须藤放置[1.5] |范围内第二小

给定一个包含 N 个整数和 Q 个查询的数组。每个查询由 L 和 R 组成。任务是打印范围 LR 中第二小的元素。如果不存在第二小的元素，则打印 -1。

例子：

Input: 
a[] = {1, 2, 2, 4} 
Queries= 2 
L = 1, R = 2 
L = 0, R = 1
Output: 
-1 
2  

方法：处理每个查询并使用以下算法打印第二小的查询。

• 将第一个和第二个最小的都初始化为 INT_MAX
• 循环遍历所有元素。
  1. 如果当前元素小于第一个元素，则更新第一个和第二个。
  2. 否则，如果当前元素小于第二个，则更新第二个

如果循环完所有元素后第二个元素仍然是 INT_MAX，则打印 -1 否则打印第二小的元素。

下面是上述方法的实现：

C++

// C++ program for
// SP - Second Smallest in Range
#include 
using namespace std;
 
// Function to find the second smallest element
// in range L-R of an array
int secondSmallest(int a[], int n, int l, int r)
{
 
    int first = INT_MAX;
    int second = INT_MAX;
    for (int i = l; i <= r; i++) {
        if (a[i] < first) {
            second = first;
            first = a[i];
        }
        else if (a[i] < second and a[i] != first) {
            second = a[i];
        }
    }
 
    if (second == INT_MAX)
        return -1;
    else
        return second;
}
 
// function to perform queries
void performQueries(int a[], int n)
{
    // 1-st query
    int l = 1;
    int r = 2;
    cout << secondSmallest(a, n, l, r) << endl;
 
    // 2nd query
    l = 0;
    r = 1;
    cout << secondSmallest(a, n, l, r);
}
 
// Driver Code
int main()
{
    int a[] = { 1, 2, 2, 4 };
    int n = sizeof(a) / sizeof(a[0]);
    performQueries(a, n);
    return 0;
}

Java

// Java program for
// SP - Second Smallest in Range
class GFG
{
// Function to find the
// second smallest element
// in range L-R of an array
static int secondSmallest(int a[], int n,
                          int l, int r)
{
int first = Integer.MAX_VALUE;
int second = Integer.MAX_VALUE;
for (int i = l; i <= r; i++)
{
    if (a[i] < first)
    {
        second = first;
        first = a[i];
    }
    else if (a[i] < second &&
             a[i] != first)
    {
        second = a[i];
    }
}
 
if (second == Integer.MAX_VALUE)
    return -1;
else
    return second;
}
 
// function to perform queries
static void performQueries(int a[], int n)
{
    // 1-st query
    int l = 1;
    int r = 2;
    System.out.println(secondSmallest(a, n, l, r));
     
    // 2nd query
    l = 0;
    r = 1;
    System.out.println(secondSmallest(a, n, l, r));
}
 
// Driver Code
public static void main(String[] args)
{
    int a[] = { 1, 2, 2, 4 };
    int n = a.length;
    performQueries(a, n);
}
}
 
// This code is contributed
// by ChitraNayal

Python

# Python program for
# SP - Second Smallest in Range
 
# Function to find the
# second smallest element
# in range L-R of an array
import sys
def secondSmallest(a, n, l, r):
 
    first = sys.maxsize
    second = sys.maxsize
    for i in range(l, r + 1):
     
        if (a[i] < first):
         
            second = first
            first = a[i]
         
        elif (a[i] < second and
              a[i] != first):
         
            second = a[i]
 
    if (second == sys.maxsize):
        return -1
    else:
        return second
 
 
# function to perform queries
def performQueries(a, n):
 
    # 1-st query
    l = 1
    r = 2
    print(secondSmallest(a, n, l, r))
 
    # 2nd query
    l = 0
    r = 1
    print(secondSmallest(a, n, l, r))
 
# Driver Code
a = [1, 2, 2, 4 ]
n = len(a)
performQueries(a, n);
     
# This code is contributed
# by Shivi_Aggarwal

C#

// C# program for
// SP - Second Smallest in Range
using System;
 
class GFG
{
// Function to find the
// second smallest element
// in range L-R of an array
static int secondSmallest(int[] a, int n,
                        int l, int r)
{
 
int first = int.MaxValue;
int second = int.MaxValue;
for (int i = l; i <= r; i++)
{
    if (a[i] < first)
    {
        second = first;
        first = a[i];
    }
    else if (a[i] < second &&
            a[i] != first)
    {
        second = a[i];
    }
}
 
if (second == int.MaxValue)
    return -1;
else
    return second;
}
 
// function to perform queries
static void performQueries(int[] a, int n)
{
    // 1-st query
    int l = 1;
    int r = 2;
    Console.WriteLine(secondSmallest(a, n, l, r));
 
    // 2nd query
    l = 0;
    r = 1;
    Console.WriteLine(secondSmallest(a, n, l, r));
}
 
// Driver Code
public static void Main()
{
    int[] a = { 1, 2, 2, 4 };
    int n = a.Length;
    performQueries(a, n);
}
}
 
// This code is contributed
// by ChitraNayal

Javascript

// Javascript program for
// SP - Second Smallest in Range
 
 
     
// Function to find the
// second smallest element
// in range L-R of an array   
function secondSmallest(a,n,l,r)
{
let first = Number.MAX_VALUE;
let second = Number.MAX_VALUE;
for (let i = l; i <= r; i++)
{
    if (a[i] < first)
    {
        second = first;
        first = a[i];
    }
    else if (a[i] < second &&
             a[i] != first)
    {
        second = a[i];
    }
}
   
if (second == Number.MAX_VALUE)
    return -1;
else
    return second;
}
 
// function to perform queries
function performQueries(a,n)
{
    // 1-st query
    let l = 1;
    let r = 2;
    document.write(secondSmallest(a, n, l, r)+"<br>");
       
    // 2nd query
    l = 0;
    r = 1;
    document.write(secondSmallest(a, n, l, r)+"<br>");
}
 
// Driver Code
let a=[1, 2, 2, 4];
let n = a.length;
performQueries(a, n);
 
     
 
 
// This code is contributed by rag2127

输出：

-1
2

时间复杂度： O(M)，其中 M = RL 是 [L, R] 范围内的元素数
注意：由于问题的约束非常少，因此蛮力解决方案将通过。可以使用段树进一步优化该解决方案。