Given an unsorted array of numbers, write a function that returns true if the array consists of consecutive numbers.
Examples:
a) If the array is {5, 2, 3, 1, 4}, then the function should return true because the array has consecutive numbers from 1 to 5.
b) If the array is {83, 78, 80, 81, 79, 82}, then the function should return true because the array has consecutive numbers from 78 to 83.
c) If the array is {34, 23, 52, 12, 3}, then the function should return false because the elements are not consecutive.
d) If the array is {7, 6, 5, 5, 3, 4}, then the function should return false because 5 and 5 are not consecutive.
Method 1 (Use Sorting)
1) Sort all the elements.
2) Do a linear scan of the sorted array. If the difference between the current element and the next element is anything other than 1, then return false. If all differences are 1, then return true.
#include <bits/stdc++.h> using namespace std;
// Function to Check if array // elements are consecutive bool areConsecutive( int arr[], int n)
{ //Sort the array
sort(arr,arr+n);
// checking the adjacent elements
for ( int i=1;i<n;i++)
{
if (arr[i]!=arr[i-1]+1)
{
return false ;
}
}
return true ;
} /* Driver program to test above functions */ int main()
{ int arr[]= {5, 4, 2, 3, 1, 6};
int n = sizeof (arr)/ sizeof (arr[0]);
if (areConsecutive(arr, n) == true )
cout<< " Array elements are consecutive " ;
else
cout<< " Array elements are not consecutive " ;
return 0;
} // This code is contributed by Aarti_Rathi |
// Java implementation of the approach import java.util.Arrays;
class AreConsecutive {
/* The function checks if the array elements are consecutive If elements are consecutive, then returns true, else returns false */ boolean areConsecutive( int arr[], int n)
{ //Sort the array
Arrays.sort(arr);
// checking the adjacent elements
for ( int i= 1 ;i<n;i++)
{
if (arr[i]!=arr[i- 1 ]+ 1 )
{
return false ;
}
}
return true ;
} public static void main(String[] args)
{
AreConsecutive consecutive = new AreConsecutive();
int arr[] = { 5 , 4 , 2 , 3 , 1 , 6 };
int n = arr.length;
if (consecutive.areConsecutive(arr, n) == true )
System.out.println( "Array elements are consecutive" );
else
System.out.println( "Array elements are not consecutive" );
}
} // This code is contributed by Aarti_Rathi |
# The function checks if the array elements # are consecutive. If elements are consecutive, # then returns true, else returns false def areConsecutive(arr, n):
# Sort the array
arr.sort()
# checking the adjacent elements
for i in range ( 1 ,n):
if (arr[i]! = arr[i - 1 ] + 1 ):
return False ;
return True ;
# Driver Code arr = [ 5 , 4 , 2 , 3 , 1 , 6 ]
n = len (arr)
if (areConsecutive(arr, n) = = True ):
print ( "Array elements are consecutive " )
else :
print ( "Array elements are not consecutive " )
# This code is contributed by Aarti_Rathi |
using System;
class GFG {
// Function to Check if array // elements are consecutive static bool areConsecutive( int []arr, int n)
{ //Sort the array
Array.Sort(arr);
// checking the adjacent elements
for ( int i=1;i<n;i++)
{
if (arr[i]!=arr[i-1]+1)
{
return false ;
}
}
return true ;
} /* Driver program to test above
functions */
public static void Main()
{
int []arr = {5, 4, 2, 3, 1, 6};
int n = arr.Length;
if (areConsecutive(arr, n) == true )
Console.Write( "Array elements "
+ "are consecutive" );
else
Console.Write( "Array elements "
+ "are not consecutive" );
}
} // This code is contributed by Aarti_Rathi |
//JS code to implement the approach // The function checks if the array elements // are consecutive. If elements are consecutive, // then returns true, else returns false function areConsecutive(arr, n)
{ // Sort the array
arr.sort();
// checking the adjacent elements
for ( var i = 1; i < n; i++)
if (arr[i]!=arr[i-1]+1)
return false ;
return true ;
} // Driver Code var arr = [5, 4, 2, 3, 1, 6];
var n = arr.length;
if (areConsecutive(arr, n) == true )
console.log( "Array elements are consecutive " );
else console.log( "Array elements are not consecutive " );
// This code is contributed by phasing17 |
Array elements are consecutive
Time Complexity: O(n log n)
Space Complexity: O(1)
Method 2 (Use visited array)
The idea is to check for the following two conditions. If the following two conditions are true, then return true.
1) max – min + 1 = n where max is the maximum element in the array, min is the minimum element in the array and n is the number of elements in the array.
