Java Array Exercises: Insert an element into an array

Java Array: Exercise-9 with Solution

Write a Java program to insert an element (specific position) into an array.

Pictorial Presentation:

Java Array Exercises: Insert an element into an array

Sample Solution:

Java Code:

import java.util.Arrays; 
public class Exercise9 {
public static void main(String[] args) {

   int[] my_array = {25, 14, 56, 15, 36, 56, 77, 18, 29, 49};

    // Insert an element in 3rd position of the array (index->2, value->5)
   int Index_position = 2;
   int newValue    = 5;

  System.out.println("Original Array : "+Arrays.toString(my_array));     
  for(int i=my_array.length-1; i > Index_position; i--){
    my_array[i] = my_array[i-1];
   my_array[Index_position] = newValue;
   System.out.println("New Array: "+Arrays.toString(my_array));

Sample Output:

Original Array : [25, 14, 56, 15, 36, 56, 77, 18, 29, 49]                                                     
New Array: [25, 14, 5, 56, 15, 36, 56, 77, 18, 29] 


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Java: Tips of the Day

Different between parseInt() and valueOf() in java?

Well, the API for Integer.valueOf(String) does indeed say that the String is interpreted exactly as if it were given to Integer.parseInt(String). However, valueOf(String) returns a new Integer() object whereas parseInt(String) returns a primitive int.

If you want to enjoy the potential caching benefits of Integer.valueOf(int), you could also use this eyesore:

Integer k = Integer.valueOf(Integer.parseInt("123"))

Now, if what you want is the object and not the primitive, then using valueOf(String) may be more attractive than making a new object out of parseInt(String) because the former is consistently present across Integer, Long, Double, etc.

Ref: https://bit.ly/3vRuIPY