CareerCup

First go through all the '0' and fill up the C array.

Next starting from the end, go to the start of array, keeping track of all open positions in C. Any time you hit a 1 in B, copy A[i] to one of the open positions. If no open positions, return 'no solution'.

Repeat above step but this time start from the start of array, keeping track of all open positions in C. Any time we hit -1, copy A[i] to one of the open positions that we are keeping track of. Remove the position once filled. Any time we hit -1 and open positions set is empty, return 'no solution'.

Construct a bipartite graph ( with the left nodes representing indices of A and right nodes indices of C) . Make the edges according to value in C.
If the max match is equal to length of element in array , solution exist.

import java.util.ArrayList;


public class three_arrays {


public static void main(String[] args) {


int[] a = {10,7,16,20,33,15,17};
int[] b = {1,-1,1,1,-1,0,-1};
int[] c = new int[7];

c = compute_a(a,b,c);


if(c==null){
System.out.println("no solution");
}
else{

for(int i=0;i<c.length;i++){

System.out.println(c[i]);

}
}

}

private static int[] compute_a(int[] a, int[] b, int[] c) {


ArrayList<Integer> ones = new ArrayList<Integer>();

for(int i=0;i<a.length;i++){

if(b[i]==0){

c[i] = a[i];

}

if(b[i]==1){

ones.add(i);

}

if(b[i]==-1){

if(ones.size()==0)
return null;
int pos = ones.remove(0);

c[i] = a[pos];
c[pos] = a[i];
}
}


if(ones.size()!=0)
return null;

return c;


}

}

Please check my code:

#include<stdio.h>

int arrange(int a[],int b[],int c[],int n)
{
	int i=0,ci=0;
	for(i=0;i<n;i++)
	{
		if(b[i] == 0)
		{
			c[i]=a[i];
		}
		if(b[i] == -1)
		{
			while(c[ci]!=0 && ci<i)
			{
				ci++;
			}
			if(c[ci]!=0 || ci>=i)
			{
				return -1;
			}
			if(b[ci]!=0)
			{
				c[ci]=a[i];
				ci++;
			}
		}
	}
	
	for(i=0;i<n;i++)
	{
		if(b[i] == 1)
		{
			while(c[ci]!=0 && ci<n)
			{
				ci++;
			}
			if(c[ci]!=0 || ci<=i)
			{
				return -1;
			}
			if(b[ci]!=0)
			{
				c[ci]=a[i];
				ci++;
			}
		}
	}
	return 0;
}


int main()
{
	int a[]={1,2,3};//{1,2,3,4,5,6,7,8};//{1,2,3,4};
	int b[]={1,0,-1};//{1,1,0,-1,1,1,-1,-1};//{1,-1,0,1};
	int c[]={0,0,0};//{0,0,0,0,0,0,0,0,0};//{0,0,0,0};
	int i=0,n=3;

	if(arrange(a,b,c,n)==-1)
	{
		printf("No Solution.\n");
	}
	else
	{	
		for(i=0;i<n;i++)
		{
			printf(" %d ",c[i]);
		}
	}
	return 0;
}

Feedbacks are welcome.

i got output as 4 7 3 8 1 2 5 6 for A={1,2,3,4,5,6,7,8} B={1,1,0,-1,1,1,-1,-1}; i hope output here to be "no solution"
i made small change on 12th line as
while( (c[ci]!=0 || b[ci]<=0) && ci<i)
i got the intended result

This works as per requirements:

public int[] makeNewArray(int A[], int B[]){
		if(A.length == 0 || B.length == 0){
			return noSolution();
		}
		int C[] = new int[A.length];
		for(int i=0;i<C.length;i++){
			C[i]=-1;
		}
		int index;
		for(int i=0;i<B.length;i++){
			if(B[i]==0){
				if(C[i]==-1){
					C[i] = A[i];
				}else{
					return noSolution();
				}
			}else{
				if(B[i]==-1){
					index = spotfromBeg(C, i);
					if(-1 != index){
						C[index]=A[i];
					}else{
						System.out.println("No Solution");
						return noSolution();
					}
				}else{
					if(B[i]==1){
						index = spotfromEnd(C, i);
						if(-1 != index){
							C[index]=A[i];
						}else{
							return noSolution();
						}
					}
				}
			}
		}
		return C;
	}
	
	public int[] noSolution(){
		System.out.println("No Solution");
		return null;
	}
	
	public int spotfromEnd(int C[],int index){
		int i;
		for(i=C.length-1;i>index && C[i]!=-1;i--){
			
		}
		if(i==index){
			return -1;
		}else{
			return i;
		}
	}
	
	public int spotfromBeg(int C[],int index){
		int i;
		for(i=0;i<index && C[i]!=-1;i++){
		
		}
		if(i==index){
			return -1;
		}else{
			return i;
		}
	}