2) All elements are distinct.
To check if all elements are distinct, we can create a visited[] array of size n. We can map the ith element of input array arr[] to the visited array by using arr[i] – min as the index in visited[].
#include<stdio.h> #include<stdlib.h> /* Helper functions to get minimum and maximum in an array */ int getMin( int arr[], int n);
int getMax( int arr[], int n);
/* The function checks if the array elements are consecutive If elements are consecutive, then returns true, else returns
false */
bool areConsecutive( int arr[], int n)
{ if ( n < 1 )
return false ;
/* 1) Get the minimum element in array */
int min = getMin(arr, n);
/* 2) Get the maximum element in array */
int max = getMax(arr, n);
/* 3) max - min + 1 is equal to n, then only check all elements */
if (max - min + 1 == n)
{
/* Create a temp array to hold visited flag of all elements.
Note that, calloc is used here so that all values are initialized
as false */
bool *visited = ( bool *) calloc (n, sizeof ( bool ));
int i;
for (i = 0; i < n; i++)
{
/* If we see an element again, then return false */
if ( visited[arr[i] - min] != false )
return false ;
/* If visited first time, then mark the element as visited */
visited[arr[i] - min] = true ;
}
/* If all elements occur once, then return true */
return true ;
}
return false ; // if (max - min + 1 != n)
} /* UTILITY FUNCTIONS */ int getMin( int arr[], int n)
{ int min = arr[0];
for ( int i = 1; i < n; i++)
if (arr[i] < min)
min = arr[i];
return min;
} int getMax( int arr[], int n)
{ int max = arr[0];
for ( int i = 1; i < n; i++)
if (arr[i] > max)
max = arr[i];
return max;
} /* Driver program to test above functions */ int main()
{ int arr[]= {5, 4, 2, 3, 1, 6};
int n = sizeof (arr)/ sizeof (arr[0]);
if (areConsecutive(arr, n) == true )
printf ( " Array elements are consecutive " );
else
printf ( " Array elements are not consecutive " );
getchar ();
return 0;
} |
class AreConsecutive
{ /* The function checks if the array elements are consecutive
If elements are consecutive, then returns true, else returns
false */
boolean areConsecutive( int arr[], int n)
{
if (n < 1 )
return false ;
/* 1) Get the minimum element in array */
int min = getMin(arr, n);
/* 2) Get the maximum element in array */
int max = getMax(arr, n);
/* 3) max - min + 1 is equal to n, then only check all elements */
if (max - min + 1 == n)
{
/* Create a temp array to hold visited flag of all elements.
Note that, calloc is used here so that all values are initialized
as false */
boolean visited[] = new boolean [n];
int i;
for (i = 0 ; i < n; i++)
{
/* If we see an element again, then return false */
if (visited[arr[i] - min] != false )
return false ;
/* If visited first time, then mark the element as visited */
visited[arr[i] - min] = true ;
}
/* If all elements occur once, then return true */
return true ;
}
return false ; // if (max - min + 1 != n)
}
/* UTILITY FUNCTIONS */
int getMin( int arr[], int n)
{
int min = arr[ 0 ];
for ( int i = 1 ; i < n; i++)
{
if (arr[i] < min)
min = arr[i];
}
return min;
}
int getMax( int arr[], int n)
{
int max = arr[ 0 ];
for ( int i = 1 ; i < n; i++)
{
if (arr[i] > max)
max = arr[i];
}
return max;
}
/* Driver program to test above functions */
public static void main(String[] args)
{
AreConsecutive consecutive = new AreConsecutive();
int arr[] = { 5 , 4 , 2 , 3 , 1 , 6 };
int n = arr.length;
if (consecutive.areConsecutive(arr, n) == true )
System.out.println( "Array elements are consecutive" );
else
System.out.println( "Array elements are not consecutive" );
}
} // This code has been contributed by Mayank Jaiswal |
# Helper functions to get Minimum and # Maximum in an array # The function checks if the array elements # are consecutive. If elements are consecutive, # then returns true, else returns false def areConsecutive(arr, n):
if ( n < 1 ):
return False
# 1) Get the Minimum element in array */
Min = min (arr)
# 2) Get the Maximum element in array */
Max = max (arr)
# 3) Max - Min + 1 is equal to n,
# then only check all elements */
if ( Max - Min + 1 = = n):
# Create a temp array to hold visited
# flag of all elements. Note that, calloc
# is used here so that all values are
# initialized as false
visited = [ False for i in range (n)]
for i in range (n):
