Better way to solve multi variable polynomial?












0












$begingroup$


$$
a + b^2 + c^3 + d^4 le S
$$



For above equation, I need to count possible solutions.



I tried following approach.



Integer inp=Integer.parseInt(br.readLine());

// get maximum possible value of b,c,d which satisfy equation

int b=(int) Math.sqrt(inp);

int d=(int) Math.sqrt(b);

int c=(int) Math.cbrt(inp);

int count=0;

for (int i = 0; i <= inp; i++) {

    for (int j = 0; j <= b; j++) {

        for (int j2 = 0; j2 <= c; j2++) {

            for (int k = 0; k <= d; k++) {

                int total=i+j*j+(int)Math.pow(j2, 3)+(int)Math.pow(k,4);

                if(total<=inp) {

                    count++;

                }

            }

        }

    }

}


Time complexity: $O(n times sqrt{n} times sqrt[3]{n} times sqrt[4]{n})$ (Not sure)



For large value, this takes time. Can someone please help me with better solution with time complexity?






java algorithm interview-questions







share|improve this question











$endgroup$








  • 1




    $begingroup$
    Are a,b,c,d elements of the natural numbers ? Can they be negative, or even complex? Please update the problem statement :)
    $endgroup$
    – RobAu
    17 hours ago


















0












$begingroup$


$$
a + b^2 + c^3 + d^4 le S
$$



For above equation, I need to count possible solutions.



I tried following approach.



Integer inp=Integer.parseInt(br.readLine());

// get maximum possible value of b,c,d which satisfy equation

int b=(int) Math.sqrt(inp);

int d=(int) Math.sqrt(b);

int c=(int) Math.cbrt(inp);

int count=0;

for (int i = 0; i <= inp; i++) {

    for (int j = 0; j <= b; j++) {

        for (int j2 = 0; j2 <= c; j2++) {

            for (int k = 0; k <= d; k++) {

                int total=i+j*j+(int)Math.pow(j2, 3)+(int)Math.pow(k,4);

                if(total<=inp) {

                    count++;

                }

            }

        }

    }

}


Time complexity: $O(n times sqrt{n} times sqrt[3]{n} times sqrt[4]{n})$ (Not sure)



For large value, this takes time. Can someone please help me with better solution with time complexity?






java algorithm interview-questions







share|improve this question











$endgroup$








  • 1




    $begingroup$
    Are a,b,c,d elements of the natural numbers ? Can they be negative, or even complex? Please update the problem statement :)
    $endgroup$
    – RobAu
    17 hours ago
















0












0








0





$begingroup$


$$
a + b^2 + c^3 + d^4 le S
$$



For above equation, I need to count possible solutions.



I tried following approach.



Integer inp=Integer.parseInt(br.readLine());

// get maximum possible value of b,c,d which satisfy equation

int b=(int) Math.sqrt(inp);

int d=(int) Math.sqrt(b);

int c=(int) Math.cbrt(inp);

int count=0;

for (int i = 0; i <= inp; i++) {

    for (int j = 0; j <= b; j++) {

        for (int j2 = 0; j2 <= c; j2++) {

            for (int k = 0; k <= d; k++) {

                int total=i+j*j+(int)Math.pow(j2, 3)+(int)Math.pow(k,4);

                if(total<=inp) {

                    count++;

                }

            }

        }

    }

}


Time complexity: $O(n times sqrt{n} times sqrt[3]{n} times sqrt[4]{n})$ (Not sure)



For large value, this takes time. Can someone please help me with better solution with time complexity?






java algorithm interview-questions







share|improve this question











$endgroup$




$$
a + b^2 + c^3 + d^4 le S
$$



For above equation, I need to count possible solutions.



I tried following approach.



