Given an array arr[] of integers, find out the maximum difference between any two elements such that larger element appears after the smaller number.
Examples :
Input : arr = {2, 3, 10, 6, 4, 8, 1}
Output : 8
Explanation : The maximum difference is between 10 and 2.Input : arr = {7, 9, 5, 6, 3, 2}
Output : 2
Explanation : The maximum difference is between 9 and 7.
Naive Approach:
We can solve this problem using two loops. In the outer loop, pick elements one by one and in the inner loop calculate the difference of the picked element with every other element in the array and compare the difference with the maximum difference calculated so far.
Below is the implementation of the above approach :
// C++ program to find Maximum difference // between two elements such that larger // element appears after the smaller number #include <bits/stdc++.h> using namespace std;
/* The function assumes that there are at least two elements in array. The
function returns a negative value if the
array is sorted in decreasing order and
returns 0 if elements are equal */
int maxDiff( int arr[], int arr_size)
{ int max_diff = arr[1] - arr[0];
for ( int i = 0; i < arr_size; i++)
{
for ( int j = i+1; j < arr_size; j++)
{
if (arr[j] - arr[i] > max_diff)
max_diff = arr[j] - arr[i];
}
}
return max_diff;
} /* Driver program to test above function */ int main()
{ int arr[] = {1, 2, 90, 10, 110};
int n = sizeof (arr) / sizeof (arr[0]);
// Function calling
cout << "Maximum difference is " << maxDiff(arr, n);
return 0;
} |
#include<stdio.h> /* The function assumes that there are at least two elements in array.
The function returns a negative value if the array is
sorted in decreasing order.
Returns 0 if elements are equal */
int maxDiff( int arr[], int arr_size)
{ int max_diff = arr[1] - arr[0];
int i, j;
for (i = 0; i < arr_size; i++)
{
for (j = i+1; j < arr_size; j++)
{
if (arr[j] - arr[i] > max_diff)
max_diff = arr[j] - arr[i];
}
}
return max_diff;
} /* Driver program to test above function */ int main()
{ int arr[] = {1, 2, 90, 10, 110};
printf ( "Maximum difference is %d" , maxDiff(arr, 5));
getchar ();
return 0;
} |
// Java program to find Maximum difference // between two elements such that larger // element appears after the smaller number import java.io.*;
public class MaximumDifference
{ /* The function assumes that there are at least two
elements in array.
The function returns a negative value if the array is
sorted in decreasing order.
Returns 0 if elements are equal */
int maxDiff( int arr[], int arr_size)
{
int max_diff = arr[ 1 ] - arr[ 0 ];
int i, j;
for (i = 0 ; i < arr_size; i++)
{
for (j = i + 1 ; j < arr_size; j++)
{
if (arr[j] - arr[i] > max_diff)
max_diff = arr[j] - arr[i];
}
}
return max_diff;
}
/* Driver program to test above functions */
public static void main(String[] args)
{
MaximumDifference maxdif = new MaximumDifference();
int arr[] = { 1 , 2 , 90 , 10 , 110 };
System.out.println( "Maximum difference is " +
maxdif.maxDiff(arr, 5 ));
}
} // This code has been contributed by Mayank Jaiswal |
# Python 3 code to find Maximum difference # between two elements such that larger # element appears after the smaller number # The function assumes that there are at # least two elements in array. The function # returns a negative value if the array is # sorted in decreasing order. Returns 0 # if elements are equal def maxDiff(arr, arr_size):
max_diff = arr[ 1 ] - arr[ 0 ]
for i in range ( arr_size ):
for j in range ( i + 1 , arr_size ):
if (arr[j] - arr[i] > max_diff):
max_diff = arr[j] - arr[i]
return max_diff
# Driver program to test above function arr = [ 1 , 2 , 90 , 10 , 110 ]
size = len (arr)
print ( "Maximum difference is" , maxDiff(arr, size))
# This code is contributed by Swetank Modi |
// C# code to find Maximum difference using System;
class GFG {
// The function assumes that there
// are at least two elements in array.
