grid47 Exploring patterns and algorithms
Sep 4, 2024
8 min read
Solution to LeetCode 640: Solve the Equation Problem
Given an equation with a variable ‘x’ and basic arithmetic operations (’+’, ‘-’), find the value of ‘x’. The equation might have a solution, no solution, or infinite solutions. Your task is to return the correct result in the form of a string.
Problem
Approach
Steps
Complexity
Input: The input consists of a string representing an equation with the variable 'x', integers, and operations (+, -). The equation contains exactly one '=' symbol, and the equation consists of integer coefficients and variables.
Example: "x+5-3+x=6+x-2"
Constraints:
• 3 <= equation.length <= 1000
• equation has exactly one '='.
• equation contains integers with absolute values between 0 and 100, and the variable 'x'.
Output: The output is a string representing the result of the equation. If there is a solution, the format is 'x=#value'. If there is no solution, return 'No solution'. If the equation has infinite solutions, return 'Infinite solutions'.
Example: "x=2"
Constraints:
• The output is always in the form 'x=#value', 'No solution', or 'Infinite solutions'.
Goal: To solve the equation for the value of 'x'. If there are multiple solutions or no solutions, return the appropriate message.
Steps:
• Parse the equation into two parts (before and after the equal sign).
• Group terms with 'x' and constant terms separately.
• Solve for 'x' by isolating it on one side of the equation.
• Handle special cases such as infinite solutions or no solution.
Goal: The input equation follows the structure described, and it is guaranteed to contain at least one term involving 'x'.
Steps:
• 3 <= equation.length <= 1000
• The input equation contains integers and the variable 'x'.
Assumptions:
• The equation will always contain at least one 'x'.
• There will always be exactly one '=' in the equation.
• Input: "x+5-3+x=6+x-2"
• Explanation: Simplifying the equation results in '2x+2=6+x-2'. After solving, we get 'x=2'.
• Input: "x=x"
• Explanation: This equation simplifies to '0=0', which is true for any value of x. Therefore, the solution is 'Infinite solutions'.
Approach: The equation needs to be parsed, and terms involving 'x' should be grouped separately. After simplifying, if there are no 'x' terms, check if the constants are equal, which indicates infinite solutions or no solution. If there are 'x' terms, solve for 'x'.
Observations:
• The equation might contain positive and negative terms involving 'x'.
• The variable 'x' can be isolated by grouping terms involving 'x' and constants separately.
• We will need to carefully handle terms on both sides of the equation and consider edge cases like 'x=x'.
Steps:
• Parse the equation into parts separated by the '=' symbol.
• Separate terms involving 'x' from constant terms.
• Sum up the coefficients of 'x' and constants on both sides.
• If there is no 'x' term, check if the constants are equal or not for infinite solutions or no solution.
• If 'x' terms are present, solve for 'x' by isolating it.
Empty Inputs:
• Empty equations are not allowed.
Large Inputs:
• Equations with large strings can still be handled by parsing and evaluating the terms in a linear manner.
Special Values:
• Special cases like 'x=x' should be handled separately to detect infinite solutions.
Constraints:
• The equation will not contain leading zeros.
string solveEquation(string eqn) {
int i =0;
int para =0, xpara =0;
int flag =1;
while(i < eqn.size()) {
int sgn =1;
int tmp =0;
if(eqn[i] =='=') {
flag =-1;
i++;
}
if(eqn[i] =='-') {
sgn =-1;
i++;
}
if(eqn[i] =='+') {
sgn =1;
i++;
}
if(isdigit(eqn[i])) {
while(i < eqn.size() && isdigit(eqn[i])) {
tmp = tmp *10+ eqn[i] -'0';
i++;
}
if(i < eqn.size() && eqn[i] =='x') {
xpara += flag * sgn * tmp;
i++;
}
else para += flag * sgn * tmp;
} else {
xpara += flag * sgn;
i++;
}
}
string res;
cout << para <<'-'<< xpara;
if(para ==0&& xpara ==0)
res ="Infinite solutions";
elseif (xpara ==0)
res ="No solution";
else res ="x="+ to_string(para/xpara*-1);
return res;
}
1 : Helper Function
string solveEquation(string eqn) {
This is the main function which solves the equation represented as a string.
