Given an array of size N. Let A and B be the minimum and maximum in the array respectively. Task is to find how many number should be added to the given array such that all the element in the range [A, B] occur at-least once in the array.
Examples:
Input : arr[] = {4, 5, 3, 8, 6}
Output : 1
Only 7 to be added in the list.
Input : arr[] = {2, 1, 3}
Output : 0
Method 1 (Sorting)
1- Sort the array.
2- Compare arr[i] == arr[i+1]-1 or not. If not, update count = arr[i+1]-arr[i]-1.
3- Return count.
C++
#include <bits/stdc++.h>
using namespace std;
int countNum(int arr[], int n)
{
int count = 0;
sort(arr, arr + n);
for (int i = 0; i < n - 1; i++)
if (arr[i] != arr[i+1] &&
arr[i] != arr[i + 1] - 1)
count += arr[i + 1] - arr[i] - 1;
return count;
}
int main()
{
int arr[] = { 3, 5, 8, 6 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << countNum(arr, n) << endl;
return 0;
}
|
Java
import java.io.*;
import java.util.*;
public class GFG {
static int countNum(int []arr, int n)
{
int count = 0;
Arrays.sort(arr);
for (int i = 0; i < n - 1; i++)
if (arr[i] != arr[i+1] &&
arr[i] != arr[i + 1] - 1)
count += arr[i + 1] - arr[i] - 1;
return count;
}
static public void main (String[] args)
{
int []arr = { 3, 5, 8, 6 };
int n = arr.length;
System.out.println(countNum(arr, n));
}
}
|
Python3
def countNum(arr, n):
count = 0
arr.sort()
for i in range(0, n-1):
if (arr[i] != arr[i+1] and
arr[i] != arr[i + 1] - 1):
count += arr[i + 1] - arr[i] - 1;
return count
arr = [ 3, 5, 8, 6 ]
n = len(arr)
print(countNum(arr, n))
|
C#
using System;
public class GFG {
static int countNum(int []arr, int n)
{
int count = 0;
Array.Sort(arr);
for (int i = 0; i < n - 1; i++)
if (arr[i] != arr[i+1] &&
arr[i] != arr[i + 1] - 1)
count += arr[i + 1] - arr[i] - 1;
return count;
}
static public void Main ()
{
int []arr = { 3, 5, 8, 6 };
int n = arr.Length;
Console.WriteLine(countNum(arr, n));
}
}
|
PHP
<?php
function countNum($arr, $n)
{
$count = 0;
sort($arr);
for ($i = 0; $i < $n - 1; $i++)
if ($arr[$i] != $arr[$i + 1] &&
$arr[$i] != $arr[$i + 1] - 1)
$count += $arr[$i + 1] -
$arr[$i] - 1;
return $count;
}
$arr = array(3, 5, 8, 6);
$n = count($arr);
echo countNum($arr, $n) ;
?>
|
Javascript
<script>
function countNum(arr, n)
{
let count = 0;
arr.sort();
for (let i = 0; i < n - 1; i++)
if (arr[i] != arr[i+1] &&
arr[i] != arr[i + 1] - 1)
count += arr[i + 1] - arr[i] - 1;
return count;
}
let arr = [ 3, 5, 8, 6 ];
let n = arr.length;
document.write(countNum(arr, n));
</script>
|
Output:
2
Time Complexity: O(n log n)
Method 2 (Use Hashing)
1- Maintain a hash of array elements.
2- Store minimum and maximum element.
3- Traverse from minimum to maximum element in hash
And count if element is not in hash.
4- Return count.
C++
#include <bits/stdc++.h>
using namespace std;
int countNum(int arr[], int n)
{
unordered_set<int> s;
int count = 0, maxm = INT_MIN, minm = INT_MAX;
for (int i = 0; i < n; i++) {
s.insert(arr[i]);
if (arr[i] < minm)
minm = arr[i];
if (arr[i] > maxm)
maxm = arr[i];
}
for (int i = minm; i <= maxm; i++)
if (s.find(arr[i]) == s.end())
count++;
return count;
}
int main()
{
int arr[] = { 3, 5, 8, 6 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << countNum(arr, n) << endl;
return 0;
}
|
Java
import java.util.HashSet;
class GFG
{
static int countNum(int arr[], int n)
{
HashSet<Integer> s = new HashSet<>();
int count = 0,
maxm = Integer.MIN_VALUE,
minm = Integer.MAX_VALUE;
for (int i = 0; i < n; i++)
{
s.add(arr[i]);
if (arr[i] < minm)
minm = arr[i];
if (arr[i] > maxm)
maxm = arr[i];
}
for (int i = minm; i <= maxm; i++)
if (!s.contains(i))
count++;
return count;
}
public static void main(String[] args)
{
int arr[] = { 3, 5, 8, 6 };
int n = arr.length;
System.out.println(countNum(arr, n));
}
}
|
Python3
def countNum(arr, n):
s = dict()
count, maxm, minm = 0, -10**9, 10**9
for i in range(n):
s[arr[i]] = 1
if (arr[i] < minm):
minm = arr[i]
if (arr[i] > maxm):
maxm = arr[i]
for i in range(minm, maxm + 1):
if i not in s.keys():
count += 1
return count
arr = [3, 5, 8, 6 ]
n = len(arr)
print(countNum(arr, n))
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static int countNum(int []arr, int n)
{
HashSet<int> s = new HashSet<int>();
int count = 0,
maxm = int.MinValue,
minm = int.MaxValue;
for (int i = 0; i < n; i++)
{
s.Add(arr[i]);
if (arr[i] < minm)
minm = arr[i];
if (arr[i] > maxm)
maxm = arr[i];
}
for (int i = minm; i <= maxm; i++)
if (!s.Contains(i))
count++;
return count;
}
public static void Main(String[] args)
{
int []arr = { 3, 5, 8, 6 };
int n = arr.Length;
Console.WriteLine(countNum(arr, n));
}
}
|
Javascript
<script>
function countNum(arr,n)
{
let s = new Set();
let count = 0,
maxm = Number.MIN_VALUE,
minm = Number.MAX_VALUE;
for (let i = 0; i < n; i++)
{
s.add(arr[i]);
if (arr[i] < minm)
minm = arr[i];
if (arr[i] > maxm)
maxm = arr[i];
}
for (let i = minm; i <= maxm; i++)
if (!s.has(i))
count++;
return count;
}
let arr=[3, 5, 8, 6 ];
let n = arr.length;
document.write(countNum(arr, n));
</script>
|
Output:
2
Time Complexity- O(max – min + 1)
This article is contributed by Sahil Chhabra. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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