// The function returns a negative
// value if the array is sorted in
// decreasing order. Returns 0 if
// elements are equal
static int maxDiff( int [] arr, int arr_size)
{
int max_diff = arr[1] - arr[0];
int i, j;
for (i = 0; i < arr_size; i++) {
for (j = i + 1; j < arr_size; j++) {
if (arr[j] - arr[i] > max_diff)
max_diff = arr[j] - arr[i];
}
}
return max_diff;
}
// Driver code
public static void Main()
{
int [] arr = { 1, 2, 90, 10, 110 };
Console.Write( "Maximum difference is " +
maxDiff(arr, 5));
}
} // This code is contributed by Sam007 |
<script> // javascript program to find Maximum difference // between two elements such that larger // element appears after the smaller number /* The function assumes that there are at least two elements in array. The
function returns a negative value if the
array is sorted in decreasing order and
returns 0 if elements are equal */
function maxDiff( arr, arr_size)
{ let max_diff = arr[1] - arr[0];
for (let i = 0; i < arr_size; i++)
{
for (let j = i+1; j < arr_size; j++)
{
if (arr[j] - arr[i] > max_diff)
max_diff = arr[j] - arr[i];
}
}
return max_diff;
} // Driver program to test above function
let arr = [1, 2, 90, 10, 110];
let n = arr.length;
// Function calling
document.write( "Maximum difference is " + maxDiff(arr, n));
// This code is contributed by jana_sayantan.
</script> |
<?php // PHP program to find Maximum difference // between two elements such that larger // element appears after the smaller number /* The function assumes that there are at least two elements in array. The function returns a negative value if the array is sorted in decreasing order and returns 0 if elements are equal */ function maxDiff( $arr , $arr_size )
{ $max_diff = $arr [1] - $arr [0];
for ( $i = 0; $i < $arr_size ; $i ++)
{ for ( $j = $i +1; $j < $arr_size ; $j ++)
{
if ( $arr [ $j ] - $arr [ $i ] > $max_diff )
$max_diff = $arr [ $j ] - $arr [ $i ];
}
} return $max_diff ;
} // Driver Code $arr = array (1, 2, 90, 10, 110);
$n = sizeof( $arr );
// Function calling echo "Maximum difference is " .
maxDiff( $arr , $n );
// This code is contributed // by Akanksha Rai(Abby_akku) |
Maximum difference is 109
Time Complexity : O(n^2)
Auxiliary Space : O(1)
Efficient Approach:
In this method, instead of taking difference of the picked element with every other element, we take the difference with the minimum element found so far. So we need to keep track of 2 things:
1) Maximum difference found so far (max_diff).
2) Minimum number visited so far (min_element).
Below is the implementation of the above idea:
// C++ program to find Maximum difference // between two elements such that larger // element appears after the smaller number #include <bits/stdc++.h> using namespace std;
/* The function assumes that there are at least two elements in array. The
function returns a negative value if the
array is sorted in decreasing order and
returns 0 if elements are equal */
int maxDiff( int arr[], int arr_size)
{ // Maximum difference found so far
int max_diff = arr[1] - arr[0];
// Minimum number visited so far
int min_element = arr[0];
for ( int i = 1; i < arr_size; i++)
{
if (arr[i] - min_element > max_diff)
max_diff = arr[i] - min_element;
if (arr[i] < min_element)
min_element = arr[i];
}
return max_diff;
} /* Driver program to test above function */ int main()
{ int arr[] = {1, 2, 90, 10, 110};
int n = sizeof (arr) / sizeof (arr[0]);
// Function calling
cout << "Maximum difference is " << maxDiff(arr, n);
return 0;
} |
#include<stdio.h> /* The function assumes that there are at least two elements in array. The function returns a negative value if the array is sorted in decreasing order. Returns 0 if elements are equal */ int maxDiff( int arr[], int arr_size)
{ int max_diff = arr[1] - arr[0];
int min_element = arr[0];
int i;
for (i = 1; i < arr_size; i++)
{ if (arr[i] - min_element > max_diff)
max_diff = arr[i] - min_element;
if (arr[i] < min_element)
min_element = arr[i];
} return max_diff;
} /* Driver program to test above function */ int main()
{ int arr[] = {1, 2, 6, 80, 100};
int size = sizeof (arr)/ sizeof (arr[0]);
printf ( "Maximum difference is %d" , maxDiff(arr, size));
getchar ();
return 0;
} |
// Java program to find Maximum difference // between two elements such that larger // element appears after the smaller number import java.io.*;
public class MaximumDifference
{ /* The function assumes that there are at least two
elements in array.
