C Programming Solution:

void countApplesAndOranges(int s, int t, int a, int b, int apples_count, int* apples, int oranges_count, int* oranges) {
    int count[2]={0,0};
    for(int i=0; i<2;i++){
        if(i==0){
            for(int j=0; j<apples_count;j++){
                if((a+apples[j])>=s && (a+apples[j])<=t ){
                    (count[i])++;
                }
            }
        }
        else{
            for(int j=0; j<oranges_count;j++){
                if((b+oranges[j])>=s && (b+oranges[j])<=t ){
                    (count[i])++;
                }
            }
        }
    }
    printf("%d\n%d",count[0],count[1]);
}

def countApplesAndOranges(s, t, a, b, apples, oranges): # Write your code here new_app_cord=[a+app for app in apples] new_orr_cord=[b+orr for orr in oranges] ca=0 co=0 for app in new_app_cord: if app>=s and app<=t: ca+=1 for orr in new_orr_cord: if orr>=s and orr<=t: co+=1 print(ca,co,sep="\n")

Java solution:

import java.io.; import java.math.; import java.security.; import java.text.; import java.util.; import java.util.concurrent.; import java.util.function.; import java.util.regex.; import java.util.stream.*; import static java.util.stream.Collectors.joining; import static java.util.stream.Collectors.toList;

class Result {

/*
 * Complete the 'countApplesAndOranges' function below.
 *
 * The function accepts following parameters:
 *  1. INTEGER s
 *  2. INTEGER t
 *  3. INTEGER a
 *  4. INTEGER b
 *  5. INTEGER_ARRAY apples
 *  6. INTEGER_ARRAY oranges
 */

public static void countApplesAndOranges(int s, int t, int a, int b, List<Integer> apples, List<Integer> oranges) {
// Write your code here
int apple_count=0;
int orange_count=0;

for(Integer apple:apples)
{
    if(apple<0)
    {
        continue;
    }
    else
    {
        if(a+apple>=s && a+apple<=t)
        {
            apple_count++;
        }
    }
}
System.out.println(apple_count);

for(Integer orange:oranges)
{
    if(orange>0)
    {
        continue;
    }
    else{
        if(b+orange>=s && b+orange<=t)
        {
            orange_count++;
        }
    }
}
System.out.println(orange_count);
return;

}

}

public class Solution { public static void main(String[] args) throws IOException { BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));

    String[] firstMultipleInput = bufferedReader.readLine().replaceAll("\\s+$", "").split(" ");

    int s = Integer.parseInt(firstMultipleInput[0]);

    int t = Integer.parseInt(firstMultipleInput[1]);

    String[] secondMultipleInput = bufferedReader.readLine().replaceAll("\\s+$", "").split(" ");

    int a = Integer.parseInt(secondMultipleInput[0]);

    int b = Integer.parseInt(secondMultipleInput[1]);

    String[] thirdMultipleInput = bufferedReader.readLine().replaceAll("\\s+$", "").split(" ");

    int m = Integer.parseInt(thirdMultipleInput[0]);

    int n = Integer.parseInt(thirdMultipleInput[1]);

    List<Integer> apples = Stream.of(bufferedReader.readLine().replaceAll("\\s+$", "").split(" "))
        .map(Integer::parseInt)
        .collect(toList());

    List<Integer> oranges = Stream.of(bufferedReader.readLine().replaceAll("\\s+$", "").split(" "))
        .map(Integer::parseInt)
        .collect(toList());

    Result.countApplesAndOranges(s, t, a, b, apples, oranges);

    bufferedReader.close();
}

}

Here is my solution in Python 3!

def countApplesAndOranges(s, t, a, b, apples, oranges):
    print(len(list(filter(lambda x: s<=x<=t, [a+p for p in apples]))))
    print(len(list(filter(lambda x: s<=x<=t, [b+p for p in oranges]))))

func countApplesAndOranges(s: Int, t: Int, a: Int, b: Int, apples: [Int], oranges: [Int]) -> Void {
    // Write your code here
    print(apples.filter { s...t ~= ($0 + a)}.count)
    print(oranges.filter { s...t ~= ($0 + b)}.count)    
}

Swifty way of doing better things.