ABC --132- A & B & C

AtCoder ABC 132 A&B&C AtCoder - 132

Since the bug of D combination cannot be removed, I will postpone it after D.

A - Fifty-Fifty

――If you count the characters, you must have two

	private void solveA() {
		String[] a = next().split("");

		Map<String, Long> memo = Arrays.stream(a).collect(Collectors.groupingBy(s -> s, Collectors.counting()));

		for (long elm : memo.values()) {
			if (elm != 2) {
				out.println("No");
				return;
			}
		}
		out.println("Yes");
	}

B - Ordinary Number

-$ p_i $ is the second smallest of the three numbers $ p_ {i−1}, p_i, p_ {i + 1} $ --Since the above is satisfied, check $ p_ {i−1} <p_i <p_ {i + 1} $ or $ p_ {i + 1} <p_i <p_ {i−1} $.

	private void solveB() {
		int n = nextInt();
		int[] p = IntStream.range(0, n).map(i -> nextInt()).toArray();

		int res = 0;
		for (int i = 0; i < p.length - 2; i++) {
			if ((p[i] < p[i + 1] && p[i + 1] < p[i + 2]) || (p[i + 2] < p[i + 1] && p[i + 1] < p[i])) {
				res++;
			}
		}

		out.println(res);
	}

C - Divide the Problems

--Sort --Extract the number of problem boundaries between ARC and ABC --The number of selectable differences in the number of boundaries

	private void solveC() {
		int n = nextInt();
		int[] d = IntStream.range(0, n).map(i -> nextInt()).sorted().toArray();

		out.println(d[n / 2] - d[n / 2 - 1]);
	}