Codeforces Round #698 (Div. 2) - A. Nezzar and Colorful Balls

Codeforces Round #698 (Div. 2)-A. Nezzar and Colorful Balls

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Time Limit: 1 second
Memory Limit: 512 megabytes

Problem Description

Nezzar has $n$ balls, numbered with integers $1,2,\dots ,n$. Numbers $a_1,a_2,\dots ,a_n$ are written on them, respectively. Numbers on those balls form a non-decreasing sequence, which means that $a_i\le a_{i+1}$ for all $1\le i<n$.

Nezzar wants to color the balls using the minimum number of colors, such that the following holds.

Note that a sequence with the length at most $1$ is considered as a strictly increasing sequence.

Please help Nezzar determine the minimum number of colors.

Input

The first line contains a single integer $t$ ($1\le t\le 100$) — the number of testcases.

The first line of each test case contains a single integer $n$ ($1\le n\le 100$).

The second line of each test case contains $n$ integers $a_1,a_2,\dots ,a_n$ ($1\le a_i\le n$). It is guaranteed that $a_1\le a_2\le \dots \le a_n$.

Output

For each test case, output the minimum number of colors Nezzar can use.

Sample Input

5
6
1 1 1 2 3 4
5
1 1 2 2 3
4
2 2 2 2
3
1 2 3
1
1

Sample Onput

3
2
4
1
1

Note

Let’s match each color with some numbers. Then:

In the first test case, one optimal color assignment is $[1,2,3,3,2,1]$.

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In the second test case, one optimal color assignment is $[1,2,1,2,1]$.

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题目大意

给n个气球，每个气球都有编号
现在给气球涂色
相同编号的气球颜色不能相同，问至少需要几种颜色。

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解题思路

很显然，看哪个编号的气球最多，就需要这么多的颜色。

题目很友好，给的编号都是按顺序的，因此就不需要再排序了。

下面用一种空间消耗比较小的方法，输入完毕即可得到答案。
用变量last记录上一个气球的编号，如果这个气球编号不同于上一个，就更新最大值M。
初始值last = -1，这样第一个气球就会与它颜色不同。

用变量locLast记录上次是第几个气球（第一个气球按第0个处理）。

AC代码

#include <bits/stdc++.h>
using namespace std;
int main()
{
    
    
	int N;
	cin >> N;
	while (N--)
	{
    
    
		int n;
		cin >> n;
		int last = -1;     //上一个气球的编号
		int M = 1;
		int locLast = -1;  //这个颜色的气球的上一个location
		for (int i = 0; i < n; i++)
		{
    
    
			int t;
			scanf("%d", &t); //这个气球的编号
			if (t != last) //这个气球编号和上一个不同
			{
    
    
				last = t;  //更新last
				M = max(M, i - locLast); //更新最大值M
				locLast = i; //更新locLast
			}
		}
		/*最后一个气球与倒数第二个气球编号相同的话，最后一个气球没有更新。
		可以理解为多加上一个编号为-1的气球，这个气球loc=n*/
		M = max(n - locLast, M); 
		printf("%d\n", M);
	}
	return 0;
}

时间复杂度 O(n)
空间复杂度 O(1)