arrange elements in array based on pivot value

Arrange elements in array based on particular value

• In a given integer array, arrange value to left and right side of array based on a particular value (say pivot).
• values less than pivot will arrange on the left side of the array and values greater than pivot will arrange on the right side of the array

Examples:

Input: { 4, 9, 7, 6, 3, 2, 8, 9, 1 }

o/p:    4 3 2 1 6 9 8 7 9 

Input: { 4, 9, 7, 6, 3, 2, 8, 5, 1 }

o/p:    4 3 2 5 1 6 8 7 9

Here pivot=6. so values less than pivot will arrange to left and values greater than pivot will arrange to the right of the array.

Program:

public class arrangeArrayParticularValue {
    public static void main(String[] args) {
        int[] arr = { 4, 9, 7, 6, 3, 2, 8, 5, 1 };
        int pivot = 6;
        // new array to store values in arranged order
        int[] b = new int[arr.length];
        // starting index for new array
        int k = 0;
        // last position of new array
        int max = arr.length - 1;
        // loop - to arrange values to left side
        for (int i = 0; i <= arr.length - 1; i++) {
            if (arr[i] < pivot) {
                b[k] = arr[i];
                k++;
            }
        }
        // after left arrange completed - then add pivot value
        b[k] = pivot;
        k++;
        // after pivot added, loop - to arrange values to right side
        for (int i = 0; i <= arr.length - 1; i++) {
            if (arr[i] > pivot) {
                b[max] = arr[i]; //adding values from last index
                max--;
            }
        }
        // printing arranged array
        for (int n : b) {
            System.out.print(n + " ");
        }
    }
}

Output:

arrange elements in an array based on pivot value

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