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Find the largest interval that contains exactly one of the given N integers.
  • Difficulty Level : Medium
  • Last Updated : 23 Apr, 2021

Given an array arr[] of N distinct integers, the task is to find the maximum element in an interval [L, R] such that the interval contains exactly one of the given N integers and 1 ≤ L ≤ R ≤ 105
 

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 
 

 

Approach: The idea is to fix the element we want our interval to contain. Now we are interested in how much we can extend our interval to left and right without overlapping with other elements. 
So, we first sort our array. Then for a fixed element, we determine its ends using the previous and next elements. We should also take care of corner cases that are when we fix our first and last intervals. This way for every element i, we find the maximum length of the interval.
Below is the implementation of the above approach: 
 

CPP




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the maximum
// size of the required interval
int maxSize(vector<int>& 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<int> 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<Integer> 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<int> 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<int> v = new List<int>(arr);
        int n = v.Count;
 
        Console.WriteLine(maxSize(v, n));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// 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));
 
</script>
Output: 
99990

 

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