两个最近的最小值之间的距离

给定一个包含 n 个整数的数组。找出数组中任意两次出现的最小整数之间的最小距离。
例子：

Input : arr[] = {5, 1, 2, 3, 4, 1, 2, 1}
Output : 2
Explanation: The minimum element 1 occurs at 
             indexes: 1, 5 and 7. So the minimum
             distance is 7-5 = 2.

Input : arr[] = {1, 2, 1}
Output : 2

蛮力方法：最简单的方法是找到最小元素的所有索引对并计算最小距离。
时间复杂度：O(n^2)，其中 n 是数组中元素的总数。
有效方法：一种有效的方法是观察索引 j 和 i 之间的距离总是小于索引 k 和 i 之间的距离，其中 k 大于 j。也就是说，我们只需要检查连续的最小元素对之间的距离，而不是所有对。下面是逐步算法：

• 查找数组中的最小元素
• 查找数组中所有出现的最小元素，并将索引插入新数组或列表或向量中。
• 检查索引列表的大小是否大于一，即最小元素至少出现两次。如果不是返回-1。
• 遍历索引列表并计算任意两个连续索引之间的最小差异。

下面是上述想法的实现：

C++

// CPP program to find Distance between
// two closest minimum
#include 
#include 
#include 
 
using namespace std;
 
// function to find Distance between
// two closest minimum
int findClosestMin(int arr[], int n)
{
    int min = INT_MAX;
 
    // find the min element in the array
    for (int i = 0; i < n; i++)
        if (arr[i] < min)
            min = arr[i];
 
    // vector to store indexes of occurrences
    // of minimum element in the array
    vector indexes;
 
    // store indexes of occurrences
    // of minimum element in the array
    for (int i = 0; i < n; i++)
        if (arr[i] == min)
            indexes.push_back(i);
 
    // if minimum element doesnot occurs atleast
    // two times, return -1.
    if (indexes.size() < 2)
        return -1;
 
    int min_dist = INT_MAX;
 
    // calculate minimum difference between
    // any two consecutive indexes
    for (int i = 1; i < indexes.size(); i++)
        if ((indexes[i] - indexes[i - 1]) < min_dist)
            min_dist = (indexes[i] - indexes[i - 1]);
 
    return min_dist;
}
 
// Driver code
int main()
{
    int arr[] = { 5, 1, 2, 3, 4, 1, 2, 1 };
    int size = sizeof(arr) / sizeof(arr[0]);
    cout << findClosestMin(arr, size);
    return 0;
}

Java

// Java program to find Distance between
// two closest minimum
import java.util.Vector;
 
class GFG {
 
// function to find Distance between
// two closest minimum
    static int findClosestMin(int arr[], int n) {
        int min = Integer.MAX_VALUE;
 
        // find the min element in the array
        for (int i = 0; i < n; i++) {
            if (arr[i] < min) {
                min = arr[i];
            }
        }
 
        // vector to store indexes of occurrences
        // of minimum element in the array
        Vector indexes = new Vector<>();
 
        // store indexes of occurrences
        // of minimum element in the array
        for (int i = 0; i < n; i++) {
            if (arr[i] == min) {
                indexes.add(i);
            }
        }
 
        // if minimum element doesnot occurs atleast
        // two times, return -1.
        if (indexes.size() < 2) {
            return -1;
        }
 
        int min_dist = Integer.MAX_VALUE;
 
        // calculate minimum difference between
        // any two consecutive indexes
        for (int i = 1; i < indexes.size(); i++) {
            if ((indexes.get(i) - indexes.get(i - 1)) < min_dist) {
                min_dist = (indexes.get(i) - indexes.get(i - 1));
            }
        }
 
        return min_dist;
    }
 
// Driver code
    public static void main(String args[]) {
        int arr[] = {5, 1, 2, 3, 4, 1, 2, 1};
        int size = arr.length;
        System.out.println(findClosestMin(arr, size));
    }
}
 
