# Find the missing digit x from the given expression

• Difficulty Level : Easy
• Last Updated : 06 May, 2022

Given an alphanumeric string, consisting of a single alphabet X, which represents an expression of the form:

A operator B = C where A, B and C denotes integers and the operator can be either of +, -, * or /

The task is to evaluate the missing digit X present in any of the integers A, B and C such that the given expression holds to be valid.

Examples:

Input: S = “3x + 12 = 46”
Output: 4
Explanation:
If we subtract 12 from 46, we will get 34.
So, on comparing 3x and 34. the value of x = 4

Input: S = “4 – 2 = x”
Output: 2
Explanation:
After solving the equation, the value of x = 2.

Approach: Follow the steps below to solve the problem:

• Split the string to extract the two operands, operator and the resultant.
• Check if X is present in the resultant or not. If so, then compute the value of the resultant by applying operations on the first operand and second operand with the operator.
• Otherwise, if X is not present in the resultant. Then check if X is present in the first operand. If so, then apply the operation on the second operand and resultant with the operator.
• Otherwise, if X is not present in the first operand also. Then check if X is present in the second operand. If so, then apply the operation on the first operand and resultant with the operator.

Below is the implementation of the above approach:

## Python3

 `# Python3 program to find missing``# digit x in expression`  `def` `MissingDigit(exp):``  ` `    ``# Split the expression to``    ``# extract operands, operator``    ``# and resultant``    ``exp ``=` `list``(exp.split())` `    ``first_operand ``=` `exp[``0``]``    ``operator ``=` `exp[``1``]``    ``second_operand ``=` `exp[``2``]``    ``resultant ``=` `exp[``-``1``]` `    ``# If x is present in resultant``    ``if` `'x'` `in` `resultant:``        ``x ``=` `resultant``        ``first_operand ``=` `int``(first_operand)``        ``second_operand ``=` `int``(second_operand)` `        ``if` `operator ``=``=` `'+'``:``            ``res ``=` `first_operand ``+` `second_operand``        ``elif` `operator ``=``=` `'-'``:``            ``res ``=` `first_operand ``-` `second_operand``        ``elif` `operator ``=``=` `'*'``:``            ``res ``=` `first_operand ``*` `second_operand``        ``else``:``            ``res ``=` `first_operand ``/``/` `second_operand` `     ``# If x in present in operands``    ``else``:` `        ``resultant ``=` `int``(resultant)` `        ``# If x in the first operand``        ``if` `'x'` `in` `first_operand:` `            ``x ``=` `first_operand``            ``second_operand ``=` `int``(second_operand)` `            ``if` `operator ``=``=` `'+'``:``                ``res ``=` `resultant ``-` `second_operand``            ``elif` `operator ``=``=` `'-'``:``                ``res ``=` `resultant ``+` `second_operand``            ``elif` `operator ``=``=` `'*'``:``                ``res ``=` `resultant ``/``/` `second_operand``            ``else``:``                ``res ``=` `resultant ``*` `second_operand` `        ``# If x is in the second operand``        ``else``:` `            ``x ``=` `second_operand``            ``first_operand ``=` `int``(first_operand)` `            ``if` `operator ``=``=` `'+'``:``                ``res ``=` `resultant``-``first_operand``            ``elif` `operator ``=``=` `'-'``:``                ``res ``=` `first_operand ``-` `resultant``            ``elif` `operator ``=``=` `'*'``:``                ``res ``=` `resultant ``/``/` `first_operand``            ``else``:``                ``res ``=` `first_operand ``/``/` `resultant` `    ``res ``=` `str``(res)``    ``k ``=` `0``    ``for` `i ``in` `x:``        ``if` `i ``=``=` `'x'``:``            ``result ``=` `res[k]``            ``break``        ``else``:``            ``k ``=` `k ``+` `1` `    ``return` `result`  `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``# input expression``    ``exp ``=` `"3x + 12 = 46"` `    ``print``(MissingDigit(exp))`

## Javascript

 ``

Output:

`4`

Time Complexity: O(L), where is the length of the equation.
Auxiliary Space: O(1)

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