Integer inp=Integer.parseInt(br.readLine());

// get maximum possible value of b,c,d which satisfy equation

int b=(int) Math.sqrt(inp);

int d=(int) Math.sqrt(b);

int c=(int) Math.cbrt(inp);

int count=0;

for (int i = 0; i <= inp; i++) {

    for (int j = 0; j <= b; j++) {

        for (int j2 = 0; j2 <= c; j2++) {

            for (int k = 0; k <= d; k++) {

                int total=i+j*j+(int)Math.pow(j2, 3)+(int)Math.pow(k,4);

                if(total<=inp) {

                    count++;

                }

            }

        }

    }

}


Time complexity: $O(n times sqrt{n} times sqrt[3]{n} times sqrt[4]{n})$ (Not sure)



For large value, this takes time. Can someone please help me with better solution with time complexity?






java algorithm interview-questions




java algorithm interview-questions






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 17 hours ago









Peter Taylor

15.9k2759




15.9k2759










asked 17 hours ago









user3857802user3857802

211




211








  • 1




    $begingroup$
    Are a,b,c,d elements of the natural numbers ? Can they be negative, or even complex? Please update the problem statement :)
    $endgroup$
    – RobAu
    17 hours ago
















  • 1




    $begingroup$
    Are a,b,c,d elements of the natural numbers ? Can they be negative, or even complex? Please update the problem statement :)
    $endgroup$
    – RobAu
    17 hours ago










1




1




$begingroup$
Are a,b,c,d elements of the natural numbers ? Can they be negative, or even complex? Please update the problem statement :)
$endgroup$
– RobAu
17 hours ago






$begingroup$
Are a,b,c,d elements of the natural numbers ? Can they be negative, or even complex? Please update the problem statement :)
$endgroup$
– RobAu
17 hours ago












1 Answer
1






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oldest

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1












$begingroup$


a + b^2 + c^3 + d^4 <= S

For above equation ,I need to count possible solutions.



I tried following approach.



Integer inp=Integer.parseInt(br.readLine());

      // get maximum possible value of b,c,d which satisfy equation

      int b=(int) Math.sqrt(inp);

        int d=(int) Math.sqrt(b);

        int c=(int) Math.cbrt(inp);

        int count=0;


...




If I were the interviewer, I would already have marked you down heavily for your choice of variable names. The problem statement defines a, b, c, d, S. The natural choice of variable names for those values are a, b, c, d, S. Using b, c, d for something else almost looks as though you were deliberately trying to make the code unmaintainable.




                        if(total<=inp) {

                            count++;

                        }


That is nearly always the wrong way to count something.



If I asked you to count solutions to $a le S$ would you write a loop? I hope not, because you can do it without any loop or any mathematical operations.



Now, how about $a + b^2 le S$? This can be done with one mathematical operation.



Cubes already grow quite fast, so



for (int c = 0; c <= maxC; c++) {

    for (int d = 0; d <= maxD; d++) {

        count += countAB(S - c*c*c - d*d*d*d)

    }

}


would be reasonably efficient. If you really want to microoptimise you can eliminate the multiplications in favour of some accumulators and addition, but that's probably not necessary for an interview question. They might ask about it in a follow-up.






share|improve this answer









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    1












    $begingroup$


    a + b^2 + c^3 + d^4 <= S

    For above equation ,I need to count possible solutions.



    I tried following approach.



    Integer inp=Integer.parseInt(br.readLine());
    
      // get maximum possible value of b,c,d which satisfy equation
    
      int b=(int) Math.sqrt(inp);
    
        int d=(int) Math.sqrt(b);
    
        int c=(int) Math.cbrt(inp);
    
        int count=0;


    ...




    If I were the interviewer, I would already have marked you down heavily for your choice of variable names. The problem statement defines a, b, c, d, S. The natural choice of variable names for those values are a, b, c, d, S. Using b, c, d for something else almost looks as though you were deliberately trying to make the code unmaintainable.




                            if(total<=inp) {
    
                            count++;
    
                        }


    That is nearly always the wrong way to count something.



    If I asked you to count solutions to $a le S$ would you write a loop? I hope not, because you can do it without any loop or any mathematical operations.



    Now, how about $a + b^2 le S$? This can be done with one mathematical operation.



    Cubes already grow quite fast, so



    for (int c = 0; c <= maxC; c++) {
    
    for (int d = 0; d <= maxD; d++) {
    
        count += countAB(S - c*c*c - d*d*d*d)
    
    }
    
}


    would be reasonably efficient. If you really want to microoptimise you can eliminate the multiplications in favour of some accumulators and addition, but that's probably not necessary for an interview question. They might ask about it in a follow-up.






    share|improve this answer









    $endgroup$


















      1












      $begingroup$


      a + b^2 + c^3 + d^4 <= S

      For above equation ,I need to count possible solutions.



      I tried following approach.



