Implement a Java program that finds two neighbouring numbers in an array with th
ID: 3565817 • Letter: I
Question
Implement a Java program that finds two neighbouring numbers in an array with the smallest distance to each other. The function should return the index of the first number. In the sequence 4 8 6 1 2 9 4 the minimum distance is 1 (between 1 and 2). The function should return the index of number 1 which is 3. For each array element take user input to form the array. Hint: Use Math.abs() to calculate the distance of two numbers. Math.abs() is used to give you the absolute value. For example the difference between 1 and 2 is 1-2 = -1. If you use Math.abs(1-2) it will give you 1 instead of -1.
Sample Run:
Enter the size of the array: 7
Please enter 7 positive integers: 4 8 6 1 2 9 4
The minimum difference is 1 and the numbers are 1 and 2. The index of 1 is 3. Bye!!!
Explanation / Answer
In the sequence 4 8 6 1 2 9 4 the minimum distance is 1 (between 1 and 2). The function should return the index 3 (of number 1).
class LangFund3{
static void smallestDistance(int [] array){
int smallest = Math.abs(array[0]-array[1]);
int index = 0;
for(int i=1; i<array.length-1; i++){
int value = Math.abs(array[i]-array[i+1]);
if(value< smallest){
smallest= value;
index = i;
}
}
System.out.println(smallest);
System.out.println(index);
}
public static void main(String [] args){
int []arr= new int[]{4,8,6,1,2,9,4};
smallestDistance(arr);
}
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