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"
Javascript solution using regex below:-
<html>
<input type = "label"/>Enter expression of the form x+4x+5-6=8+x
<input type = "text" id="equation" maxlength="100" placeholder="Enter expression"/><br/>
<input type = "button" value="compute" onclick="solveEquation(document.getElementById('equation').value)"/>
</html>
<script>
/**
var lhs = arr[0],
rhs = arr[1];
var lhsxcoeff, lhsconstant, rhsxcoeff, rhsconstant;
// if LHS equals RHS
if (lhs === rhs)
return "Infinite solutions";
// parse and extract lhs x coefficient and constant
[lhsxcoeff, lhsconstant] = parse(lhs);
// parse and extract rhs x coefficient and constant
[rhsxcoeff, rhsconstant] = parse(rhs);
// compute the final x coefficient by substracting RHS x coeff. from LHS x coeff.
var finalXcoeff = lhsxcoeff - rhsxcoeff;
// if the RHS and LHS coeffcients and constants match respectively
if (lhsxcoeff === rhsxcoeff && lhsconstant === rhsconstant)
return "Infinite solutions";
// if there is no final x coefficient left say for e.g. in equations like x = x + 2;
if (finalXcoeff === 0)
return "No solution";
// final case compute the value of x
if (finalXcoeff >= 1 || finalXcoeff < 0) {
return 'x=' + ((rhsconstant - lhsconstant) / finalXcoeff);
}
};
// parse equation (LHS and RHS) using regex
function parse(expression) {
var arr = expression.match(/-?(\d*x|\d+)/g);
var finalXCoefficient = 0;
var constantSum = 0;
for (var i = 0; i < arr.length; i++) {
if (arr[i].match(/x/g)) {
var xcoefficient = arr[i].substring(0, arr[i].length - 1);
if (xcoefficient === '-')
xcoefficient = -1;
if (xcoefficient === '')
xcoefficient = 1;
finalXCoefficient += parseInt(xcoefficient);
} else
constantSum += parseInt(arr[i]);
}
return [finalXCoefficient, constantSum];
}
</script>
Explanation:
The time complexity of the above algorithm is in the order of O(N).
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