Create Maximum Number in JavaScript (Time Complexity: O((m+n)^3))

You are given two integer arrays nums1 and nums2 of lengths m and n respectively. nums1 and nums2 represent the digits of two numbers. You are also given an integer k.

Create the maximum number of length k <= m + n from digits of the two numbers. The relative order of the digits from the same array must be preserved.

Return an array of the k digits representing the answer.

Example 1:

Input: nums1 = [3,4,6,5], nums2 = [9,1,2,5,8,3], k = 5
Output: [9,8,6,5,3]

Example 2:

Input: nums1 = [6,7], nums2 = [6,0,4], k = 5
Output: [6,7,6,0,4]

Example 3:

Input: nums1 = [3,9], nums2 = [8,9], k = 3
Output: [9,8,9]

Constraints:

m == nums1.length
n == nums2.length
1 <= m, n <= 500
0 <= nums1[i], nums2[i] <= 9
1 <= k <= m + n

Understanding the Problem

The core challenge of this problem is to create the maximum number of length k by selecting digits from two arrays while maintaining the relative order of digits from each array. This problem is significant in scenarios where we need to merge sorted lists or arrays while maintaining the order and maximizing the resultant value.

Potential pitfalls include not maintaining the relative order of digits from the same array and not considering all possible combinations of digits from both arrays.

Approach

To solve this problem, we need to consider the following steps:

Generate all possible subarrays of different lengths from both arrays.
Merge these subarrays to form the maximum number of length k.
Compare and select the maximum number from all possible combinations.

Let's start with a naive solution and then optimize it.

Naive Solution

The naive solution involves generating all possible subarrays of different lengths from both arrays and then merging them to form the maximum number. This approach is not optimal due to its high time complexity.

Optimized Solution

We can optimize the solution by using a greedy approach to generate the maximum subarray of a given length from an array. Then, we merge these subarrays to form the maximum number of length k.

Algorithm

Here is a step-by-step breakdown of the optimized algorithm:

Define a function to generate the maximum subarray of a given length from an array.
Define a function to merge two subarrays to form the maximum number while maintaining the relative order.
Iterate through all possible lengths of subarrays from both arrays and use the above functions to generate and merge subarrays.
Compare and select the maximum number from all possible combinations.

Code Implementation

/**
 * Function to generate the maximum subarray of a given length from an array
 * @param {number[]} nums - The input array
 * @param {number} k - The length of the subarray
 * @return {number[]} - The maximum subarray of length k
 */
function maxSubArray(nums, k) {
    const stack = [];
    let drop = nums.length - k;
    for (let num of nums) {
        while (drop && stack.length && stack[stack.length - 1] < num) {
            stack.pop();
            drop--;
        }
        stack.push(num);
    }
    return stack.slice(0, k);
}

/**
 * Function to merge two subarrays to form the maximum number
 * @param {number[]} nums1 - The first subarray
 * @param {number[]} nums2 - The second subarray
 * @return {number[]} - The merged maximum number
 */
function merge(nums1, nums2) {
    const result = [];
    while (nums1.length || nums2.length) {
        if (nums1 > nums2) {
            result.push(nums1.shift());
        } else {
            result.push(nums2.shift());
        }
    }
    return result;
}

/**
 * Main function to create the maximum number of length k from two arrays
 * @param {number[]} nums1 - The first input array
 * @param {number[]} nums2 - The second input array
 * @param {number} k - The length of the maximum number
 * @return {number[]} - The maximum number of length k
 */
function maxNumber(nums1, nums2, k) {
    let max = [];
    for (let i = Math.max(0, k - nums2.length); i <= Math.min(k, nums1.length); i++) {
        const candidate = merge(maxSubArray(nums1, i), maxSubArray(nums2, k - i));
        if (candidate > max) {
            max = candidate;
        }
    }
    return max;
}

// Example usage:
console.log(maxNumber([3, 4, 6, 5], [9, 1, 2, 5, 8, 3], 5)); // Output: [9, 8, 6, 5, 3]
console.log(maxNumber([6, 7], [6, 0, 4], 5)); // Output: [6, 7, 6, 0, 4]
console.log(maxNumber([3, 9], [8, 9], 3)); // Output: [9, 8, 9]

Complexity Analysis

The time complexity of the optimized solution is O((m+n)^3) due to the nested loops and comparisons. The space complexity is O(k) for storing the result.

Edge Cases

Potential edge cases include:

Arrays with all identical elements.
Arrays with only one element.
Arrays where k is equal to the sum of the lengths of both arrays.

Each algorithm handles these edge cases by ensuring the relative order of digits and considering all possible combinations.

Testing

To test the solution comprehensively, include a variety of test cases:

Simple cases with small arrays.
Complex cases with larger arrays and different values of k.
Edge cases as mentioned above.

Use testing frameworks like Jest or Mocha for automated testing.

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Break down the problem into smaller subproblems.
Think about the constraints and how they affect the solution.
Use a greedy approach to maximize the result.
Practice similar problems to improve problem-solving skills.

Conclusion

In this blog post, we discussed how to create the maximum number of length k from two arrays while maintaining the relative order of digits. We explored a naive solution and an optimized solution, provided a detailed algorithm, and analyzed the complexity. Understanding and solving such problems is crucial for improving problem-solving skills and preparing for coding interviews.

Additional Resources

For further reading and practice, consider the following resources: