找出恰好包含给定 N 个整数之一的最大区间。

给定一个由N 个不同整数组成的数组arr[] ，任务是找到区间[L, R] 中的最大元素，使得该区间恰好包含给定N 个整数之一，并且1 ≤ L ≤ R ≤ 10 ⁵

Input: arr[] = {5, 10, 200} 
Output: 99990 
All possible intervals are [1, 9], [6, 199] and [11, 100000]. 
[11, 100000] has the maximum integers i.e. 99990.
Input: arr[] = {15000, 25000, 40000, 70000, 80000} 
Output: 44999 
 

方法：这个想法是修复我们希望我们的间隔包含的元素。现在我们感兴趣的是我们可以在不与其他元素重叠的情况下将我们的间隔向左和向右扩展多少。
所以，我们首先对我们的数组进行排序。然后对于固定元素，我们使用前一个和下一个元素确定其端点。我们还应该处理我们修复第一个和最后一个间隔时的极端情况。这样对于每个元素 i，我们找到了间隔的最大长度。
下面是上述方法的实现：

CPP

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the maximum
// size of the required interval
int maxSize(vector& v, int n)
{
    // Insert the borders for array
    v.push_back(0);
    v.push_back(100001);
    n += 2;
 
    // Sort the elements in ascending order
    sort(v.begin(), v.end());
 
    // To store the maximum size
    int mx = 0;
    for (int i = 1; i < n - 1; i++) {
 
        // To store the range [L, R] such that
        // only v[i] lies within the range
        int L = v[i - 1] + 1;
        int R = v[i + 1] - 1;
 
        // Total integers in the range
        int cnt = R - L + 1;
        mx = max(mx, cnt);
    }
 
    return mx;
}
 
// Driver code
int main()
{
    vector v = { 200, 10, 5 };
    int n = v.size();
 
    cout << maxSize(v, n);
 
    return 0;
}

Java

// Java implementation of the approach
import static java.lang.Integer.max;
import java.util.*;
 
class GFG
{
 
    // Function to return the maximum
    // size of the required interval
    static int maxSize(Vector v, int n)
    {
        // Insert the borders for array
        v.add(0);
        v.add(100001);
        n += 2;
 
        // Sort the elements in ascending order
        Collections.sort(v);
 
        // To store the maximum size
        int mx = 0;
        for (int i = 1; i < n - 1; i++)
        {
 
            // To store the range [L, R] such that
            // only v[i] lies within the range
            int L = v.get(i - 1) + 1;
            int R = v.get(i + 1) - 1;
 
            // Total integers in the range
            int cnt = R - L + 1;
            mx = max(mx, cnt);
        }
 
        return mx;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        Integer arr[] = {200, 10, 5};
        Vector v = new Vector(Arrays.asList(arr));
        int n = v.size();
 
        System.out.println(maxSize(v, n));
    }
}
 
// This code is contributed by Princi Singh

Python

# Python3 implementation of the approach
 
# Function to return the maximum
# size of the required interval
def maxSize(v, n):
 
    # Insert the borders for array
    v.append(0)
    v.append(100001)
    n += 2
 
    # Sort the elements in ascending order
    v = sorted(v)
 
    # To store the maximum size
    mx = 0
    for i in range(1, n - 1):
 
        # To store the range [L, R] such that
        # only v[i] lies within the range
        L = v[i - 1] + 1
        R = v[i + 1] - 1
 
        # Total integers in the range
        cnt = R - L + 1
        mx = max(mx, cnt)
     
 
    return mx
 
 
# Driver code
v = [ 200, 10, 5]
n = len(v)
 
print(maxSize(v, n))
 
# This code is contributed by mohit kumar 29

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Function to return the maximum
    // size of the required interval
    static int maxSize(List v, int n)
    {
        // Insert the borders for array
        v.Add(0);
        v.Add(100001);
        n += 2;
 
        // Sort the elements in ascending order
        v.Sort();
 
        // To store the maximum size
        int mx = 0;
        for (int i = 1; i < n - 1; i++)
        {
 
            // To store the range [L, R] such that
            // only v[i] lies within the range
            int L = v[i - 1] + 1;
            int R = v[i + 1] - 1;
 
            // Total integers in the range
            int cnt = R - L + 1;
            mx = Math.Max(mx, cnt);
        }
 
        return mx;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {200, 10, 5};
        List v = new List(arr);
        int n = v.Count;
 
        Console.WriteLine(maxSize(v, n));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript implementation of the approach
 
// Function to return the maximum
// size of the required interval
function maxSize(v, n)
{
    // Insert the borders for array
    v.push(0);
    v.push(100001);
    n += 2;
 
    // Sort the elements in ascending order
    v.sort((a, b) => a - b);
 
    // To store the maximum size
    let mx = 0;
    for (let i = 1; i < n - 1; i++) {
 
        // To store the range [L, R] such that
        // only v[i] lies within the range
        let L = v[i - 1] + 1;
        let R = v[i + 1] - 1;
 
        // Total integers in the range
        let cnt = R - L + 1;
        mx = Math.max(mx, cnt);
    }
 
    return mx;
}
 
// Driver code
    let v = [ 200, 10, 5 ];
    let n = v.length;
 
    document.write(maxSize(v, n));
 

输出：

99990

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