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 = 5
Output: 98653
Explanation: 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 = 5
Output: 67604
Explanation: 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(2N)
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[].
- After the above steps, the array ans[] in the resultant subsequence.
- 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:
- 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)) |
[9, 8, 6, 5, 3]
Time Complexity: O(K*(M + N))
Auxiliary Space: O(K)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
