Dada una ecuación lineal, la tarea es encontrar el valor de la variable utilizada. La ecuación contiene solo la operación ‘+’, ‘-‘, la variable y su coeficiente.
- Si no hay solución para la ecuación, devuelve «Sin solución».
- Si hay soluciones infinitas para la ecuación, devuelve «Soluciones infinitas».
- Si hay exactamente una solución para la ecuación, asegúrese de que el valor de x sea un número entero.
Ejemplos:
Input : "x + 5 - 3 + x = 6 + x - 2" Output : "x = 2" Input : "x = x" Output : "Infinite solutions" Input: "2x = x" Output: "x = 0" Input: "x = x + 2" Output: "No solution"
Enfoque: La idea es utilizar dos punteros para actualizar dos parámetros: el coeficiente de la variable utilizada y la suma total. En el lado izquierdo y derecho de ‘=’, use signos opuestos para cada número que está a cargo de una variable de signo, que cambiará una vez que se vea ‘=’.
Ahora, en el caso de una solución única, la relación entre el total efectivo y el coeficiente da el resultado requerido. En el caso de soluciones infinitas, tanto el total efectivo como el coeficiente resultan ser cero, por ejemplo, x + 1 = x + 1. En caso de que no haya solución, el coeficiente de x resulta ser cero, pero el total efectivo es distinto de cero. .
C++
// CPP program to solve the given equation
#include <bits/stdc++.h>
using namespace std;
// Function to solve the given equation
string solveEquation(string equation)
{
int n = equation.size(), sign = 1, coeff = 0;
int total = 0, i = 0;
// Traverse the equation
for (int j = 0; j < n; j++) {
if (equation[j] == '+' || equation[j] == '-') {
if (j > i)
total += sign * stoi(equation.substr(i, j - i));
i = j;
}
// For cases such as: x, -x, +x
else if (equation[j] == 'x') {
if ((i == j) || equation[j - 1] == '+')
coeff += sign;
else if (equation[j - 1] == '-')
coeff -= sign;
else
coeff += sign * stoi(equation.substr(i, j - i));
i = j + 1;
}
// Flip sign once '=' is seen
else if (equation[j] == '=') {
if (j > i)
total += sign * stoi(equation.substr(i, j - i));
sign = -1;
i = j + 1;
}
}
// There may be a number left in the end
if (i < n)
total += sign * stoi(equation.substr(i));
// For infinite solutions
if (coeff == 0 && total == 0)
return "Infinite solutions";
// For no solution
if (coeff == 0 && total)
return "No solution";
// x = total sum / coeff of x
// '-' sign indicates moving
// numeric value to right hand side
int ans = -total / coeff;
return "x=" + to_string(ans);
}
// Driver code
int main()
{
string equation = "x+5-3+x=6+x-2";
cout << solveEquation(equation);
return 0;
}
Java
// Java program to solve
// the given equation
import java.io.*;
class GFG
{
// Function to solve
// the given equation
static String solveEquation(String equation)
{
int n = equation.length(),
sign = 1, coeff = 0;
int total = 0, i = 0;
// Traverse the equation
for (int j = 0; j < n; j++)
{
if (equation.charAt(j) == '+' ||
equation.charAt(j) == '-')
{
if (j > i)
total += sign *
Integer.parseInt(
equation.substring(i, j));
i = j;
}
// For cases such
// as: x, -x, +x
else if (equation.charAt(j) == 'x')
{
if ((i == j) ||
equation.charAt(j - 1) == '+')
coeff += sign;
else if (equation.charAt(j - 1) == '-')
coeff -= sign;
else
coeff += sign *
Integer.parseInt(
equation.substring(i, j));
i = j + 1;
}
// Flip sign once
// '=' is seen
else if (equation.charAt(j) == '=')
{
if (j > i)
total += sign *
Integer.parseInt(
equation.substring(i, j));
sign = -1;
i = j + 1;
}
}
// There may be a
// number left in the end
if (i < n)
total = total +
sign *
Integer.parseInt(
equation.substring(i));
// For infinite
// solutions
if (coeff == 0 &&
total == 0)
return "Infinite solutions";
// For no solution
if (coeff == 0 &&
total != 0)
return "No solution";
// x = total sum / coeff
// of x '-' sign indicates
// moving numeric value to
// right hand side
int ans = -total / coeff;
return (Integer.toString(ans));
}
// Driver code
public static void main(String args[])
{
String equation = new String("x+5-3+x=6+x-2");
System.out.print("x = " +
solveEquation(equation));
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)
Python3
# Python program to solve
# the given equation
# def to solve
# the given equation
def solveEquation(equation) :
n = len(equation)
sign = 1
coeff = 0
total = 0
i = 0
# Traverse the equation
for j in range(0, n) :
if (equation[j] == '+' or
equation[j] == '-') :
if (j > i) :
total = (total + sign *
int(equation[i: j]))
i = j
# For cases such
# as: x, -x, +x
elif (equation[j] == 'x') :
if ((i == j) or
equation[j - 1] == '+') :
coeff += sign
elif (equation[j - 1] == '-') :
coeff = coeff - sign
else :
coeff = (coeff + sign *
int(equation[i: j]))
i = j + 1
# Flip sign once
# '=' is seen
elif (equation[j] == '=') :
if (j > i) :
total = (total + sign *
int(equation[i: j]))
sign = -1
i = j + 1
# There may be a number
# left in the end
if (i < n) :
total = (total + sign *
int(equation[i: len(equation)]))
# For infinite solutions
if (coeff == 0 and
total == 0) :
return "Infinite solutions"
# For no solution
if (coeff == 0 and total) :
return "No solution"
# x = total sum / coeff of x
# '-' sign indicates moving
# numeric value to right hand side
ans = -total / coeff
return int(ans)
# Driver code
equation = "x+5-3+x=6+x-2"
print ("x = {}" .
