Java Exercises: Get the number of element that are smaller than the number of another
Java Basic: Exercise-172 with Solution
Write a Java program to get the number of element in a given array of integers that are smaller than the integer of another given array of integers.
Pictorial Presentation:

Sample Solution:
Java Code:
import java.util.ArrayList;
import java.util.Arrays;
public class Solution {
public static void main(String[] args) {
int[] main_arra = {1, 2, 3, 4, 5, 6, 7, 8};
int[]query_arra = {1, 4, 8};
ArrayList < Integer > result = count_smaller_number(main_arra, query_arra);
for (int i = 0; i < result.size(); i++) {
System.out.println(result.get(i));
}
}
public static ArrayList < Integer > count_smaller_number(int[] main_arra, int[] query_arra) {
ArrayList < Integer > result = new ArrayList < > ();
Arrays.sort(main_arra);
for (int i = 0; i < query_arra.length; i++) {
result.add(temp(main_arra, query_arra[i]));
}
return result;
}
private static int temp(int[] main_arra, int num) {
int ctr = 0;
for (int i = 0; i < main_arra.length; i++) {
if (main_arra[i] < num) {
ctr++;
} else {
break;
}
}
return ctr;
}
}
Sample Output:
0 3 7
Flowchart:

Java Code Editor:
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Previous: Write a Java program to accept two string and test if the second string contains the first one.
Next: Write a Java program to find the median of the number inside the window (size k) at each moving in a given array of intergers with duplicate numbers. Move the window from the start of the array.
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Java: Tips of the Day
Java: Anagrams
Generates all anagrams of a string.
public static List<String> anagrams(String input) { if (input.length() <= 2) { return input.length() == 2 ? Arrays.asList(input, input.substring(1) + input.substring(0, 1)) : Collections.singletonList(input); } return IntStream.range(0, input.length()) .mapToObj(i -> new SimpleEntry<>(i, input.substring(i, i + 1))) .flatMap(entry -> anagrams(input.substring(0, entry.getKey()) + input.substring(entry.getKey() + 1)) .stream() .map(s -> entry.getValue() + s)) .collect(Collectors.toList()); }
Ref: https://bit.ly/3rvAdAK
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