Can someone help me with these two Infosys HackWithInfy practice questions?

for the first question… what if we traverse the array from both directions and output minimum of the two?

MINIMUM PLUSES:
I 've tried this approach but, it passed only 62% of test cases:)

class Solution {
public static int minimum_pluses(String A,String Y){
       if(A.equals(Y))
        return -1;
    char ch[]=A.toCharArray();
    int rev=0,check=0,counter=0;
    int n=ch.length;
 
    int minimum=Integer.MAX_VALUE;
    for( int i=0; i<n; i++){
        int no=(ch[i]-'0');
        rev=rev*10+no;
        check=rev;
        counter=0;
       for( int j=i+1; j<n; j++){
             counter++;
            if((check+(ch[j]-'0'))==Integer.parseInt(Y)){
                minimum=Math.min(minimum,counter);
                break;
            }
           check+=(ch[j]-'0');
       
       }     
      
    }
   
    
    return minimum==Integer.MAX_VALUE? -1:minimum;

}

public static void main(String []args){
    Scanner scan = new Scanner(System.in);
    String A;
    A=scan.next();
    int result=Integer.MAX_VALUE;
    String s[]=A.split("=");
    result = minimum_pluses(s[0],s[1]);

    System.out.print(result);
 }
}

`

Answer of the second one:
import java.util.*;
public class Main {
public static void main(String args[]) {
Scanner scn=new Scanner(System.in);
String x=scn.next();//L.H.S of the equation
int y=scn.nextInt();//R.H.S of the equation
int res=min(x,y);
if(res>=10000)
{
System.out.println(-1);
}
else
{
System.out.println(res);
}
}

public static int min(String str,int num) {
    if(num<0)
    {
        return 10000;
    }
    if(Integer.parseInt(str)==num)
    {
        return 0;
    }
    if(str.length()==1)
    {
        if(Integer.parseInt(str)==num)
        {
            return 0;
        }
        else
        {
            return 10000;
        }
    }
    int ans=10000;
    for(int i=1;i<str.length();i++)
    {
        int n=Integer.parseInt(str.substring(0,i));
        String sub=str.substring(i);
        ans=Math.min(ans,min(sub,num-n)+1);
    }
    return ans;
}

}

For 1st problem, you can find max length subarray possible with sum as (totalSumOfArrayElements - X), then if found such subarray then answer will be (lengthOfArray - maxLenSatisfying).
Using 2 pointer algo you can find max length subarray satisying condition in O(n) time complexity.

"""2nd one Answer"""

def permute(s):
result = [[s]]
for i in range(1, len(s)):
first = [s[:i]]
rest = s[i:]
for p in permute(rest):
result.append(first + p)
return [[int(j) for j in i] for i in result]
def problem(s):
x,y=s.split("=")
data=permute(x)
newdata=[]
for i in range(1,len(x)+1,1):
for j in data:
if i==len(j):
newdata.append(j)
for i in newdata:
if sum(i)==int(y):
print(“str 1”,i)
return
print(“str -1”)
def check_constraint(s):
if (not (1<=len(s)<=10^3)):
print(-1)
elif (s.split("=")[0]==s.split("=")[1]):
print(1)
elif (not (len(s.split("=")[0])>=len(s.split("=")[1]))):
print(-1)
else:
problem(s)
A=input()
check_constraint(A)

Are hackwithinfy questions really hard like this every year?