Patching Array

To solve this coding challenge, we need to determine the minimum number of patches (additions) required to make sure that any number in the range \[1, n\] can be formed by the sum of some elements in the sorted integer array
nums
. We'll look closely at each element in
nums
and keep track of the largest number we can form using elements from
nums
along with the added patches. The strategy can be summarized as follows:
  1. Initialize a variable to track the largest number we can form with the current elements and patches.
  2. Iterate through the array
    nums
    and use elements that do not exceed the largest number plus one.
  3. Whenever there is a gap (i.e., the current number in the array is greater than the largest number plus one), we patch by adding this largest number plus one to the set of elements.
  4. Repeat until we can form all numbers up to
    n
    .
    5. Let’s walk through this problem with an example to understand the detailed steps and then craft a pseudocode solution.

    Example Walkthrough

    Consider
    nums = [1, 3]
    and
    n = 6
    .
    Initial Setup
  5. covered
    (i.e., the largest number we can form) is initially
    0
    .
  6. We will iterate through
    nums
    and patch whenever necessary.
    7. Iteration Steps:
  7. First Iteration:
    • nums[0] = 1
    • Since
      1 <= 0 + 1
      , we can add
      1
      to the covered range.
    • Update
      covered
      to
      1
      (
      0 + 1
      ).
  8. Second Iteration:
    • nums[1] = 3
    • Since
      3 > 1 + 1
      , patch required to cover
      2
      .
    • Add
      2
      to the patched array, now
      covered
      becomes
      3
      (
      1 + 2
      ).
    • No need to increment
      i
      , because we haven't consumed
      3
      yet.
  9. Second Iteration (continued):
    • Now,
      3 <= 3
      , we can add
      3
      to the covered range.
    • Update
      covered
      to
      6
      (
      3 + 3
      ).
    At the end of these iterations, we can cover all numbers from
    1
    to
    6
    using
    [1, 2, 3]
    (with one patch
    2
    ). Hence, the output is
    1
    .

    Detailed Steps in Pseudocode

                                                
# Initialize the count of patches needed
patches = 0
# Initialize the largest number that can be covered with current set of numbers and patches
covered = 0
# Initialize index to iterate through nums
index = 0

# While the largest covered number is less than n
while covered < n:
# Check if the current element in nums can extend the range
if index < length of nums and nums[index] <= covered + 1:
# Extend the covered range by the current element in nums
covered = covered + nums[index]
# Move to the next element in nums
index = index + 1
else:
# Need to patch, extend the range by adding covered + 1
patches = patches + 1
covered = covered + (covered + 1)

# Return the total number of patches needed
return patches

    Explanation

  10. Initialization:
    • patches
      is set to
      0
      to count the number of patches.
    • covered
      is set to
      0
      to keep track of the range of numbers that can be formed.
    • index
      is set to
      0
      to traverse the
      nums
      array.
  11. Loop until
    covered
    is less than
    n
    :
    • If the current element in
      nums
      at
      index
      can extend our range (
      nums[index] <= covered + 1
      ), add it to
      covered
      and move to the next element.
    • If the current element cannot extend our range, add a new patch (
      covered + 1
      ), increment the count of
      patches
      , and update
      covered
      .
  12. Termination:
    • When
      covered
      is no longer less than
      n
      , it guarantees that we can form every number in the range
      [1, n]
      .
This thorough methodology ensures that we always minimally patch the array wherever there is a gap preventing us from covering range
[1, n]
.