Max count of N using digits of M such that 2 and 5, and, 6 and 9 can be treated as same respectively

Given an integer N, and the string integer M, the task is to find the total count to make N by using the digits of string M. Also, digit 2 can be treated as digit 5, and digit 6 can be treated as digit 9 and vice versa and each digit from the string M can be used at most once.

Examples:

Input: N = 6, M = "245769"
Output: 2
Explanation: Digits 5 and 6 are used to form the number 56. iop[The digits 2 and 9 are used to form the number 56. As 2 is treated as 5 and 9 is treated as 6.

Input: N = 25, M = "55"
Output: 1

Approach: The given problem can be solved by Hashing. Follow the steps below to solve the problem:

• Create an empty hashmap, say map to store the frequency of the digits of the given string M.
• Create a variable, say, len to store the length of the string.
• Traverse the given string S using the variable i and Iterate until the value of i is less than len and perform the following steps:
  • If the character S[i] is equal to '5', then change it to '2'.
  • If the character S[i] is equal to '9', then change it to '6'.
  • If the character is present in the mymap, then change the frequency as mymap.put(x, map.get(x)+1).
  • Otherwise, insert the character in the map with frequency 1 as mymap.put(x, 1).
  • After adding the frequency to the map, increment i and continue to the next iteration.
• Create an empty hashmap, say rems to store the digits of the number N.
• Iterate until the value of N is greater than 0, and perform the following steps:
  • Create a variable, say rem to store the last digit of N by using the modulus operator as N%10.
  • If rem is equal to 5, then change it to 2.
  • If rem is equal to 9, then change it to 6.
  • If the rem is present in the rems map, then increase the frequency by 1as rems.put(rem, rems.get(rem)+1).
  • Otherwise, insert it to the rems map as rems.put(rem, 1).
  • Divide N by 10.
• Create a variable, say cnt to store the maximum count of the number N that can be formed using the given digits of string M.
• Traverse through the map rems, and perform the following steps:
  • Let each object in the map is ele.
  • Check if the key from ele is present in the frequency map of string mymap.
  • If not present, the return 0 (The number N cannot be formed if a digit from N is not present in string M).
  • Calculate the count by dividing the frequency of the key in mymap with the frequency in rems map as mymap.get(key)/ele.getValue().
  • Update the minimum value from all iterations in cnt. 
• After completing the above steps, print the value of cnt as the result.

Below is the implementation of the above approach:

2

Time Complexity: O(N)
Auxiliary Space: O(1)