Minimum number of consecutive sequences that can be formed in an array
Last Updated :
24 Jun, 2022
Given an array of integers. The task is to find the minimum number of consecutive sequences that can be formed using the elements of the array.
Examples:
Input: arr[] = { -3, -2, -1, 0, 2 }
Output: 2
Consecutive sequences are (-3, -2, -1, 0), (2).
Input: arr[] = { 3, 4, 0, 2, 6, 5, 10 }
Output: 3
Consecutive sequences are (0), {2, 3, 4, 5, 6} and {10}
Approach:
- Sort the array.
- Iterate the array, and check if current element is just 1 smaller than the next element.
- If it is then increment the count by 1.
- Return the final count of consecutive sequences.
Below is the implementation of above approach :
C++
#include <bits/stdc++.h>
using namespace std;
int countSequences( int arr[], int n)
{
int count = 1;
sort(arr, arr + n);
for ( int i = 0; i < n - 1; i++)
if (arr[i] + 1 != arr[i + 1])
count++;
return count;
}
int main()
{
int arr[] = { 1, 7, 3, 5, 10 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << countSequences(arr, n);
return 0;
}
|
Java
import java.util.Arrays;
import java.io.*;
class GFG {
static int countSequences( int arr[], int n)
{
int count = 1 ;
Arrays.sort(arr);
for ( int i = 0 ; i < n - 1 ; i++)
if (arr[i] + 1 != arr[i + 1 ])
count++;
return count;
}
public static void main (String[] args) {
int arr[] = { 1 , 7 , 3 , 5 , 10 };
int n = arr.length;
System.out.println( countSequences(arr, n));
}
}
|
Python3
def countSequences(arr, n) :
count = 1
arr.sort()
for i in range ( n - 1 ) :
if (arr[i] + 1 ! = arr[i + 1 ]) :
count + = 1
return count
if __name__ = = "__main__" :
arr = [ 1 , 7 , 3 , 5 , 10 ]
n = len (arr)
print (countSequences(arr, n))
|
C#
using System;
class GFG {
static int countSequences( int []arr, int n)
{
int count = 1;
Array.Sort(arr);
for ( int i = 0; i < n - 1; i++)
if (arr[i] + 1 != arr[i + 1])
count++;
return count;
}
static public void Main (String []args) {
int []arr = { 1, 7, 3, 5, 10 };
int n = arr.Length;
Console.WriteLine( countSequences(arr, n));
}
}
|
PHP
<?php
function countSequences( $arr , $n )
{
$count = 1;
sort( $arr );
for ( $i = 0; $i < $n - 1; $i ++)
if ( $arr [ $i ] + 1 != $arr [ $i + 1])
$count ++;
return $count ;
}
$arr = array ( 1, 7, 3, 5, 10 );
$n = count ( $arr );
echo countSequences( $arr , $n );
?>
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Javascript
<script>
function countSequences(arr, n)
{
let count = 1;
arr.sort( function (a, b){ return a - b});
for (let i = 0; i < n - 1; i++)
if (arr[i] + 1 != arr[i + 1])
count++;
return count;
}
let arr = [ 1, 7, 3, 5, 10 ];
let n = arr.length;
document.write(countSequences(arr, n));
</script>
|
Time Complexity: O(n log n), where n is the size of the array.
Auxiliary Space: O(1)
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