The function returns a negative value if the array is
sorted in decreasing order.
Returns 0 if elements are equal */
int maxDiff( int arr[], int arr_size)
{
int max_diff = arr[ 1 ] - arr[ 0 ];
int min_element = arr[ 0 ];
int i;
for (i = 1 ; i < arr_size; i++)
{
if (arr[i] - min_element > max_diff)
max_diff = arr[i] - min_element;
if (arr[i] < min_element)
min_element = arr[i];
}
return max_diff;
}
/* Driver program to test above functions */
public static void main(String[] args)
{
MaximumDifference maxdif = new MaximumDifference();
int arr[] = { 1 , 2 , 90 , 10 , 110 };
int size = arr.length;
System.out.println( "MaximumDifference is " +
maxdif.maxDiff(arr, size));
}
} // This code has been contributed by Mayank Jaiswal |
# Python 3 code to find Maximum difference # between two elements such that larger # element appears after the smaller number # The function assumes that there are # at least two elements in array. # The function returns a negative # value if the array is sorted in # decreasing order. Returns 0 if # elements are equal def maxDiff(arr, arr_size):
max_diff = arr[ 1 ] - arr[ 0 ]
min_element = arr[ 0 ]
for i in range ( 1 , arr_size ):
if (arr[i] - min_element > max_diff):
max_diff = arr[i] - min_element
if (arr[i] < min_element):
min_element = arr[i]
return max_diff
# Driver program to test above function arr = [ 1 , 2 , 6 , 80 , 100 ]
size = len (arr)
print ( "Maximum difference is" ,
maxDiff(arr, size))
# This code is contributed by Swetank Modi |
// C# code to find Maximum difference using System;
class GFG {
// The function assumes that there
// are at least two elements in array.
// The function returns a negative
// value if the array is sorted in
// decreasing order.Returns 0 if
// elements are equal
static int maxDiff( int [] arr, int arr_size)
{
int max_diff = arr[1] - arr[0];
int min_element = arr[0];
int i;
for (i = 1; i < arr_size; i++) {
if (arr[i] - min_element > max_diff)
max_diff = arr[i] - min_element;
if (arr[i] < min_element)
min_element = arr[i];
}
return max_diff;
}
// Driver code
public static void Main()
{
int [] arr = { 1, 2, 90, 10, 110 };
int size = arr.Length;
Console.Write( "MaximumDifference is " +
maxDiff(arr, size));
}
} // This code is contributed by Sam007 |
<script> // Javascript code to find Maximum difference
// between two elements such that larger
// element appears after the smaller number
// The function assumes that there
// are at least two elements in array.