# If we see an element again,
# then return false */
if (visited[arr[i] - Min ] ! = False ):
return False
# If visited first time, then mark
# the element as visited */
visited[arr[i] - Min ] = True
# If all elements occur once,
# then return true */
return True
return False # if (Max - Min + 1 != n)
# Driver Code arr = [ 5 , 4 , 2 , 3 , 1 , 6 ]
n = len (arr)
if (areConsecutive(arr, n) = = True ):
print ( "Array elements are consecutive " )
else :
print ( "Array elements are not consecutive " )
# This code is contributed by mohit kumar |
using System;
class GFG {
/* The function checks if the array elements
are consecutive If elements are consecutive,
then returns true, else returns false */
static bool areConsecutive( int []arr, int n)
{
if (n < 1)
return false ;
/* 1) Get the minimum element in array */
int min = getMin(arr, n);
/* 2) Get the maximum element in array */
int max = getMax(arr, n);
/* 3) max - min + 1 is equal to n, then
only check all elements */
if (max - min + 1 == n)
{
/* Create a temp array to hold visited
flag of all elements. Note that, calloc
is used here so that all values are
initialized as false */
bool []visited = new bool [n];
int i;
for (i = 0; i < n; i++)
{
/* If we see an element again, then
return false */
if (visited[arr[i] - min] != false )
return false ;
/* If visited first time, then mark
the element as visited */
visited[arr[i] - min] = true ;
}
/* If all elements occur once, then
return true */
return true ;
}
return false ; // if (max - min + 1 != n)
}
/* UTILITY FUNCTIONS */
static int getMin( int []arr, int n)
{
int min = arr[0];
for ( int i = 1; i < n; i++)
{
if (arr[i] < min)
min = arr[i];
}
return min;
}
static int getMax( int []arr, int n)
{
int max = arr[0];
for ( int i = 1; i < n; i++)
{
if (arr[i] > max)
max = arr[i];
}
return max;
}
/* Driver program to test above functions */
public static void Main()
{
int []arr = {5, 4, 2, 3, 1, 6};
int n = arr.Length;
if (areConsecutive(arr, n) == true )
Console.Write( "Array elements are"
+ " consecutive" );
else
Console.Write( "Array elements are"
+ " not consecutive" );
}
} // This code is contributed by nitin mittal. |
<script> /* The function checks if the array elements are consecutive
If elements are consecutive, then returns true, else returns
false */
function areConsecutive(arr,n)
{
if (n < 1)
return false ;
/* 1) Get the minimum element in array */
let min = getMin(arr, n);
/* 2) Get the maximum element in array */
let max = getMax(arr, n);
/* 3) max - min + 1 is equal to n, then only check all elements */
if (max - min + 1 == n)
{
/* Create a temp array to hold visited flag of all elements.
Note that, calloc is used here so that all values are initialized
as false */
let visited = new Array(n);
for (let i=0;i<n;i++)
{
visited[i]= false ;
}
let i;
for (i = 0; i < n; i++)
{
/* If we see an element again, then return false */
if (visited[arr[i] - min] != false )
{
return false ;
}
/* If visited first time, then mark the element as visited */
visited[arr[i] - min] = true ;
}
/* If all elements occur once, then return true */
return true ;
}
return false ; // if (max - min + 1 != n)
}
/* UTILITY FUNCTIONS */
function getMin(arr, n)
{
let min = arr[0];
for (let i = 1; i < n; i++)
{
if (arr[i] < min)
min = arr[i];
}
return min;
}
function getMax(arr,n)
{
let max = arr[0];
for (let i = 1; i < n; i++)
{
if (arr[i] > max)
max = arr[i];
}
return max;
}
/* Driver program to test above functions */
let arr=[5, 4, 2, 3, 1, 6]
let n = arr.length;
if (areConsecutive(arr, n))
{
document.write( "Array elements are consecutive" );
}
else
{
document.write( "Array elements are not consecutive" );
}
// This code is contributed by avanitrachhadiya2155
</script> |
<?php // PHP Program for above approach // The function checks if the array elements // are consecutive. If elements are consecutive, // then returns true, else returns false function areConsecutive( $arr , $n )
{ if ( $n < 1 )
return false;
// 1) Get the minimum element in array
$min = getMin( $arr , $n );
// 2) Get the maximum element in array
$max = getMax( $arr , $n );
// 3) $max - $min + 1 is equal to $n,
// then only check all elements
if ( $max - $min + 1 == $n )
{
// Create a temp array to hold
// visited flag of all elements.
