Sum of all differences between Maximum and Minimum of increasing Subarrays

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:
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:
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: 
 



Below is the implementation of the above approach:
 

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// 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;
}
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// 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
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# 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
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// 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
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Output: 
5


 

Time Complexity: O(N) 
Auxiliary Space: O(1)
 

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