import java.util.*;
public class Main {
public static void main(String args[]) {
// Your Code Here
Scanner ti=new Scanner(System.in);
int n=ti.nextInt();
int[] arr=new int[n];
for(int i=0;i<n;i++){
arr[i]=ti.nextInt();
}
int s=0;
int m=0;
int l=n-1;
while(m<=l){
if(arr[m]==0){
int temp=arr[s];
arr[s]=arr[m];
arr[m]=temp;
m++;
s++;
}
if(arr[m]==1){
m++;
}
else{
int temp=arr[l];
arr[l]=arr[m];
arr[m]=temp;
l--;
}
}
}
}
Test cases do not passes why
Hey @yashkdot_da870b580af21a9d. You are getting wrong answers bcoz after sorting the array you need to print the array. Secondly Your logic is having some deficiencies that’s why a test case like :
5
2
0
2
1
0
is failing. So you can refer to my logic for benchmarking your logic.
For the question the approach to be followed is to define regions of zero, one and two as there are only these three elements in the array. The region 1 will be of zero as it the least among the three then the region for 1 and in the end the third region of 2’s.
Logic
In this question there will be three regions defined for three different elements. A pointer that keeps a track of where the region for 1 starts and a pointer for where the region for 2 starts. Since 2 is to be placed in the end so the region for two starts in the end and the pointer for 2 decreases as we place a new two. If current element is zero then place it with the pointer where 1 starts and increase the pointer by 1 as zero’s region is expanded.