$visited = array ();
for ( $i = 0; $i < $n ; $i ++)
{
$visited [ $i ] = false;
}
for ( $i = 0; $i < $n ; $i ++)
{
// If we see an element again,
// then return false
if ( $visited [ $arr [ $i ] - $min ] != false )
return false;
// If visited first time, then mark
// the element as visited
$visited [ $arr [ $i ] - $min ] = true;
}
// If all elements occur once,
// then return true
return true;
}
return false; // if ($max - $min + 1 != $n)
} // UTILITY FUNCTIONS function getMin( $arr , $n )
{ $min = $arr [0];
for ( $i = 1; $i < $n ; $i ++)
if ( $arr [ $i ] < $min )
$min = $arr [ $i ];
return $min ;
} function getMax( $arr , $n )
{ $max = $arr [0];
for ( $i = 1; $i < $n ; $i ++)
if ( $arr [ $i ] > $max )
$max = $arr [ $i ];
return $max ;
} // Driver Code $arr = array (5, 4, 2, 3, 1, 6);
$n = count ( $arr );
if (areConsecutive( $arr , $n ) == true)
echo "Array elements are consecutive " ;
else echo "Array elements are not consecutive " ;
// This code is contributed by rathbhupendra ?> |
Array elements are consecutive
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 3 (Mark visited array elements as negative)
This method is O(n) time complexity and O(1) extra space, but it changes the original array, and it works only if all numbers are positive. We can get the original array by adding an extra step though. It is an extension of method 2, and it has the same two steps.
1) max – min + 1 = n where max is the maximum element in the array, min is the minimum element in the array and n is the number of elements in the array.
2) All elements are distinct.
In this method, the implementation of step 2 differs from method 2. Instead of creating a new array, we modify the input array arr[] to keep track of visited elements. The idea is to traverse the array and for each index i (where 0 ? i < n), make arr[arr[i] – min]] as a negative value. If we see a negative value again then there is repetition.
#include<stdio.h> #include<stdlib.h> /* Helper functions to get minimum and maximum in an array */ int getMin( int arr[], int n);
int getMax( int arr[], int n);
/* The function checks if the array elements are consecutive If elements are consecutive, then returns true, else returns
false */
bool areConsecutive( int arr[], int n)
{ if ( n < 1 )
return false ;
/* 1) Get the minimum element in array */
int min = getMin(arr, n);
/* 2) Get the maximum element in array */
int max = getMax(arr, n);
/* 3) max - min + 1 is equal to n then only check all elements */
if (max - min + 1 == n)
{
int i;
for (i = 0; i < n; i++)
{
int j;
if (arr[i] < 0)
j = -arr[i] - min;
else
j = arr[i] - min;
// if the value at index j is negative then
// there is repetition
if (arr[j] > 0)
arr[j] = -arr[j];
else
return false ;
}
/* If we do not see a negative value then all elements
are distinct */
return true ;
}
return false ; // if (max - min + 1 != n)
} /* UTILITY FUNCTIONS */ int getMin( int arr[], int n)
{ int min = arr[0];
for ( int i = 1; i < n; i++)
if (arr[i] < min)
min = arr[i];
return min;
} int getMax( int arr[], int n)
{ int max = arr[0];
for ( int i = 1; i < n; i++)
if (arr[i] > max)
max = arr[i];
return max;
} /* Driver program to test above functions */ int main()
{ int arr[]= {1, 4, 5, 3, 2, 6};
int n = sizeof (arr)/ sizeof (arr[0]);
if (areConsecutive(arr, n) == true )
printf ( " Array elements are consecutive " );
else
printf ( " Array elements are not consecutive " );
getchar ();
return 0;
} |
class AreConsecutive
{ /* The function checks if the array elements are consecutive
If elements are consecutive, then returns true, else returns
false */
boolean areConsecutive( int arr[], int n)
{
if (n < 1 )
return false ;
/* 1) Get the minimum element in array */
int min = getMin(arr, n);
/* 2) Get the maximum element in array */
int max = getMax(arr, n);
/* 3) max-min+1 is equal to n then only check all elements */
if (max - min + 1 == n)
{
int i;
for (i = 0 ; i < n; i++)
{
int j;
if (arr[i] < 0 )
