Leetcode problems explained - 330 - Patching Array

Leetcode problems solved and explained for beginners

In this series I’m going to solve Leetcode and explain, step by step my solution and the reason behind it.

Today’s problem: 330. Patching Array

Given a sorted integer array and a target integer ( n ), how can you modify the array so that you can form every integer between 1 and ( n ) using the sum of some elements from the array?

In other words: Imagine you’re in a kitchen, and your goal is to make dishes that require a certain range of ingredients, specifically every combination of ingredients from 1 to n. You have a box of sorted spices, each marked with a number indicating how many dishes it can contribute to. But, there might be some gaps in what you can make with them—from a dish needing just 1 spice to one needing n spices altogether.

You need to determine the fewest number of additional spices (let’s call these “patches“) you need to buy so that you can confidently cook every single dish requiring anywhere between 1 and n spices. The spices you buy should effectively fill in the gaps left by your current spice collection.

This modification involves adding the minimum number of integers (or patches). We’ll break down an efficient solution to this problem, step by step, making it accessible even if you’re just starting out with coding!

The Problem Statement

Suppose you are given a sorted array nums of integers and another integer ( n ). Your task is to add the fewest number of integers to nums so that you can create every integer from 1 to ( n ) as the sum of some elements of the modified array.

Example:

Input: nums = [1, 3] , ( n = 6 )
Output: 1 (You need to add the number 2 to cover all sums from 1 to 6)

Breaking Down the Solution

Let’s dissect an efficient approach using a simple example. We’ll use the array nums = [1, 3] and ( n = 6 ) to illustrate the process.

Key Variables

sum = 0: This will keep track of the maximum sum we can achieve with the current elements and patches.
additions = 0: A counter to track the number of patches (sauces) we add.
i = 0: Index to iterate through the array nums.

The Algorithm

Here’s a JavaScript function that implements our strategy:

function minPatches(nums, n) {
    let additions = 0, i = 0, sum = 0;

    while (sum < n) {
        if (i < nums.length && nums[i] <= sum + 1) {
            sum += nums[i], i++;
        }
        else {
            sum += sum + 1;
            additions++;
        }
    }
    return additions;
}

Step-by-Step Explanation

1 – Initialize variables

sum is initialized to 0 — it represents the maximum sum that can be achieved with the given nums and any patches.
additions starts at 0, counting how many numbers we need to add to nums.
i will traverse through each element in nums.

2 – Start Looping Until Coverage:

The loop continues as long as sum is less than ( n ). If sum is ( n ) or more, it means we can already achieve every number up to ( n ).

3 – Use Existing Numbers or Add Patches:

If the current number nums[i] can be used to extend the sum without missing any numbers (nums[i] <= sum + 1), we add it to sum.
If not, we need to patch. The optimal number is always sum + 1 because it exactly fills the gap needed to continue.

4 – Update sum and additions

Whether we use a number from nums or add a patch, we update sum accordingly.
If a patch is added, increment the additions counter.

5 – Complete and return

Once the loop concludes, additions will hold the total number of patches added to achieve the goal.

Why This Approach Is Efficient

Minimal Operations: The function smartly chooses between using an existing number and adding a minimal patch to ensure all numbers up to ( n ) can be formed.
Optimal Patching: By adding sum + 1, we guarantee no gaps in achievable sums, minimizing the total patches required.
Greedy Strategy: This solution uses a greedy algorithm, which always takes the best immediate choice to solve part of the problem, often leading to an overall optimal solution.

Time Efficiency

While Loop Condition (sum < n): The loop continues until the sum of achievable numbers meets or exceeds n.

Array Utilization (i < nums.length && nums[i] <= sum + 1): Each element in the array is considered once. This part of the condition contributes O(m) time complexity where m is the number of elements in nums.

Doubling Through Patches: Whenever a patch is added, sum effectively doubles or more. This means the value of sum grows exponentially towards n.

Exponential Growth and Doubling Effect: The doubling effect of adding sum + 1 (when nums[i] is unavailable or too large) is crucial because it means the number of patches k needed grows logarithmically relative to n. This is because each patch added exponentially increases the range of achievable sums, significantly reducing the frequency of needed patches as sum grows

Mathematical Implication: If you start at 1 and keep doubling, you will reach or exceed any number n in log2(n) steps. Thus, the addition of patches, which effectively doubles the sum, implies that the number of patches (or doubling steps) required is logarithmic relative to n.

Space Complexity

O(1)

Since the space required does not grow with the size of the input (n or the length of nums), the space complexity is constant, O(1). This means the algorithm needs a fixed amount of memory regardless of the input size.

Running the Function

Here’s how you can run the above function with an example:

console.log(minPatches([1, 3], 6)); // Output: 1

Conclusion

Understanding this problem helps in grasping how greedy algorithms can be applied to real-world scenarios effectively. By iterating through the solution and making the best possible decision at each step, we ensure that our program runs efficiently and correctly. The array patching problem isn’t just a test of coding skill but also of problem-solving strategy and understanding

the underlying logic and mathematics.

Feel free to experiment with other examples to see how the function behaves and possibly refine it further! Happy coding!