Last updated on October 5th, 2024 at 09:27 pm

Here, We see Create Maximum Number LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Create Maximum Number LeetCode Solution

Problem Statement

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]

1. Create Maximum Number LeetCode Solution C++

class Solution {
  public: vector < int > maxNumber(vector < int > & nums1, vector < int > & nums2, int k) {
    int n1 = nums1.size(), n2 = nums2.size();
    vector < int > best;
    for (int k1 = max(k - n2, 0); k1 <= min(k, n1); ++k1)
      best = max(best, maxNumber(maxNumber(nums1, k1),
        maxNumber(nums2, k - k1)));
    return best;
  }
  vector < int > maxNumber(vector < int > nums, int k) {
    int drop = nums.size() - k;
    vector < int > out;
    for (int num: nums) {
      while (drop && out.size() && out.back() < num) {
        out.pop_back();
        drop--;
      }
      out.push_back(num);
    }
    out.resize(k);
    return out;
  }
  vector < int > maxNumber(vector < int > nums1, vector < int > nums2) {
    vector < int > out;
    while (nums1.size() + nums2.size()) {
      vector < int > & now = nums1 > nums2 ? nums1 : nums2;
      out.push_back(now[0]);
      now.erase(now.begin());
    }
    return out;
  }
};

2. Create Maximum Number LeetCode Solution Java

class Solution {
  public int[] maxNumber(int[] nums1, int[] nums2, int k) {
    int n = nums1.length;
    int m = nums2.length;
    int[] ans = new int[k];
    for (int i = Math.max(0, k - m); i <= k && i <= n; ++i) {
      int[] candidate = merge(maxArray(nums1, i), maxArray(nums2, k - i), k);
      if (greater(candidate, 0, ans, 0)) ans = candidate;
    }
    return ans;
  }
  private int[] merge(int[] nums1, int[] nums2, int k) {
    int[] ans = new int[k];
    for (int i = 0, j = 0, r = 0; r < k; ++r)
      ans[r] = greater(nums1, i, nums2, j) ? nums1[i++] : nums2[j++];
    return ans;
  }
  public boolean greater(int[] nums1, int i, int[] nums2, int j) {
    while (i < nums1.length && j < nums2.length && nums1[i] == nums2[j]) {
      i++;
      j++;
    }
    return j == nums2.length || (i < nums1.length && nums1[i] > nums2[j]);
  }
  public int[] maxArray(int[] nums, int k) {
    int n = nums.length;
    int[] ans = new int[k];
    for (int i = 0, j = 0; i < n; ++i) {
      while (n - i + j > k && j > 0 && ans[j - 1] < nums[i]) j--;
      if (j < k) ans[j++] = nums[i];
    }
    return ans;
  }
}

3. Create Maximum Number Solution JavaScript

var maxNumber = function(nums1, nums2, k) {
	const maxArray = (nums, k) => {
		const stack = [];
		let popCount = nums.length - k;
		for (const num of nums) {
			while (popCount > 0 && stack.length && stack[stack.length - 1] < num) {
				stack.pop();
				popCount--;
			}
			stack.push(num);
		}
		return stack.slice(0, k);
	};
	const mergeArrays = (arr1, arr2) => {
		const merged = [];
		while (arr1.length || arr2.length) {
			const bigger = arr1 > arr2 ? arr1 : arr2;
			merged.push(bigger.shift());
		}
		return merged;
	};
	let max = [];
	for (let i = Math.max(0, k - nums2.length); i <= Math.min(k, nums1.length); i++) {
		const merged = mergeArrays(maxArray(nums1, i), maxArray(nums2, k - i));
		if (merged > max) max = merged;
	}
	return max;
};

4. Create Maximum Number Solution Python

class Solution(object):
    def maxNumber(self, nums1, nums2, k):
        def prep(nums, k):
            drop = len(nums) - k
            out = []
            for num in nums:
                while drop and out and out[-1] < num:
                    out.pop()
                    drop -= 1
                out.append(num)
            return out[:k]
        def merge(a, b):
            return [max(a, b).pop(0) for _ in a + b]
        return max(
            merge(prep(nums1, i), prep(nums2, k - i))
            for i in range(k + 1)
            if i <= len(nums1) and k - i <= len(nums2)
        )