j = -arr[i] - min;
else
j = arr[i] - min;
// if the value at index j is negative then
// there is repetition
if (arr[j] > 0 )
arr[j] = -arr[j];
else
return false ;
}
/* If we do not see a negative value then all elements
are distinct */
return true ;
}
return false ; // if (max-min+1 != n)
}
/* UTILITY FUNCTIONS */
int getMin( int arr[], int n)
{
int min = arr[ 0 ];
for ( int i = 1 ; i < n; i++)
{
if (arr[i] < min)
min = arr[i];
}
return min;
}
int getMax( int arr[], int n)
{
int max = arr[ 0 ];
for ( int i = 1 ; i < n; i++)
{
if (arr[i] > max)
max = arr[i];
}
return max;
}
/* Driver program to test above functions */
public static void main(String[] args)
{
AreConsecutive consecutive = new AreConsecutive();
int arr[] = { 5 , 4 , 2 , 3 , 1 , 6 };
int n = arr.length;
if (consecutive.areConsecutive(arr, n) == true )
System.out.println( "Array elements are consecutive" );
else
System.out.println( "Array elements are not consecutive" );
}
} // This code is contributed by Mayank Jaiswal |
# Helper functions to get minimum and # maximum in an array # The function checks if the array # elements are consecutive. If elements # are consecutive, then returns true, # else returns false def areConsecutive(arr, n):
if ( n < 1 ):
return False
# 1) Get the minimum element in array
min = getMin(arr, n)
# 2) Get the maximum element in array
max = getMax(arr, n)
# 3) max - min + 1 is equal to n
# then only check all elements
if ( max - min + 1 = = n):
for i in range (n):
if (arr[i] < 0 ):
j = - arr[i] - min
else :
j = arr[i] - min
# if the value at index j is negative
# then there is repetition
if (arr[j] > 0 ):
arr[j] = - arr[j]
else :
return False
# If we do not see a negative value
# then all elements are distinct
return True
return False # if (max - min + 1 != n)
# UTILITY FUNCTIONS def getMin(arr, n):
min = arr[ 0 ]
for i in range ( 1 , n):
if (arr[i] < min ):
min = arr[i]
return min
def getMax(arr, n):
max = arr[ 0 ]
for i in range ( 1 , n):
if (arr[i] > max ):
max = arr[i]
return max
# Driver Code if __name__ = = "__main__" :
arr = [ 1 , 4 , 5 , 3 , 2 , 6 ]
n = len (arr)
if (areConsecutive(arr, n) = = True ):
print ( " Array elements are consecutive " )
else :
print ( " Array elements are not consecutive " )
# This code is contributed by ita_c |
using System;
class GFG {
/* The function checks if the array
elements are consecutive If elements
are consecutive, then returns true,
else returns false */
static bool areConsecutive( int []arr, int n)
{
if (n < 1)
return false ;
/* 1) Get the minimum element in
array */
int min = getMin(arr, n);
/* 2) Get the maximum element in
array */
int max = getMax(arr, n);
/* 3) max-min+1 is equal to n then
only check all elements */
if (max - min + 1 == n)
{
int i;
for (i = 0; i < n; i++)
{
int j;
if (arr[i] < 0)
j = -arr[i] - min;
else
j = arr[i] - min;
// if the value at index j
// is negative then
// there is repetition
if (arr[j] > 0)
arr[j] = -arr[j];
else
return false ;
}
/* If we do not see a negative
value then all elements
are distinct */
return true ;
}
// if (max-min+1 != n)
return false ;
}
/* UTILITY FUNCTIONS */
static int getMin( int []arr, int n)
{
int min = arr[0];
for ( int i = 1; i < n; i++)
{
if (arr[i] < min)
min = arr[i];
}
return min;
}
static int getMax( int []arr, int n)
{
int max = arr[0];
for ( int i = 1; i < n; i++)
{
if (arr[i] > max)
max = arr[i];
}
return max;
}
/* Driver program to test above
functions */
public static void Main()
{
int []arr = {5, 4, 2, 3, 1, 6};
int n = arr.Length;
if (areConsecutive(arr, n) == true )
Console.Write( "Array elements "
+ "are consecutive" );
else
Console.Write( "Array elements "
+ "are not consecutive" );
}
} // This code is contributed by nitin mittal. |
<script> /* The function checks if the array elements are consecutive
If elements are consecutive, then returns true, else returns
false */
function areConsecutive(arr,n)
{
if (n < 1)
return false ;
/* 1) Get the minimum element in array */