2 : Variable Initialization
int i =0;
Initializes the variable 'i' to track the index as we iterate over the equation.
3 : Variable Initialization
int para =0, xpara =0;
Initializes 'para' and 'xpara' to store the constants and coefficients of 'x' respectively.
4 : Variable Initialization
int flag =1;
Initializes the 'flag' variable to keep track of the sign for each term.
5 : Loop
while(i < eqn.size()) {
Begins a loop that will iterate over the entire equation string.
6 : Variable Initialization
int sgn =1;
Initializes the 'sgn' variable to 1, representing a positive sign for the current term.
7 : Variable Initialization
int tmp =0;
Initializes 'tmp' to accumulate the value of the current number in the equation.
8 : Conditional Check
if(eqn[i] =='=') {
Checks if the current character is the equal sign, indicating the start of the second half of the equation.
9 : Update Variables
flag =-1;
Changes the flag to -1, indicating that the terms after the equal sign will have a negative sign.
10 : Increment Index
i++;
Increments the index to move past the '=' character.
11 : Conditional Check
if(eqn[i] =='-') {
Checks if the current character is a minus sign, indicating a negative term.
12 : Update Variables
sgn =-1;
Sets the 'sgn' variable to -1, indicating that the current term will be negative.
13 : Increment Index
i++;
Increments the index to move past the minus sign.
14 : Conditional Check
if(eqn[i] =='+') {
Checks if the current character is a plus sign, indicating a positive term.
15 : Update Variables
sgn =1;
Sets the 'sgn' variable to 1, indicating that the current term will be positive.
16 : Increment Index
i++;
Increments the index to move past the plus sign.
17 : Conditional Check
if(isdigit(eqn[i])) {
Checks if the current character is a digit, indicating the start of a number.
18 : Loop through Digits
while(i < eqn.size() && isdigit(eqn[i])) {
Begins a loop to accumulate the digits of the current number.
19 : Accumulate Number
tmp = tmp *10+ eqn[i] -'0';
Accumulates the current digit into the 'tmp' variable.
20 : Increment Index
i++;
Increments the index to move to the next character.
21 : Conditional Check
if(i < eqn.size() && eqn[i] =='x') {
Checks if the next character is 'x', indicating that the current number is a coefficient of 'x'.
22 : Update Coefficient
xpara += flag * sgn * tmp;
Updates the coefficient of 'x' based on the current sign and flag.
23 : Increment Index
i++;
Increments the index to move past the 'x'.
24 : Else Condition
else para += flag * sgn * tmp;
If the number is not associated with 'x', it is added to the 'para' variable.
25 : Else Condition
} else {
Handles the case where the character is neither a digit nor 'x'.
26 : Update Coefficient
xpara += flag * sgn;
If the character is not a digit, it implies that 'x' has a coefficient of 1 (or -1).
27 : Increment Index
i++;
Increments the index to move past the current character.
28 : Final Calculation
string res;
Declares a string variable 'res' to store the result.
29 : Print Debug Information
cout << para <<'-'<< xpara;
Prints the values of 'para' and 'xpara' for debugging purposes.
30 : Conditional Check
if(para ==0&& xpara ==0)
Checks if both 'para' and 'xpara' are zero, which indicates infinite solutions.
31 : Assign Result
res ="Infinite solutions";
Assigns the string 'Infinite solutions' to 'res'.
32 : Conditional Check
elseif (xpara ==0)
Checks if 'xpara' is zero, which indicates no solution.
33 : Assign Result
res ="No solution";
Assigns the string 'No solution' to 'res'.
34 : Assign Result
else res ="x="+ to_string(para/xpara*-1);
If a solution exists, calculates and assigns the value of 'x' to 'res'.
35 : Return Result
return res;
Returns the final result.
Best Case: O(n), where n is the length of the equation.
Average Case: O(n), as we parse through the equation only once.
Worst Case: O(n), where n is the length of the equation.
Description: The time complexity is linear since we only need to parse the equation once.
Best Case: O(1), if the equation is small.
Worst Case: O(1), as we do not need additional data structures to store intermediate results.
Description: The space complexity is constant since we are only using a few variables to store intermediate results.
