Given a number, find the nearest perfect square

Using divide and conquer approach, or we can say using modified binarysearch technique we can calculate nearest perfect square of a given number

public class ClosestPerfectSquare {

    public int getClosestPerfectSquare(int x) {
        int min = findMinSquareRoot(x, 1, x);

        int a = min * min;
        int b = (min + 1) * (min + 1);

        if (x - a < b - x) {
            return a;
        } else if (x - a > b - x) {
            return b;
        } else {
            return x;
        }
    }

    private int findMinSquareRoot(int x, int left, int right) {
        if (left > right) {
            return -1;
        }
        if (left * left <= x && x <= (left + 1) * (left + 1)) {
            return left;
        }

        int mid = left + (right - left) / 2;
        int midSquare = mid * mid;
        int midOneSquare = (mid + 1) * (mid + 1);

        if (x > midOneSquare) {
            return findMinSquareRoot(x, mid + 1, right);
        } else if (x < midSquare) {
            return findMinSquareRoot(x, left + 1, mid);
        } else {
            return mid;
        }
    }

    public static void main(String... args) throws Exception {
        for (int i = 1; i <= 100; i++) {
            int result = new ClosestPerfectSquare().getClosestPerfectSquare(i);
            System.out.println(String.format("%3d : %3d : %3d", i, result, (result - i)));
        }
    }
}