      Integer inp=Integer.parseInt(br.readLine());
      
      // get maximum possible value of b,c,d which satisfy equation
      
      int b=(int) Math.sqrt(inp);
      
        int d=(int) Math.sqrt(b);
      
        int c=(int) Math.cbrt(inp);
      
        int count=0;


      ...




      If I were the interviewer, I would already have marked you down heavily for your choice of variable names. The problem statement defines a, b, c, d, S. The natural choice of variable names for those values are a, b, c, d, S. Using b, c, d for something else almost looks as though you were deliberately trying to make the code unmaintainable.




                              if(total<=inp) {
      
                            count++;
      
                        }


      That is nearly always the wrong way to count something.



      If I asked you to count solutions to $a le S$ would you write a loop? I hope not, because you can do it without any loop or any mathematical operations.



      Now, how about $a + b^2 le S$? This can be done with one mathematical operation.



      Cubes already grow quite fast, so



      for (int c = 0; c <= maxC; c++) {
      
    for (int d = 0; d <= maxD; d++) {
      
        count += countAB(S - c*c*c - d*d*d*d)
      
    }
      
}


      would be reasonably efficient. If you really want to microoptimise you can eliminate the multiplications in favour of some accumulators and addition, but that's probably not necessary for an interview question. They might ask about it in a follow-up.






      share|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$


        a + b^2 + c^3 + d^4 <= S

        For above equation ,I need to count possible solutions.



        I tried following approach.



        Integer inp=Integer.parseInt(br.readLine());
        
      // get maximum possible value of b,c,d which satisfy equation
        
      int b=(int) Math.sqrt(inp);
        
        int d=(int) Math.sqrt(b);
        
        int c=(int) Math.cbrt(inp);
        
        int count=0;


        ...




        If I were the interviewer, I would already have marked you down heavily for your choice of variable names. The problem statement defines a, b, c, d, S. The natural choice of variable names for those values are a, b, c, d, S. Using b, c, d for something else almost looks as though you were deliberately trying to make the code unmaintainable.




                                if(total<=inp) {
        
                            count++;
        
                        }


        That is nearly always the wrong way to count something.



        If I asked you to count solutions to $a le S$ would you write a loop? I hope not, because you can do it without any loop or any mathematical operations.



        Now, how about $a + b^2 le S$? This can be done with one mathematical operation.



        Cubes already grow quite fast, so



        for (int c = 0; c <= maxC; c++) {
        
    for (int d = 0; d <= maxD; d++) {
        
        count += countAB(S - c*c*c - d*d*d*d)
        
    }
        
}


        would be reasonably efficient. If you really want to microoptimise you can eliminate the multiplications in favour of some accumulators and addition, but that's probably not necessary for an interview question. They might ask about it in a follow-up.






        share|improve this answer









        $endgroup$




        a + b^2 + c^3 + d^4 <= S

        For above equation ,I need to count possible solutions.



        I tried following approach.



        Integer inp=Integer.parseInt(br.readLine());
        
      // get maximum possible value of b,c,d which satisfy equation
        
      int b=(int) Math.sqrt(inp);
        
        int d=(int) Math.sqrt(b);
        
        int c=(int) Math.cbrt(inp);
        
        int count=0;


        ...




        If I were the interviewer, I would already have marked you down heavily for your choice of variable names. The problem statement defines a, b, c, d, S. The natural choice of variable names for those values are a, b, c, d, S. Using b, c, d for something else almost looks as though you were deliberately trying to make the code unmaintainable.




                                if(total<=inp) {
        
                            count++;
        
                        }


        That is nearly always the wrong way to count something.



        If I asked you to count solutions to $a le S$ would you write a loop? I hope not, because you can do it without any loop or any mathematical operations.



        Now, how about $a + b^2 le S$? This can be done with one mathematical operation.



        Cubes already grow quite fast, so



        for (int c = 0; c <= maxC; c++) {
        
    for (int d = 0; d <= maxD; d++) {
        
        count += countAB(S - c*c*c - d*d*d*d)
        
    }
        
}


        would be reasonably efficient. If you really want to microoptimise you can eliminate the multiplications in favour of some accumulators and addition, but that's probably not necessary for an interview question. They might ask about it in a follow-up.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 17 hours ago









        Peter TaylorPeter Taylor

        15.9k2759




        15.9k2759






























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