If you are new to learning programming and do not know where to start, please try Problem A “Welcome to AtCoder” from practice contest. There you can find a sample code for each language.
Also, if you are not familiar with problems in programming contests, we recommend you to try some problems in “AtCoder Beginners Selection” (https://atcoder.jp/contests/abs).

Let us try determining whether “there exists an integer \(i\) between \(1\) and \(N-2\), inclusive, such that \(A_i=A_{i+1}=A_{i+2}\),” as written after “more formally.”

Many programming languages employ \(0\)-based indexing, that is, the first element of an array \(A\) is specified by \(A[0]\) instead of \(A[1]\), so the condition is translated into code as “whether there exists an integer \(i\) between \(0\) (inclusive) and \(N-2\) (exclusive), such that \(A_i=A_{i+1}=A_{i+2}\).”

This can be done by the for statement.

There are several ways to actually implement this algorithm. For example, while iterating \(i\) with a for statement, one can:

• print Yes and terminate the program once a conforming \(i\) is found. If the program is not terminated after completing the for statement, it means that the answer is No, so print No.
• implement the decision problem as a function or a method. Once such an \(i\) is found, return Yes (or something equivalent, like True). If the function proceeds after completing the for statement, it means that the answer is No, so return No (or something equivalent, like False).

This way, your implementation will be short and concise.

The following is sample code in C++ and Python for each implementation strategy.

Sample code (C++)

#include <bits/stdc++.h>
using namespace std;
int main() {
	int n;
	cin >> n;
	vector<int> a(n);
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}
	for (int i = 0; i < n - 2; i++) {
		if (a[i] == a[i + 1] && a[i + 1] == a[i + 2]) {
			cout << "Yes" << endl;
			return 0;
		}
	}
	cout << "No" << endl;
	return 0;
}

Sample code (Python)

n = int(input())
a = list(map(int, input().split()))
for i in range(n - 2):
    if a[i] == a[i + 1] == a[i + 2]:
        print("Yes")
        exit()
print("No")

Sample code (C++)

#include <bits/stdc++.h>
using namespace std;
bool judge(int n, vector<int> a) {
	for (int i = 0; i < n - 2; i++) {
		if (a[i] == a[i + 1] && a[i + 1] == a[i + 2]) {
			return true;
		}
	}
	return false;
}
int main() {
	int n;
	cin >> n;
	vector<int> a(n);
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}
	if (judge(n, a)) {
		cout << "Yes" << endl;
	} else {
		cout << "No" << endl;
	}
	return 0;
}

Sample code (Python)

def judge(n, a):
    for i in range(n - 2):
        if a[i] == a[i + 1] == a[i + 2]:
            return True
    return False


n = int(input())
a = list(map(int, input().split()))
if judge(n, a):
    print("Yes")
else:
    print("No")