let min = getMin(arr, n);
/* 2) Get the maximum element in array */
let max = getMax(arr, n);
/* 3) max-min+1 is equal to n then only check all elements */
if (max - min + 1 == n)
{
let i;
for (i = 0; i < n; i++)
{
let j;
if (arr[i] < 0)
j = -arr[i] - min;
else
j = arr[i] - min;
// if the value at index j is negative then
// there is repetition
if (arr[j] > 0)
arr[j] = -arr[j];
else
return false ;
}
/* If we do not see a negative value then all elements
are distinct */
return true ;
}
return false ; // if (max-min+1 != n)
}
/* UTILITY FUNCTIONS */
function getMin(arr,n)
{
let min = arr[0];
for (let i = 1; i < n; i++)
{
if (arr[i] < min)
min = arr[i];
}
return min;
}
function getMax(arr,n)
{
let max = arr[0];
for (let i = 1; i < n; i++)
{
if (arr[i] > max)
max = arr[i];
}
return max;
}
/* Driver program to test above functions */
let arr=[5, 4, 2, 3, 1, 6];
let n = arr.length;
if (areConsecutive(arr, n) == true )
document.write( "Array elements are consecutive" );
else
document.write( "Array elements are not consecutive" );
// This code is contributed by unknown2108
</script> |
<?php /* The function checks if the array elements are consecutive If elements are consecutive, then returns true, else returns false */ function areConsecutive( $arr , $n )
{ if ( $n < 1 )
return false;
/* 1) Get the minimum element in array */
$min = getMin( $arr , $n );
/* 2) Get the maximum element in array */
$max = getMax( $arr , $n );
/* 3) max - min + 1 is equal to n then
only check all elements */
if ( $max - $min + 1 == $n )
{
$i ;
for ( $i = 0; $i < $n ; $i ++)
{
$j ;
if ( $arr [ $i ] < 0)
$j = - $arr [ $i ] - $min ;
else
$j = $arr [ $i ] - $min ;
// if the value at index j is
// negative then there is
// repetition
if ( $arr [ $j ] > 0)
$arr [ $j ] = - $arr [ $j ];
else
return false;
}
/* If we do not see a negative value
then all elements are distinct */
return true;
}
return false; // if (max - min + 1 != n)
} /* UTILITY FUNCTIONS */ function getMin( $arr , $n )
{ $min = $arr [0];
for ( $i = 1; $i < $n ; $i ++)
if ( $arr [ $i ] < $min )
$min = $arr [ $i ];
return $min ;
} function getMax( $arr , $n )
{ $max = $arr [0];
for ( $i = 1; $i < $n ; $i ++)
if ( $arr [ $i ] > $max )
$max = $arr [ $i ];
return $max ;
} /* Driver program to test above functions */ $arr = array (1, 4, 5, 3, 2, 6);
$n = count ( $arr );
if (areConsecutive( $arr , $n ) == true)
echo " Array elements are consecutive " ;
else
echo " Array elements are not consecutive " ;
// This code is contributed by anuj_67. ?> |
Array elements are consecutive
Note that this method might not work for negative numbers. For example, it returns false for {2, 1, 0, -3, -1, -2}.
Time Complexity: O(n)
Auxiliary Space: O(1)
Method 4 (Using XOR property)
This method is O(n) time complexity and O(1) extra space, does not changes the original array, and it works every time.
- As elements should be consecutive, let’s find minimum element or maximum element in array.
- Now if we take xor of two same elements it will result in zero (a^a = 0).
-
Suppose array is {-2, 0, 1, -3, 4, 3, 2, -1}, now if we xor all array elements with minimum element and keep increasing minimum element, the resulting xor will become 0 only if elements are consecutive
//Code is contributed by Dhananjay Dhawale @chessnoobdj #include <iostream> #include <algorithm> using namespace std;
bool areConsecutive( int arr[], int n)
{ int min_ele = *min_element(arr, arr+n), num = 0;
for ( int i=0; i<n; i++){
num ^= min_ele^arr[i];
min_ele += 1;
}
if (num == 0)
return 1;
return 0;
} /* Driver program to test above functions */ int main()
{ int arr[]= {1, 4, 5, 3, 2, 6};
int n = sizeof (arr)/ sizeof (arr[0]);
if (areConsecutive(arr, n) == true )
printf ( " Array elements are consecutive " );
else
printf ( " Array elements are not consecutive " );
getchar ();
return 0;
} |
//Code is contributed by Dhananjay Dhawale @chessnoobdj #include<stdio.h> #include<stdlib.h> /* UTILITY FUNCTIONS */ int getMin( int arr[], int n)