// The function returns a negative
// value if the array is sorted in
// decreasing order.Returns 0 if
// elements are equal
function maxDiff(arr, arr_size)
{
let max_diff = arr[1] - arr[0];
let min_element = arr[0];
let i;
for (i = 1; i < arr_size; i++) {
if (arr[i] - min_element > max_diff)
max_diff = arr[i] - min_element;
if (arr[i] < min_element)
min_element = arr[i];
}
return max_diff;
}
let arr = [ 1, 2, 90, 10, 110 ];
let size = arr.length;
document.write( "Maximum difference is " + maxDiff(arr, size));
</script> |
<?php // PHP program to find Maximum // difference between two elements // such that larger element appears // after the smaller number // The function assumes that there // are at least two elements in array. // The function returns a negative // value if the array is sorted in // decreasing order and returns 0 // if elements are equal function maxDiff( $arr , $arr_size )
{ // Maximum difference found so far
$max_diff = $arr [1] - $arr [0];
// Minimum number visited so far
$min_element = $arr [0];
for ( $i = 1; $i < $arr_size ; $i ++)
{
if ( $arr [ $i ] - $min_element > $max_diff )
$max_diff = $arr [ $i ] - $min_element ;
if ( $arr [ $i ] < $min_element )
$min_element = $arr [ $i ];
}
return $max_diff ;
} // Driver Code $arr = array (1, 2, 90, 10, 110);
$n = count ( $arr );
// Function calling echo "Maximum difference is " .
maxDiff( $arr , $n );
// This code is contributed by Sam007 ?> |
Maximum difference is 109
Time Complexity : O(n)
Auxiliary Space : O(1)
Approach Using Kadane’s Algorithm:
First find the difference between the adjacent elements of the array and store all differences in an auxiliary array diff[] of size n-1. Now this problems turns into finding the maximum sum subarray of this difference array. Thanks to Shubham Mittal for suggesting this solution.
Below is the implementation :
// C++ program to find Maximum difference // between two elements such that larger // element appears after the smaller number #include <bits/stdc++.h> using namespace std;
/* The function assumes that there are at least two elements in array. The
function returns a negative value if the
array is sorted in decreasing order and
returns 0 if elements are equal */
int maxDiff ( int arr[], int n)
{ // Initialize diff, current sum and max sum
int diff = arr[1]-arr[0];
int curr_sum = diff;
int max_sum = curr_sum;
for ( int i=1; i<n-1; i++)
{
// Calculate current diff
diff = arr[i+1]-arr[i];
// Calculate current sum
if (curr_sum > 0)
curr_sum += diff;
else
curr_sum = diff;
// Update max sum, if needed
if (curr_sum > max_sum)
max_sum = curr_sum;
}
return max_sum;
} /* Driver program to test above function */ int main()
{ int arr[] = {80, 2, 6, 3, 100};
int n = sizeof (arr) / sizeof (arr[0]);
// Function calling
cout << "Maximum difference is " << maxDiff(arr, n);
return 0;
} |
// Java program to find Maximum // difference between two elements // such that larger element appears // after the smaller number import java.io.*;
public class GFG
{ /* The function assumes that there are at least two elements in array. The function returns a negative value if the array is sorted in decreasing order and returns 0 if elements are equal */ static int maxDiff ( int arr[], int n)
{ // Initialize diff, current
// sum and max sum
int diff = arr[ 1 ] - arr[ 0 ];
int curr_sum = diff;
int max_sum = curr_sum;
for ( int i = 1 ; i < n - 1 ; i++)
{
// Calculate current diff
diff = arr[i + 1 ] - arr[i];
// Calculate current sum
if (curr_sum > 0 )
curr_sum += diff;
else
curr_sum = diff;
// Update max sum, if needed
if (curr_sum > max_sum)
max_sum = curr_sum;
}
return max_sum;
} // Driver Code public static void main(String[] args)
{ int arr[] = { 80 , 2 , 6 , 3 , 100 };
int n = arr.length;
// Function calling System.out.print( "Maximum difference is " +
maxDiff(arr, n));
} } // This code is contributed by Smitha |
