אמיר ותמר

/**
     * A game with two players, each can pick a number from the sides of the array
     * The winner is the one who finished with the highest sum
     * the method ensures that the first player (Amir) will never loose
     *
     * @param k the game board
     *          Time complexity - O(n) - because the loop iterates the array only once
     */
    public static void win(int[] k) {
        //finding the highest score path between the odd indexes and the even indexes
        int countOdd = 0;
        int countEven = 0;
        boolean amirTookOddPath;
        for (int i = 0; i < k.length; i++) {
            if (i % 2 == 0)
                countEven += k[i];
            else
                countOdd += k[i];
        }
        amirTookOddPath = countOdd >= countEven ? true : false;

        int min = 0, max = k.length - 1;
        //the odd path is the winning path
        while (min < max) {
            //Amir's path - the highest count of values between the odd indexes and the even indexes
            if (amirTookOddPath ? min % 2 != 0 : min % 2 == 0) {
                System.out.println("Amir took " + k[min]);
                min++;
            } else {
                System.out.println("Amir took " + k[max]);
                max--;
            }
            //Tamar's path - the opposite path of Amir, always choosing the highest value between the two
            if (k[max] > k[min]) {
                System.out.println("Tamar took " + k[max]);
                max--;
            } else {
                System.out.println("Tamar took " + k[min]);
                min++;
            }
        }
        System.out.println("Final score:");
        System.out.println("Amir total " + (amirTookOddPath? countOdd : countEven));
        System.out.println("Tamar total " + (amirTookOddPath? countEven : countOdd));
    }