{ int min = arr[0];
for ( int i = 1; i < n; i++)
if (arr[i] < min)
min = arr[i];
return min;
} int areConsecutive( int arr[], int n)
{ int min_ele = getMin(arr, n), num = 0;
for ( int i=0; i<n; i++){
num ^= min_ele^arr[i];
min_ele += 1;
}
if (num == 0)
return 1;
return 0;
} /* Driver program to test above functions */ int main()
{ int arr[]= {1, 4, 5, 3, 2, 6};
int n = sizeof (arr)/ sizeof (arr[0]);
if (areConsecutive(arr, n) == 1)
printf ( " Array elements are consecutive " );
else
printf ( " Array elements are not consecutive " );
getchar ();
return 0;
} |
// Java implementation of the approach import java.util.Arrays;
import java.util.Collections;
class AreConsecutive {
boolean areConsecutive( int arr[], int n)
{ int min_ele = Arrays.stream(arr).min().getAsInt();
int num = 0 ;
for ( int i= 0 ; i<n; i++){
num = num ^ min_ele ^ arr[i];
min_ele += 1 ;
}
if (num == 0 )
return true ;
return false ;
} public static void main(String[] args)
{
AreConsecutive consecutive = new AreConsecutive();
int arr[] = { 5 , 4 , 2 , 3 , 1 , 6 };
int n = arr.length;
if (consecutive.areConsecutive(arr, n) == true )
System.out.println( "Array elements are consecutive" );
else
System.out.println( "Array elements are not consecutive" );
}
} // This code is contributed by Aarti_Rathi |
# Function to Check if array # elements are consecutive def areConsecutive(arr, n):
min_ele = arr.index( min (arr))
num = 0
for i in range ( 0 , n):
num ^ = arr[min_ele] ^ arr[i]
arr[min_ele] + = 1
if num = = 0 :
return True
return False
# Driver program to test above # functions if __name__ = = "__main__" :
arr = [ 1 , 4 , 5 , 3 , 2 , 6 ]
n = len (arr)
if areConsecutive(arr, n) = = True :
print ( " Array elements are consecutive " , end = ' ' )
else :
print ( " Array elements are not consecutive " , end = ' ' )
# This code is contributed by Aarti_Rathi |
using System;
using System.Linq;
class GFG {
// Function to Check if array // elements are consecutive static bool areConsecutive( int []arr, int n)
{ int min_ele = arr.Min();
int num = 0;
for ( int i=0; i<n; i++){
num ^= min_ele^arr[i];
min_ele += 1;
}
if (num == 0)
return true ;
return false ;
} /* Driver program to test above
functions */
public static void Main()
{
int []arr = {5, 4, 2, 3, 1, 6};
int n = arr.Length;
if (areConsecutive(arr, n) == true )
Console.Write( "Array elements "
+ "are consecutive" );
else
Console.Write( "Array elements "
+ "are not consecutive" );
}
} // This code is contributed by Aarti_Rathi |
// Javascript Program var areConsecutive = function (arr)
{ var min_ele = Math.min.apply(Math, arr);
var num = 0;
for ( var i = 0; i < arr.length; i++){
num = num ^ min_ele ^ arr[i];
min_ele += 1;
}
if (num == 0)
return 1;
return 0;
} /* Driver program to test above functions */ arr = [1, 2, 3, 4, 5, 6]; if (areConsecutive(arr) == 1){
console.log( " Array elements are consecutive " );}
else console.log( " Array elements are not consecutive " );
// This code is contributed by Sajal Aggarwal. |
Array elements are consecutive
Time Complexity: O(n)
Auxiliary Space: O(1)
Approach:
We can solve this problem using a mathematical formula. If the array has consecutive numbers, then the sum of the elements should be equal to n*(n+1)/2, where n is the size of the array. We can use this formula to check if the array has consecutive numbers.
Steps of approach:
- Initialize a variable sum to 0, minVal to the first element of the array, and maxVal to the first element of the array.
- Traverse the array and Add the current element to sum.
- If the current element is less than minVal, update minVal to the current element.
- If the current element is greater than maxVal, update maxVal to the current element.
- Check if the sum of the elements is equal to n*(n+1)/2 and the difference between the minimum and maximum values is equal to n-1, where n is the size of the array.
- If the above condition is true, return true, else return false.