# Python3 program to find Maximum difference # between two elements such that larger # element appears after the smaller number # The function assumes that there are # at least two elements in array. The # function returns a negative value if # the array is sorted in decreasing # order and returns 0 if elements are equal def maxDiff (arr, n):
# Initialize diff, current
# sum and max sum
diff = arr[ 1 ] - arr[ 0 ]
curr_sum = diff
max_sum = curr_sum
for i in range ( 1 , n - 1 ):
# Calculate current diff
diff = arr[i + 1 ] - arr[i]
# Calculate current sum
if (curr_sum > 0 ):
curr_sum + = diff
else :
curr_sum = diff
# Update max sum, if needed
if (curr_sum > max_sum):
max_sum = curr_sum
return max_sum
# Driver Code if __name__ = = '__main__' :
arr = [ 80 , 2 , 6 , 3 , 100 ]
n = len (arr)
# Function calling
print ( "Maximum difference is" ,
maxDiff(arr, n))
# This code is contributed # by 29AjayKumar |
// C# program to find Maximum // difference between two elements // such that larger element appears // after the smaller number using System;
class GFG
{ /* The function assumes that there are at least two elements in array. The function returns a negative value if the array is sorted in decreasing order and returns 0 if elements are equal */ static int maxDiff ( int [] arr, int n)
{ // Initialize diff, current
// sum and max sum
int diff = arr[1] - arr[0];
int curr_sum = diff;
int max_sum = curr_sum;
for ( int i = 1; i < n - 1; i++)
{
// Calculate current diff
diff = arr[i + 1] - arr[i];
// Calculate current sum
if (curr_sum > 0)
curr_sum += diff;
else
curr_sum = diff;
// Update max sum, if needed
if (curr_sum > max_sum)
max_sum = curr_sum;
}
return max_sum;
} // Driver Code public static void Main()
{ int [] arr = {80, 2, 6, 3, 100};
int n = arr.Length;
// Function calling Console.WriteLine( "Maximum difference is " +
maxDiff(arr, n));
} } // This code is contributed // by Akanksha Rai(Abby_akku) |
<script> // Javascript program to find Maximum // difference between two elements // such that larger element appears // after the smaller number /* The function assumes that there are at least two elements in array. The function returns a negative value if the array is sorted in decreasing order and returns 0 if elements are equal */ function maxDiff (arr, n)
{ // Initialize diff, current
// sum and max sum
let diff = arr[1] - arr[0];
let curr_sum = diff;
let max_sum = curr_sum;
for (let i = 1; i < n - 1; i++)
{
// Calculate current diff
diff = arr[i + 1] - arr[i];
// Calculate current sum
if (curr_sum > 0)
curr_sum += diff;
else
curr_sum = diff;
// Update max sum, if needed
if (curr_sum > max_sum)
max_sum = curr_sum;
}
return max_sum;
} // Driver Code let arr = [ 80, 2, 6, 3, 100 ]; let n = arr.length; // Function calling document.write( "Maximum difference is " +
maxDiff(arr, n));
// This code is contributed by rag2127 </script> |
<?php // PHP program to find Maximum difference // between two elements such that larger // element appears after the smaller number /* The function assumes that there are at least two elements in array. The function returns a negative value if the array is sorted in decreasing order and returns 0 if elements are equal */ function maxDiff ( $arr , $n )
{ // Initialize diff, current sum
// and max sum
$diff = $arr [1] - $arr [0];
$curr_sum = $diff ;
$max_sum = $curr_sum ;
for ( $i = 1; $i < $n - 1; $i ++)
{
// Calculate current diff
$diff = $arr [ $i + 1] - $arr [ $i ];
// Calculate current sum
if ( $curr_sum > 0)
$curr_sum += $diff ;
else
$curr_sum = $diff ;
// Update max sum, if needed
if ( $curr_sum > $max_sum )
$max_sum = $curr_sum ;
}
return $max_sum ;
} // Driver Code $arr = array (80, 2, 6, 3, 100);
$n = sizeof( $arr );
// Function calling echo "Maximum difference is " ,
maxDiff( $arr , $n );
// This code is contributed // by Sach_code ?> |
Maximum difference is 98
Time Complexity : O(n)
Auxiliary Space : O(1)
Below is a variation of this problem:
Maximum difference of sum of elements in two rows in a matrix
Please write comments if you find any bug in above codes/algorithms, or find other ways to solve the same problem