LeetCode //C - 330. Patching Array

330. Patching Array

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 < = n u m s [ i ] < = 1 0 4 1 <= nums[i] <= 10^4 1<=nums[i]<=104
• nums is sorted in ascending order.
• 1 < = n < = 2 31 − 1 1 <= n <= 2^{31} - 1 1<=n<=231−1

From: LeetCode

Link: 330. Patching Array

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Solution:

Ideas:

• miss: This variable represents the smallest sum that cannot be achieved with the current elements of the array.
• i: This is the index to traverse the array nums.
• patches: This is the counter for the number of patches needed.

Code:

int minPatches(int* nums, int numsSize, int n) {
    long long miss = 1;  // The smallest sum that we cannot achieve yet
    int i = 0, patches = 0;
    
    while (miss <= n) {
        if (i < numsSize && nums[i] <= miss) {
            miss += nums[i];
            i++;
        } else {
            miss += miss;
            patches++;
        }
    }
    
    return patches;
}