// This code is contributed by PrinciRaj19992

Python3

# Python3 program to find Distance
# between two closest minimum
import sys
 
# function to find Distance between
# two closest minimum
def findClosestMin(arr, n):
     
    #assigning maximum value in python
    min = sys.maxsize
     
     
    for i in range(0, n):
        if (arr[i] < min):
            min = arr[i]
 
    # list in python to store indexes
    # of occurrences of minimum element
    # in the array
    indexes = []
 
    # store indexes of occurrences
    # of minimum element in the array
    for i in range(0, n):
        if (arr[i] == min):
            indexes.append(i)
 
    # if minimum element doesnot occurs
    #  atleast two times, return -1.
    if (len(indexes) < 2):
        return -1
 
    min_dist = sys.maxsize
 
    # calculate minimum difference between
    # any two consecutive indexes
    for i in range(1, len(indexes)):
        if ((indexes[i] - indexes[i - 1]) < min_dist):
            min_dist = (indexes[i] - indexes[i - 1]);
 
    return min_dist;
 
# Driver code
arr = [ 5, 1, 2, 3, 4, 1, 2, 1 ]
ans = findClosestMin(arr, 8)
print (ans)
 
# This code is contributed by saloni1297.

C#

// C# program to find Distance between
// two closest minimum
using System;
using System.Collections.Generic;
public class GFG {
  
// function to find Distance between
// two closest minimum
    static int findClosestMin(int []arr, int n) {
        int min = int.MaxValue;
  
        // find the min element in the array
        for (int i = 0; i < n; i++) {
            if (arr[i] < min) {
                min = arr[i];
            }
        }
  
        // vector to store indexes of occurrences
        // of minimum element in the array
        List indexes = new List();
  
        // store indexes of occurrences
        // of minimum element in the array
        for (int i = 0; i < n; i++) {
            if (arr[i] == min) {
                indexes.Add(i);
            }
        }
  
        // if minimum element doesnot occurs atleast
        // two times, return -1.
        if (indexes.Count < 2) {
            return -1;
        }
        int min_dist = int.MaxValue;
  
        // calculate minimum difference between
        // any two consecutive indexes
        for (int i = 1; i < indexes.Count; i++) {
            if ((indexes[i] - indexes[i-1]) < min_dist) {
                min_dist = (indexes[i] - indexes[i-1]);
            }
        }
  
        return min_dist;
    }
  
// Driver code
    public static void Main() {
        int []arr = {5, 1, 2, 3, 4, 1, 2, 1};
        int size = arr.Length;
        Console.WriteLine(findClosestMin(arr, size));
    }
}
  
// This code is contributed by PrinciRaj19992

Javascript

    // Javascript program to find Distance between two closest minimum
     
    // function to find Distance between
    // two closest minimum
    function findClosestMin(arr, n) {
        let min = Number.MAX_VALUE;
    
        // find the min element in the array
        for (let i = 0; i < n; i++) {
            if (arr[i] < min) {
                min = arr[i];
            }
        }
    
        // vector to store indexes of occurrences
        // of minimum element in the array
        let indexes = [];
    
        // store indexes of occurrences
        // of minimum element in the array
        for (let i = 0; i < n; i++) {
            if (arr[i] == min) {
                indexes.push(i);
            }
        }
    
        // if minimum element doesnot occurs atleast
        // two times, return -1.
        if (indexes.length < 2) {
            return -1;
        }
        let min_dist = Number.MAX_VALUE;
    
        // calculate minimum difference between
        // any two consecutive indexes
        for (let i = 1; i < indexes.length; i++) {
            if ((indexes[i] - indexes[i-1]) < min_dist) {
                min_dist = (indexes[i] - indexes[i-1]);
            }
        }
    
        return min_dist;
    }
     
    let arr = [5, 1, 2, 3, 4, 1, 2, 1];
    let size = arr.length;
    document.write(findClosestMin(arr, size));
     
    // This code is contributed by suresh07.

输出：

2

时间复杂度：O(n)
辅助空间：O(n)