public int minPatches(int[] nums, int n) { long x = 1; int ans = 0; for (int i = 0; x <= n;) { if (i < nums.length && nums[i] <= x) { x += nums[i++]; } else { ++ans; x <<= 1; } } return ans; } }(code-box)

class Solution { public: int minPatches(vector<int>& nums, int n) { long long x = 1; int ans = 0; for (int i = 0; x <= n;) { if (i < nums.size() && nums[i] <= x) { x += nums[i++]; } else { ++ans; x <<= 1; } } return ans; } };(code-box)

func minPatches(nums []int, n int) (ans int) { x := 1 for i := 0; x <= n; { if i < len(nums) && nums[i] <= x { x += nums[i] i++ } else { ans++ x <<= 1 } } return }(code-box)

function minPatches(nums: number[], n: number): number { let x = 1; let ans = 0; for (let i = 0; x <= n; ) { if (i < nums.length && nums[i] <= x) { x += nums[i++]; } else { ++ans; x *= 2; } } return ans; }(code-box)