# Leetcode: Solve the Equation

Solve the Equation Similar Problems:

Solve a given equation and return the value of x in the form of string “x=#value”. The equation contains only ‘+’, ‘-‘ operation, the variable x and its coefficient.

If there is no solution for the equation, return “No solution”.

If there are infinite solutions for the equation, return “Infinite solutions”.

If there is exactly one solution for the equation, we ensure that the value of x is an integer.

Example 1:

```Input: "x+5-3+x=6+x-2"
Output: "x=2"
```

Example 2:

```Input: "x=x"
Output: "Infinite solutions"
```

Example 3:

```Input: "2x=x"
Output: "x=0"
```

Example 4:

```Input: "2x+3x-6x=x+2"
Output: "x=-1"
```

Example 5:

```Input: "x=x+2"
Output: "No solution"
```

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

```## Blog link: https://code.dennyzhang.com/solve-the-equation
## Basic Ideas:
##
##  2x+3x-6x=x+2
##  ["2x", "+3x", "-6x", "-x", "-2"]
##
## Assumptions: equation has valid format
##
## Complexity: Time O(n), Space O(n)
class Solution:
def solveEquation(self, equation):
"""
:type equation: str
:rtype: str
"""
queue = []
reverse = False
myStr, sign = '', ''
for ch in equation:
if ch == '=':
queue.append("%s%s" % ('-' if sign == '-' else '', myStr))
myStr, sign = '', ''

reverse = True
continue

if ch in '+-':
if myStr == '':
if (ch == '-' and reverse is False) or \
(ch == '+' and reverse is True):
sign = '-'
else:
sign = '+'
continue
if sign == '' and reverse is True: sign = '-'
queue.append("%s%s" % ('-' if sign == '-' else '', myStr))

# get new item
myStr, sign = '', ''
if (ch == '-' and reverse is False) or \
(ch == '+' and reverse is True):
sign = '-'
else:
sign = '+'
continue
myStr += ch

if myStr != '':
if sign == '' and reverse is True: sign = '-'
queue.append("%s%s" % ('-' if sign == '-' else '', myStr))

# print(queue)
x, v = 0, 0
for item in queue:
if item[-1] == 'x':
if item == '-x':
x -= 1
elif item == 'x':
x += 1
else:
x += int(item[0:-1])
else:
v += int(item)
v = -v
if x == 0:
if v != 0: return "No solution"
else: return "Infinite solutions"
else:
return "x=%d" % (int(v/x))

# s = Solution()
# print(s.solveEquation("-x=-1")) # x=1
# print(s.solveEquation("2x+3x-6x=x+2")) # x=-1
# print(s.solveEquation("x=x+2")) # "No solution"
# print(s.solveEquation("x=x")) # "Infinite solutions"
```

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