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.
Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.
Java:
public int minPatches(int[] nums, int n) {
long miss = 1;
int count = 0;
int i = 0;
while(miss <= n){
if(i<nums.length && nums[i] <= miss){
miss = miss + nums[i];
i++;
}else{
miss += miss;
count++;
}
}
return count;
}
Python:
class Solution(object):
def minPatches(self, nums, n):
"""
:type nums: List[int]
:type n: int
:rtype: int
"""
patch, miss, i = 0, 1, 0
while miss <= n:
if i < len(nums) and nums[i] <= miss:
miss += nums[i]
i += 1
else:
miss += miss
patch += 1 return patch
C++:
class Solution {
public:
int minPatches(vector<int>& nums, int n) {
long miss = 1, res = 0, i = 0;
while (miss <= n) {
if (i < nums.size() && nums[i] <= miss) {
miss += nums[i++];
} else {
miss += miss;
++res;
}
}
return res;
}
};
C++:
class Solution {
public:
int minPatches(vector<int>& nums, int n) {
long miss = 1, k = nums.size(), i = 0;
while (miss <= n) {
if (i >= nums.size() || nums[i] > miss) {
nums.insert(nums.begin() + i, miss);
}
miss += nums[i++];
}
return nums.size() - k;
}
};