Below is the implementation of the above approach:
#include <bits/stdc++.h> using namespace std;
bool areConsecutive( int arr[], int n)
{ int sum = 0, minVal = arr[0], maxVal = arr[0];
// Calculate the sum of the elements and find the minimum and maximum values
for ( int i = 0; i < n; i++)
{
sum += arr[i];
if (arr[i] < minVal)
{
minVal = arr[i];
}
if (arr[i] > maxVal)
{
maxVal = arr[i];
}
}
// Check if the sum of the elements is equal to n*(n+1)/2 and the difference between the minimum and maximum values is equal to n-1
if (sum == (n * (n + 1)) / 2 && maxVal - minVal == n - 1)
{
return true ;
}
else
{
return false ;
}
} int main()
{ int arr[] = {5, 2, 3, 1, 4};
int n = sizeof (arr) / sizeof (arr[0]);
if (areConsecutive(arr, n))
{
cout << "Array elements are consecutive" << endl;
}
else
{
cout << "Array elements are not consecutive" << endl;
}
return 0;
} |
import java.util.Arrays;
public class GFG {
static boolean areConsecutive( int [] arr, int n) {
int sum = 0 ;
int minVal = arr[ 0 ];
int maxVal = arr[ 0 ];
// Calculate the sum of the elements and find the
// minimum and maximum values
for ( int i = 0 ; i < n; i++) {
sum += arr[i];
if (arr[i] < minVal) {
minVal = arr[i];
}
if (arr[i] > maxVal) {
maxVal = arr[i];
}
}
// Check if the sum of the elements is equal to n*(n+1)/2 and the difference between the minimum and maximum values is equal to n-1
if (sum == (n * (n + 1 )) / 2 && maxVal - minVal == n - 1 ) {
return true ;
} else {
return false ;
}
}
public static void main(String[] args) {
int [] arr = { 5 , 2 , 3 , 1 , 4 };
int n = arr.length;
if (areConsecutive(arr, n)) {
System.out.println( "Array elements are consecutive" );
} else {
System.out.println( "Array elements are not consecutive" );
}
}
} |
def are_consecutive(arr):
n = len (arr)
_sum = 0
min_val = arr[ 0 ]
max_val = arr[ 0 ]
# Calculate the sum of the elements and find the minimum and maximum values
for i in range (n):
_sum + = arr[i]
if arr[i] < min_val:
min_val = arr[i]
if arr[i] > max_val:
max_val = arr[i]
# Check if the sum of the elements is equal to n*(n+1)/2 and the difference between the minimum and maximum values is equal to n-1
if _sum = = (n * (n + 1 )) / / 2 and max_val - min_val = = n - 1 :
return True
else :
return False
def main():
arr = [ 5 , 2 , 3 , 1 , 4 ]
if are_consecutive(arr):
print ( "Array elements are consecutive" )
else :
print ( "Array elements are not consecutive" )
if __name__ = = "__main__" :
main()
# This article is contributed by Abhay |
using System;
class GFG
{ // Function to check if the elements in the array are consecutive
static bool AreConsecutive( int [] arr, int n)
{
int sum = 0, minVal = arr[0], maxVal = arr[0];
// Calculate the sum of the elements
// and find the minimum and maximum values
for ( int i = 0; i < n; i++)
{
sum += arr[i];
if (arr[i] < minVal)
{
minVal = arr[i];
}
if (arr[i] > maxVal)
{
maxVal = arr[i];
}
}
// Check if the sum of the elements is equal to n*(n+1)/2
// and the difference between the minimum
// and maximum values is equal to n-1
if (sum == (n * (n + 1)) / 2 && maxVal - minVal == n - 1)
{
return true ;
}
else
{
return false ;
}
}
static void Main()
{
int [] arr = { 5, 2, 3, 1, 4 };
int n = arr.Length;
if (AreConsecutive(arr, n))
{
Console.WriteLine( "Array elements are consecutive" );
}
else
{
Console.WriteLine( "Array elements are not consecutive" );
}
}
} |
function areConsecutive(arr, n) {
let sum = 0;
let minVal = arr[0];
let maxVal = arr[0];
// Calculate the sum of the elements and find the
// minimum and maximum values
for (let i = 0; i < n; i++) {
sum += arr[i];
if (arr[i] < minVal) {
minVal = arr[i];
}
if (arr[i] > maxVal) {
maxVal = arr[i];
}
}
// Check if the sum of the elements is equal to n*(n+1)/2 and the difference between the minimum and maximum values is equal to n-1
if (sum === (n * (n + 1)) / 2 && maxVal - minVal === n - 1) {
return true ;
} else {
return false ;
}
} const arr = [5, 2, 3, 1, 4]; const n = arr.length; if (areConsecutive(arr, n))
console.log( "Array elements are consecutive" );
else console.log( "Array elements are not consecutive" );
|
Output:
Array elements are consecutive
Time Complexity: O(n) as it involves only a single traversal of the array.
Space Complexity: O(1), as we are not using any extra space.
Please suggest if someone has a better solution which is more efficient in terms of space and time.
Related Articles: Check if array elements are consecutive in O(n) time and O(1) space (Handles Both Positive and negative numbers)