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Sum of all differences between Maximum and Minimum of increasing Subarrays
  • Difficulty Level : Hard
  • Last Updated : 01 Apr, 2021

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:  

  • 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++




// 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




// 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




# 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#




// 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

Javascript




<script>
 
// Javascript 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
function sum_of_differences(arr, N)
{
     
    // Stores the sum
    let sum = 0;
 
    let 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
let arr = [ 6, 1, 2, 5, 3, 4 ];
 
let N = arr.length;
 
document.write(sum_of_differences(arr, N));
 
// This code is contributed by divyesh072019
 
</script>
Output: 
5

 

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

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