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:
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
Approach #1: C++. [easy understand]
class Solution {
public:
int minPatches(vector<int>& nums, int n) {
int size = nums.size();
long pre_sum = 0;
int ans = 0;
for (int i = 0; i < size && pre_sum < n; ++i) {
int v = nums[i];
while (v > pre_sum + 1 && pre_sum < n) {
pre_sum += (pre_sum + 1);
ans += 1;
}
pre_sum += v;
}
if (pre_sum < n) {
while (pre_sum < n) {
pre_sum += (pre_sum + 1);
ans += 1;
}
}
return ans;
}
};
Analysis:
We can use pre_sum as base to extend the sum. If current element nums[i] > pre_sum, we should extend the pre_sum by add (pre_sum + 1).
Approach #2: C++. [This is more clear.]
int minPatches(vector<int>& nums, int n) {
long miss = 1, added = 0, i = 0;
while (miss <= n) {
if (i < nums.size() && nums[i] <= miss) {
miss += nums[i++];
} else {
miss += miss;
added++;
}
}
return added;
}
Analysis: