/*
 *  RapidMiner
 *
 *  Copyright (C) 2001-2014 by RapidMiner and the contributors
 *
 *  Complete list of developers available at our web site:
 *
 *       http://rapidminer.com
 *
 *  This program is free software: you can redistribute it and/or modify
 *  it under the terms of the GNU Affero General Public License as published by
 *  the Free Software Foundation, either version 3 of the License, or
 *  (at your option) any later version.
 *
 *  This program is distributed in the hope that it will be useful,
 *  but WITHOUT ANY WARRANTY; without even the implied warranty of
 *  MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
 *  GNU Affero General Public License for more details.
 *
 *  You should have received a copy of the GNU Affero General Public License
 *  along with this program.  If not, see http://www.gnu.org/licenses/.
 */
package com.rapidminer.tools.math;

import java.util.LinkedList;
import java.util.List;

/**
 * This simple implementation returns a peak if the specified number of
 * neighbors is smaller than the current value.
 * 
 * @author Ingo Mierswa
 */
public class SimplePeakFinder implements PeakFinder {

	private int numberOfNeighbours = 5;

	public SimplePeakFinder(int neighbours) {
		this.numberOfNeighbours = neighbours;
	}

	/** Returns a list with peaks. */
	public List<Peak> getPeaks(Peak[] series) {
		List<Peak> result = new LinkedList<Peak>();
		for (int i = 0; i < series.length; i++) {
			if (isPeak(series, i)) {
				result.add(series[i]);
			}
		}
		return result;
	}

	/**
	 * Returns true if the value for index is an extremum of the given type
	 * between the given numbers of neighbours.
	 */
	private boolean isPeak(Peak[] series, int index) {
		boolean ok = false;
		int current = index;
		int okValues = 0;

		// values to left
		while (!ok) {
			current--;
			if (current < 0)
				return false;
			if (!isOk(series, current, index))
				return false;
			okValues++;
			if (okValues > numberOfNeighbours)
				ok = true;
		}
		ok = false;
		current = index;
		okValues = 0;
		// values to right
		while (!ok) {
			current++;
			if (current >= series.length)
				return false;
			if (!isOk(series, current, index))
				return false;
			okValues++;
			if (okValues > numberOfNeighbours)
				ok = true;
		}
		return true;
	}

	/**
	 * In the minimum case this method returns true, if the current value is
	 * bigger than the index value.
	 */
	private boolean isOk(Peak[] series, int current, int index) {
		if (series[current].getMagnitude() <= series[index].getMagnitude())
			return false;
		else
			return true;
	}
}