/**
 * An iterative solution to the Stock Market Problem from Xeno Fish
 * 
 * The idea behind this iterative solution is that we only need to consider subarrays of the original array
 * where the subarray begins at some minimum value and continues until the next minimum value. For example,
 * consider the array: 
 * 
 * 								[min.....new min...end]
 * 
 * Any solution to the problem is entirely contained in either the subarray: 
 * 
 * 								[min...new min-1]
 * 
 * or the subarray:
 * 								[new min...end]
 * 
 * In other words, it is never better to have a buy date and sell date that spans across the subarrays  
 * @author alchambers
 *
 */
public class Iterative implements StockIfc {

	@Override
	public StockInfo solve(int[] array) {
		StockInfo answer = new StockInfo();
		answer.profit = Integer.MIN_VALUE;
		
		int currentBuy = 0;
		int currentSell = 0;			
		for(int i = 1; i < array.length; i++){

			// We've found a new minimum...we're starting a new subarray
			if(array[i] < array[currentBuy]){
				currentBuy = i;
			}
			
			// Search within the subarray for a better profit
			else if(array[currentSell] - array[currentBuy] > answer.profit){
				answer.profit = array[currentSell] - array[currentBuy];
				answer.buy_index = currentBuy;
				answer.sell_index = currentSell;
			}
		}	
		return answer;			
	}
}