# Program to Find Out the Number of Corrections to be Done to Fix an Equation in Python

Suppose we have a string s which represents an equation of the form x+y=z. We have to find the minimum number of digits that we need to add into s so the equation becomes true.

So, if the input is like s = '2+6=7', then the output will be 2.

We can turn the equation into "21+6=27" by inserting "1" and "2". So the total number of corrections required is 2.

To solve this, we will follow these steps −

• split s into parts based on "+" character, put left part into A and right part into rest

• split rest into parts based on "=" character, put left part into B and right part into C

• return dp(size of A - 1, size of B - 1, size of C - 1, 0)

• Define a function dp() . This will take i, j, k, carry

• if i <= -1 and j <= -1 and k <= -1, then

• return 0 if carry is same as 0 otherwise 1

• last1 := (A[i]) if i >= 0 otherwise 0

• last2 := (B[j]) if j >= 0 otherwise 0

• last3 := (C[k]) if k >= 0 otherwise 0

• prefix1 := (A[from index 0 to i + 1]) if i >= 0 otherwise 0

• prefix2 := (B[from index 0 to j + 1]) if j >= 0 otherwise 0

• prefix3 := (C[from index 0 to k + 1]) if k >= 0 otherwise 0

• if i <= -1 and j <= -1, then

• rhs := prefix3 - carry

• if rhs <= 0, then

• return |rhs|

• if i is same as -1 or j is same as -1, then

• return size of string rhs

• otherwise,

• return False

• if k <= -1, then

• return size of str(prefix1 + prefix2 + carry)

• ans := infinity

• carry2, lhs := return quotient and remainder dividing(carry + last1 + last2) by 10

• if lhs is same as last3, then

• ans := dp(i - 1, j - 1, k - 1, carry2)

• req := last3 - carry - last2

• extra_zeros := maximum of 0, -1 - i

• carry2 := 1 if req < 0 otherwise 0

• ans := minimum of ans, 1 + extra_zeros + dp(maximum of -1, i, j - 1, k - 1, carry2)

• req := last3 - carry - last1

• extra_zeros := maximum of 0, -1 - j

• carry2 := 1 if req < 0 otherwise 0

• ans = minimum of (ans, 1 + extra_zeros + dp(i - 1, max(-1, j), k - 1, carry2))

• carry2, lhs := return quotient and remainder dividing(last1 + last2 + carry) by 10

• ans := minimum of ans, 1 + dp(i - 1, j - 1, k, carry2)

• return ans

• from the main method return dp(size of A – 1, size of B – 1, size of C – 1, 0)

## Example

Let us see the following implementation to get better understanding −

Live Demo

class Solution:
def solve(self, s):
A, rest = s.split("+")
B, C = rest.split("=")
def dp(i, j, k, carry):
if i <= -1 and j <= -1 and k <= -1:
return 0 if carry == 0 else 1
last1 = int(A[i]) if i >= 0 else 0
last2 = int(B[j]) if j >= 0 else 0
last3 = int(C[k]) if k >= 0 else 0
prefix1 = int(A[: i + 1]) if i >= 0 else 0
prefix2 = int(B[: j + 1]) if j >= 0 else 0
prefix3 = int(C[: k + 1]) if k >= 0 else 0
if i <= -1 and j <= -1:
rhs = prefix3 - carry
if rhs <= 0:
return abs(rhs)
if i == -1 or j == -1:
return len(str(rhs))
else:
assert False
if k <= -1:
return len(str(prefix1 + prefix2 + carry))
ans = float("inf")
carry2, lhs = divmod(carry + last1 + last2, 10)
if lhs == last3:
ans = dp(i - 1, j - 1, k - 1, carry2)
req = last3 - carry - last2
extra_zeros = max(0, -1 - i)
carry2 = 1 if req < 0 else 0
ans = min(ans, 1 + extra_zeros + dp(max(-1, i), j - 1, k - 1, carry2))
req = last3 - carry - last1
extra_zeros = max(0, -1 - j)
carry2 = 1 if req < 0 else 0
ans = min(ans, 1 + extra_zeros + dp(i - 1, max(-1, j), k - 1, carry2))
carry2, lhs = divmod(last1 + last2 + carry, 10)
ans = min(ans, 1 + dp(i - 1, j - 1, k, carry2))
return ans
return dp(len(A) - 1, len(B) - 1, len(C) - 1, 0)

ob = Solution()
print (ob.solve('2+6=7'))

## Input

'2+6=7'

## Output

2