Program to find minimum digits sum of deleted digits in Python

Suppose we have two strings s and t of digits, we have to find a way to remove digits in the strings so that: 1. Two strings are same 2. The sum of the digits that are deleted is minimized Finally return the minimized sum.

So, if the input is like s = "41272" t = "172", then the output will be 6, as we can remove "4" and "2" from the first string to get "172".

To solve this, we will follow these steps −

• Define a function lcs() . This will take a, b, m, n

• table := a 2d matrix of size (n + 1) x (m + 1) and fill with 0

• for i in range 1 to m, do

  • for j in range 1 to n, do

    • if a[i - 1] is same as b[j - 1], then

      • table[i, j] := table[i - 1, j - 1] + 2 *(ASCII of a[i - 1] - 48)

    • otherwise,

      • table[i, j] = maximum of table[i - 1, j] and table[i, j - 1])

• return table[m, n]

• From the main method do the following

• m := size of a, n := size of b

• c := 0

• for i in range 0 to m, do

  • c := c + ASCII of a[i] - 48

• for i in range 0 to n, do

  • c := c + ASCII of b[i] - 48

• result := c - lcs(a, b, m, n)

• return result

Example (Python)

Let us see the following implementation to get better understanding −

 Live Demo

class Solution:    def lcs(self, a, b, m, n):       table = [[0 for i in range(n + 1)] for j in range(m + 1)]       for i in range(1, m + 1):          for j in range(1, n + 1):             if a[i - 1] == b[j - 1]:                table[i][j] = table[i - 1][j - 1] + 2 * (ord(a[i - 1]) - 48)             else:                table[i][j] = max(table[i - 1][j], table[i][j - 1])       return table[m][n]    def solve(self, a, b):       m = len(a)       n = len(b)       c = 0       for i in range(m):          c += ord(a[i]) - 48       for i in range(n):          c += ord(b[i]) - 48       result = c - self.lcs(a, b, m, n)       return result ob = Solution() s = "41272" t = "172" print(ob.solve(s, t))

Input

"41272", "172"

Output

6