330. Patching Array
Difficulty: Hard
Topics: Array
, Greedy
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
- Only one valid answer exists.
Solution:
We need to ensure that every number in the range [1, n]
can be formed by the sum of some elements in the given array. We can use a greedy algorithm to determine the minimum number of patches required.
Solution Explanation:
-
Key Insight:
- Start with the smallest missing number
miss
which starts at 1. - Iterate through the array
nums
. If the current number innums
is less than or equal tomiss
, thenmiss
can be covered by adding this number. Otherwise, we need to patchmiss
to the array, effectively doubling the range that can be covered.
- Start with the smallest missing number
-
Algorithm:
- Initialize
miss = 1
andpatches = 0
. - Loop through the array while
miss <= n
:- If the current number in
nums
can covermiss
, move to the next number innums
. - If the current number is greater than
miss
, patchmiss
and double the range covered by settingmiss = miss * 2
.
- If the current number in
- Continue until the entire range
[1, n]
is covered.
- Initialize
-
Complexity:
- The time complexity is O(m + log n), where
m
is the size of the arraynums
.
- The time complexity is O(m + log n), where
Let's implement this solution in PHP: 330. Patching Array
<?php
/**
* @param String $n
* @return String
*/
function minPatches($nums, $n) {
...
...
...
/**
* go to ./solution.php
*/
}
// Example usage:
// Example 1
$nums1 = [1, 3];
$n1 = 6;
echo minPatches($nums1, $n1); // Output: 1
// Example 2
$nums2 = [1, 5, 10];
$n2 = 20;
echo minPatches($nums2, $n2); // Output: 2
// Example 3
$nums3 = [1, 2, 2];
$n3 = 5;
echo minPatches($nums3, $n3); // Output: 0
?>
Explanation:
-
Initial Setup: We initialize
miss
as 1, meaning we want to ensure we can generate the number 1. -
While Loop: We loop until
miss
exceedsn
(the maximum number we need to cover). In each iteration:- If the current number in
nums
can help covermiss
, we add it tomiss
and move to the next number. - If not, we "patch" the array by effectively adding
miss
to it (without changing the array itself) and then updatemiss
to cover the new range.
- If the current number in
- Output: The total number of patches required is returned.
This algorithm ensures that we cover the entire range [1, n]
with the minimum number of patches.
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