format(solveEquation(equation)))
# This code is contributed by
# Manish Shaw(manishshaw1)
C#
// C# program to solve
// the given equation
using System;
class GFG
{
// Function to solve
// the given equation
static string solveEquation(string equation)
{
int n = equation.Length,
sign = 1, coeff = 0;
int total = 0, i = 0;
// Traverse the equation
for (int j = 0; j < n; j++)
{
if (equation[j] == '+' ||
equation[j] == '-')
{
if (j > i)
total += sign *
Int32.Parse(
equation.Substring(i, j - i));
i = j;
}
// For cases such
// as: x, -x, +x
else if (equation[j] == 'x')
{
if ((i == j) ||
equation[j - 1] == '+')
coeff += sign;
else if (equation[j - 1] == '-')
coeff -= sign;
else
coeff += sign *
Int32.Parse(
equation.Substring(i, j - i));
i = j + 1;
}
// Flip sign once
// '=' is seen
else if (equation[j] == '=')
{
if (j > i)
total += sign *
Int32.Parse(
equation.Substring(i, j - i));
sign = -1;
i = j + 1;
}
}
// There may be a
// number left in the end
if (i < n)
total += sign *
Int32.Parse(
equation.Substring(i));
// For infinite
// solutions
if (coeff == 0 && total == 0)
return "Infinite solutions";
// For no solution
if (coeff == 0 && total != 0)
return "No solution";
// x = total sum / coeff
// of x '-' sign indicates
// moving numeric value to
// right hand side
int ans = -total / coeff;
return "x = " + ans.ToString();
}
// Driver code
static void Main()
{
string equation = "x+5-3+x=6+x-2";
Console.Write(solveEquation(equation));
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)
PHP
<?php
// PHP program to solve
// the given equation
// Function to solve
// the given equation
function solveEquation($equation)
{
$n = strlen($equation);
$sign = 1; $coeff = 0;
$total = 0; $i = 0;
// Traverse the equation
for ($j = 0; $j < $n; $j++)
{
if ($equation[$j] == '+' ||
$equation[$j] == '-')
{
if ($j > $i)
$total += $sign *
intval(substr($equation,
$i, $j - $i));
$i = $j;
}
// For cases such
// as: x, -x, +x
else if ($equation[$j] == 'x')
{
if (($i == $j) ||
$equation[$j - 1] == '+')
$coeff += $sign;
else if ($equation[$j - 1] == '-')
$coeff -= $sign;
else
$coeff += $sign *
intval(substr($equation,
$i, $j - $i));
$i = $j + 1;
}
// Flip sign once
// '=' is seen
else if ($equation[$j] == '=')
{
if ($j > $i)
$total += $sign *
intval(substr($equation,
$i, $j - $i));
$sign = -1;
$i = $j + 1;
}
}
// There may be a number
// left in the end
if ($i < $n)
$total += $sign *
intval(substr($equation, $i));
// For infinite solutions
if ($coeff == 0 &&
$total == 0)
return "Infinite solutions";
// For no solution
if ($coeff == 0 && $total)
return "No solution";
// x = total sum / coeff of x
// '-' sign indicates moving
// numeric value to right hand side
$ans = -$total / $coeff;
return "x = " . $ans;
}
// Driver code
$equation = "x+5-3+x=6+x-2";
echo (solveEquation($equation));
// This code is contributed by
// Manish Shaw(manishshaw1)
?>
Producción:
x = 2
Complejidad temporal: O(n), donde n es la longitud de la string de ecuaciones.
Espacio Auxiliar : O(1)