Given an array arr[] consisting of N integers, the task is to find the sum of the differences between maximum and minimum element of all strictly increasing subarrays from the given array. All subarrays need to be in their longest possible form, i.e. if a subarray [i, j] form a strictly increasing subarray, then it should be considered as a whole and not [i, k] and [k+1, j] for some i <= k <= j.
A subarray is said to be strictly increasing if for every ith index in the subarray, except the last index, arr[i+1] > arr[i]
Examples:
Input: arr[ ] = {7, 1, 5, 3, 6, 4}
Output: 7
Explanation:
All possible increasing subarrays are {7}, {1, 5}, {3, 6} and {4}
Therefore, sum = (7 – 7) + (5 – 1) + (6 – 3) + (4 – 4) = 7
Input: arr[ ] = {1, 2, 3, 4, 5, 2}
Output: 4
Explanation:
All possible increasing subarrays are {1, 2, 3, 4, 5} and {2}
Therefore, sum = (5 – 1) + (2 – 2) = 4
Approach:
Follow the steps below to solve the problem:
- Traverse the array and for each iteration, find the rightmost element up to which the current subarray is strictly increasing.
- Let i be the starting element of the current subarray, and j index up to which the current subarray is strictly increasing. The maximum and minimum values of this subarray will be arr[j] and arr[i] respectively. So, add (arr[j] – arr[i]) to the sum.
- Continue iterating for the next subarray from (j+1)th index.
- After complete traversal of the array, print the final value of sum.
Below is the implementation of the above approach:
// C++ Program to find the sum of // differences of maximum and minimum // of strictly increasing subarrays #include <bits/stdc++.h> using namespace std;
// Function to calculate and return the // sum of differences of maximum and // minimum of strictly increasing subarrays int sum_of_differences( int arr[], int N)
{ // Stores the sum
int sum = 0;
int i, j, flag;
// Traverse the array
for (i = 0; i < N - 1; i++) {
if (arr[i] < arr[i + 1]) {
flag = 0;
for (j = i + 1; j < N - 1; j++) {
// If last element of the
// increasing sub-array is found
if (arr[j] >= arr[j + 1]) {
// Update sum
sum += (arr[j] - arr[i]);
i = j;
flag = 1;
break ;
}
}
// If the last element of the array
// is reached
if (flag == 0 && arr[i] < arr[N - 1]) {
// Update sum
sum += (arr[N - 1] - arr[i]);
break ;
}
}
}
// Return the sum
return sum;
} // Driver Code int main()
{ int arr[] = { 6, 1, 2, 5, 3, 4 };
int N = sizeof (arr) / sizeof (arr[0]);
cout << sum_of_differences(arr, N);
return 0;
} |
// Java program to find the sum of // differences of maximum and minimum // of strictly increasing subarrays class GFG{
// Function to calculate and return the // sum of differences of maximum and // minimum of strictly increasing subarrays static int sum_of_differences( int arr[], int N)
{ // Stores the sum
int sum = 0 ;
int i, j, flag;
// Traverse the array
for (i = 0 ; i < N - 1 ; i++)
{
if (arr[i] < arr[i + 1 ])
{
flag = 0 ;
for (j = i + 1 ; j < N - 1 ; j++)
{
// If last element of the
// increasing sub-array is found
if (arr[j] >= arr[j + 1 ])
{
// Update sum
sum += (arr[j] - arr[i]);
i = j;
flag = 1 ;
break ;
}
}
// If the last element of the array
// is reached
if (flag == 0 && arr[i] < arr[N - 1 ])
{
// Update sum
sum += (arr[N - 1 ] - arr[i]);
break ;
}
}
}
// Return the sum
return sum;
} // Driver Code public static void main (String []args)
{ int arr[] = { 6 , 1 , 2 , 5 , 3 , 4 };
int N = arr.length;
System.out.print(sum_of_differences(arr, N));
} } // This code is contributed by chitranayal |
# Python3 program to find the sum of # differences of maximum and minimum # of strictly increasing subarrays # Function to calculate and return the # sum of differences of maximum and # minimum of strictly increasing subarrays def sum_of_differences(arr, N):
# Stores the sum
sum = 0
# Traverse the array
i = 0
while (i < N - 1 ):
if arr[i] < arr[i + 1 ]:
flag = 0
for j in range (i + 1 , N - 1 ):
# If last element of the
# increasing sub-array is found
if arr[j] > = arr[j + 1 ]:
# Update sum
sum + = (arr[j] - arr[i])
i = j
flag = 1
break
# If the last element of the array
# is reached
if flag = = 0 and arr[i] < arr[N - 1 ]:
# Update sum
sum + = (arr[N - 1 ] - arr[i])
break
i + = 1
# Return the sum
return sum
# Driver Code arr = [ 6 , 1 , 2 , 5 , 3 , 4 ]
N = len (arr)
print (sum_of_differences(arr, N))
# This code is contributed by yatinagg |
// C# program to find the sum of // differences of maximum and minimum // of strictly increasing subarrays using System;
class GFG{
// Function to calculate and return the // sum of differences of maximum and // minimum of strictly increasing subarrays static int sum_of_differences( int []arr, int N)
{ // Stores the sum
int sum = 0;
int i, j, flag;
// Traverse the array
for (i = 0; i < N - 1; i++)
{
if (arr[i] < arr[i + 1])
{
flag = 0;
for (j = i + 1; j < N - 1; j++)
{
// If last element of the
// increasing sub-array is found
if (arr[j] >= arr[j + 1])
{
// Update sum
sum += (arr[j] - arr[i]);
i = j;
flag = 1;
break ;
}
}
// If the last element of the array
// is reached
if (flag == 0 && arr[i] < arr[N - 1])
{
// Update sum
sum += (arr[N - 1] - arr[i]);
break ;
}
}
}
// Return the sum
return sum;
} // Driver Code public static void Main ( string []args)
{ int []arr = { 6, 1, 2, 5, 3, 4 };
int N = arr.Length;
Console.Write(sum_of_differences(arr, N));
} } // This code is contributed by rock_cool |
5
Time Complexity: O(N)
Auxiliary Space: O(1)
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