Number of canoes needed

Algorithm Pseudocode

1. Sort the array in ascending order. O(n log n)
2. Start with indices on either ends i, and j.
3. Compare the sum of values at i and j to 150. If less than or equal, decrement both indices and increment numBoats.
4. If greater, check if heavier boy qualifies (is <= 150 lbs). If so, increment numBoats. If not, do nothing. In either case decrement j.

Complexity

Time: O(n log n) due to sorting at the beginning.

public class Canoe {

	final static int MAX_WEIGHT = 150;

	public static void main(String[] args) {

		// Use sorting algorithm O(n log n)
		int[] weightList = {45, 67, 84, 100, 120, 134, 150, 167, 180, 200, 213};

		System.out.println(numBoatsNeeded(weightList));

	}

	public static int numBoatsNeeded(int[] weightList) {
		int numBoats = 0;
		int i = 0;
		int j = weightList.length-1;

		while (i <= j) {

			// When only one child remains
			if (i == j) {

				if (weightList[i] <= MAX_WEIGHT) {
					numBoats++;
				}

				break;
			}

			/**
			 * Check sum of both ends, and if sum is less than max weight, 
			 * place both on same boat.
			 */
			if (weightList[i] + weightList[j] <= MAX_WEIGHT) {

				// Handled both lighter/heavier kid, so decrement
				i++;
				j--;
				numBoats++;

			// Check if heavier rider must fit in his own boat.
			} else {

				if (weightList[j] <= MAX_WEIGHT) {
					numBoats++;
				}

				// Handled heavier kid
				j--;
			}

				
		}

		return numBoats;

	}
}

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