Given two arrays **arr1[]** and **arr2[]** of length **M** and **N** consisting of digits **[0, 9]** representing two numbers and an integer **K**(**K ≤ M + N**), the task is to find the maximum K-digit number possible by selecting subsequences from the given arrays such that the relative order of the digits is the same as in the given array.

**Examples:**

Input:arr1[] = {3, 4, 6, 5}, arr2[] = {9, 1, 2, 5, 8, 3}, K = 5Output:98653Explanation:The maximum number that can be formed out of the given arrays arr1[] and arr2[] of length K is 98653.

Input:arr1[] = {6, 7}, arr2[] = {6, 0, 4}, K = 5Output:67604Explanation:The maximum number that can be formed out of the given arrays arr1[] and arr2[] of length K is 67604.

**Naive Approach:** The idea is to generate all subsequences of length **s1** from **arr1[]** and all subsequences of length **(K – s1)** from the array **arr2[]** over all values of **s1** in the range **[0, K]** and keep track of the maximum number so formed by merging both arrays in every iteration.

**Time Complexity:** O(2^{N})**Auxiliary Space:** O(1)

**Efficient Approach:** To optimize the above approach, the idea is to get the maximum number from the array **arr1[]** and of length **s1 **and maximum number from the array **arr2[]** and of length **(K – s1)**. Then, merge the two arrays to get the maximum number of length **K**. Follow the steps below to solve the given problem:

- Iterate over the range
**[0, K]**using the variable**i**and generate all possible decreasing subsequences of length**i**preserving the same order as in the array**arr1[]**and subsequences of length**(K – i)**following the same order as in the array**arr2[]**. - For generating decreasing subsequence of length
**L**of any array**arr[]**in the above step do the following:- Initialize an array
**ans[]**to store the subsequences of length**L**preserving the same order as in**arr[]**and Traverse the array**arr[]**and do the following:- Till the last element is less than the current element, then remove the last element from array
**ans[]**. - If the length of
**ans[]**is less than**L**then insert the current element in the**ans[]**.

- Till the last element is less than the current element, then remove the last element from array
- After the above steps, the array
**ans[]**in the resultant subsequence.

- Initialize an array
- While generating the subsequences of all possible length in
**Step 1**using the approach discussed in**Step 2**generate the maximum number by merging the two subsequences formed. - After the above steps, print that subsequence which gives maximum number after merging.

Below is the implementation of the above approach:

## Python3

`# Python program for the above approach ` ` ` `# Function to find maximum K-digit number ` `# possible from subsequences of the ` `# two arrays nums1[] and nums2[] ` `def` `maxNumber(nums1, nums2, k): ` ` ` ` ` `# Store lengths of the arrays ` ` ` `l1, l2 ` `=` `len` `(nums1), ` `len` `(nums2) ` ` ` ` ` `# Store the resultant subsequence ` ` ` `rs ` `=` `[] ` ` ` ` ` `# Function to calculate maximum ` ` ` `# number from nums of length c_len ` ` ` `def` `helper(nums, c_len): ` ` ` ` ` `# Store the resultant maximum ` ` ` `ans ` `=` `[] ` ` ` ` ` `# Length of the nums array ` ` ` `ln ` `=` `len` `(nums) ` ` ` ` ` `# Traverse the nums array ` ` ` `for` `idx, val ` `in` `enumerate` `(nums): ` ` ` `while` `ans ` `and` `ans[` `-` `1` `] < val ` `and` `ln` `-` `idx > c_len` `-` `len` `(ans): ` ` ` ` ` `# If true, then pop ` ` ` `# out the last element ` ` ` `ans.pop(` `-` `1` `) ` ` ` ` ` `# Check the length with ` ` ` `# required length ` ` ` `if` `len` `(ans) < c_len: ` ` ` ` ` `# Append the value to ans ` ` ` `ans.append(val) ` ` ` ` ` `# Return the ans ` ` ` `return` `ans ` ` ` ` ` `# Traverse and pick the maximum ` ` ` `for` `s1 ` `in` `range` `(` `max` `(` `0` `, k` `-` `l2), ` `min` `(k, l1)` `+` `1` `): ` ` ` ` ` `# p1 and p2 stores maximum number possible ` ` ` `# of length s1 and k - s1 from ` ` ` `# the arrays nums1[] & nums2[] ` ` ` `p1, p2 ` `=` `helper(nums1, s1), helper(nums2, k` `-` `s1) ` ` ` ` ` `# Update the maximum number ` ` ` `rs ` `=` `max` `(rs, [` `max` `(p1, p2).pop(` `0` `) ` `for` `_ ` `in` `range` `(k)]) ` ` ` ` ` `# Return the result ` ` ` `return` `rs ` ` ` ` ` `# Driver Code ` ` ` `arr1 ` `=` `[` `3` `, ` `4` `, ` `6` `, ` `5` `] ` `arr2 ` `=` `[` `9` `, ` `1` `, ` `2` `, ` `5` `, ` `8` `, ` `3` `] ` ` ` `K ` `=` `5` ` ` `# Function Call ` `print` `(maxNumber(arr1, arr2, K)) ` |

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**Output:**

[9, 8, 6, 5, 3]

**Time Complexity:** O(K*(M + N))**Auxiliary Space:** O(K)

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