优先队列STL

Training 2 - C题

You are given an array of n integer numbers a0, a1, …, an - 1. Find the distance between two closest (nearest) minimums in it. It is guaranteed that in the array a minimum occurs at least two times.

Input

The first line contains positive integer n (2 ≤ n ≤ 105) — size of the given array. The second line contains n integers a0, a1, …, an - 1 (1 ≤ ai ≤ 109) — elements of the array. It is guaranteed that in the array a minimum occurs at least two times.

Output

Print the only number — distance between two nearest minimums in the array.

Examples

Input

2
3 3

Output

1

Input

3
5 6 5

Output

2

Input

9
2 1 3 5 4 1 2 3 1

Output

3

#pragma warning (disable:4996)
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <vector>
#include <queue>
#define inf 0X7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e5 + 5;

struct Node
{
	int val;
	int pos;
	friend bool operator<(Node a, Node b)
	{
		if (a.val == b.val)
			return a.pos > b.pos;
		return a.val > b.val;
	}
};
priority_queue<Node> a;

int main()
{
	int n;
	scanf("%d", &n);
	int num = 0;
	int ans = inf;
	for (int i = 0; i < n; i++)
	{
		int val;
		scanf("%d", &val);
		Node t;
		t.val = val;
		t.pos = i;
		a.push(t);
	}

	num = a.top().val;
	int lastpos = a.top().pos;
	a.pop();
	while (a.size() != 0 && a.top().val == num)
	{
		ans = min(ans, a.top().pos - lastpos);
		lastpos = a.top().pos;
		a.pop();
	}
	printf("%d\n", ans);
	return 0;
}

思路：
用优先队列存储，将最小值取出计算当前的距离并更新答案。