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Calculate absolute difference between minimum and maximum sum of pairs in an array
  • Last Updated : 29 Apr, 2021

Given an array arr[] consisting of N integers, the task is to find the absolute difference between the minimum and maximum sum of any pairs of elements (arr[i], arr[j]) such that (i < j) and (arr[i] < arr[j]).

Examples:

Input: arr[] = {1, 2, 4, 7}
Output: 8
Explanation: All possible pairs are:

  • (1, 2) → Sum = 3
  • (1, 4) → Sum = 5
  • (1, 7) → Sum = 8
  • (2, 4) → Sum = 6
  • (2, 7) → Sum = 9
  • (4, 7) → Sum = 11

Therefore, the difference between maximum and minimum sum = 11 – 3 = 8.

Input: arr[] = {2, 5, 3}
Output: 2



Approach: The idea to solve the given problem is to create two auxiliary arrays to store minimum and maximum of suffixes of every index of suffixes of the given array and then find the required absolute difference. 
Follow the steps below to solve the problem:

  • Initialize two variables, say maxSum and minSum, to store the maximum and minimum sum of a pair of elements from the given array according to the given conditions.
  • Initialize two arrays, say suffixMax[] and suffixMin[] of size N, to store the maximum and minimum of suffixes for each index of the array arr[].
  • Traverse the given array arr[] in reverse and update suffixMin[] and suffixMax[] at each index.
  • Now, iterate over the range [0, N – 1] and perform the following steps:
    • If the current element arr[i] is less than suffixMax[i], then update the value of maxSum as the maximum of maxSum and (arr[i] + suffixMax[i]).
    • If the current element arr[i] is less than suffixMin[i], then update the value of minSum as the minimum of minSum and (arr[i] + suffixMin[i]).
  • After completing the above steps, print the value (maxSum – minSum) as the resultant difference.

Below is the implementation of the above approach: 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the difference between
// the maximum and minimum sum of a pair
// (arr[i], arr[j]) from the array such
// that i < j and arr[i] < arr[j]
int GetDiff(int A[], int N)
{
    // Stores the maximum from
    // the suffix of the array
    int SuffMaxArr[N];
 
    // Set the last element
    SuffMaxArr[N - 1] = A[N - 1];
 
    // Traverse the remaining array
    for (int i = N - 2; i >= 0; --i) {
 
        // Update the maximum from suffix
        // for the remaining indices
        SuffMaxArr[i] = max(SuffMaxArr[i + 1],
                            A[i + 1]);
    }
 
    // Stores the maximum sum of any pair
    int MaximumSum = INT_MIN;
 
    // Calculate the maximum sum
    for (int i = 0; i < N - 1; i++) {
        if (A[i] < SuffMaxArr[i])
            MaximumSum
                = max(MaximumSum,
                      A[i] + SuffMaxArr[i]);
    }
 
    // Stores the maximum sum of any pair
    int MinimumSum = INT_MAX;
 
    // Stores the minimum of suffixes
    // from the given array
    int SuffMinArr[N];
 
    // Set the last element
    SuffMinArr[N - 1] = INT_MAX;
 
    // Traverse the remaining array
    for (int i = N - 2; i >= 0; --i) {
 
        // Update the maximum from suffix
        // for the remaining indices
        SuffMinArr[i] = min(SuffMinArr[i + 1],
                            A[i + 1]);
    }
 
    // Calculate the minimum sum
    for (int i = 0; i < N - 1; i++) {
 
        if (A[i] < SuffMinArr[i]) {
            MinimumSum = min(MinimumSum,
                             A[i] + SuffMinArr[i]);
        }
    }
 
    // Return the resultant difference
    return abs(MaximumSum - MinimumSum);
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 4, 1, 3, 7, 5, 6 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << GetDiff(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to find the difference between
// the maximum and minimum sum of a pair
// (arr[i], arr[j]) from the array such
// that i < j and arr[i] < arr[j]
static int GetDiff(int A[], int N)
{
     
    // Stores the maximum from
    // the suffix of the array
    int SuffMaxArr[] = new int[N];
 
    // Set the last element
    SuffMaxArr[N - 1] = A[N - 1];
 
    // Traverse the remaining array
    for(int i = N - 2; i >= 0; --i)
    {
         
        // Update the maximum from suffix
        // for the remaining indices
        SuffMaxArr[i] = Math.max(SuffMaxArr[i + 1],
                                          A[i + 1]);
    }
 
    // Stores the maximum sum of any pair
    int MaximumSum = Integer.MIN_VALUE;
 
    // Calculate the maximum sum
    for(int i = 0; i < N - 1; i++)
    {
        if (A[i] < SuffMaxArr[i])
            MaximumSum = Math.max(MaximumSum,
                                  A[i] + SuffMaxArr[i]);
    }
 
    // Stores the maximum sum of any pair
    int MinimumSum = Integer.MAX_VALUE;
 
    // Stores the minimum of suffixes
    // from the given array
    int SuffMinArr[] = new int[N];
 
    // Set the last element
    SuffMinArr[N - 1] = Integer.MAX_VALUE;
 
    // Traverse the remaining array
    for(int i = N - 2; i >= 0; --i)
    {
         
        // Update the maximum from suffix
        // for the remaining indices
        SuffMinArr[i] = Math.min(SuffMinArr[i + 1],
                                          A[i + 1]);
    }
 
    // Calculate the minimum sum
    for(int i = 0; i < N - 1; i++)
    {
        if (A[i] < SuffMinArr[i])
        {
            MinimumSum = Math.min(MinimumSum,
                                  A[i] + SuffMinArr[i]);
        }
    }
 
    // Return the resultant difference
    return Math.abs(MaximumSum - MinimumSum);
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 2, 4, 1, 3, 7, 5, 6 };
    int N = arr.length;
 
    // Function Call
    System.out.println(GetDiff(arr, N));
}
}
 
// This code is contributed by Kingash

Python3




# Python 3 program for the above approach
import sys
 
# Function to find the difference between
# the maximum and minimum sum of a pair
# (arr[i], arr[j]) from the array such
# that i < j and arr[i] < arr[j]
def GetDiff(A, N):
   
    # Stores the maximum from
    # the suffix of the array
    SuffMaxArr = [0 for i in range(N)]
 
    # Set the last element
    SuffMaxArr[N - 1] = A[N - 1]
 
    # Traverse the remaining array
    i = N-2
    while(i >= 0):
        # Update the maximum from suffix
        # for the remaining indices
        SuffMaxArr[i] = max(SuffMaxArr[i + 1], A[i + 1])
        i -= 1
 
    # Stores the maximum sum of any pair
    MaximumSum = -sys.maxsize-1
 
    # Calculate the maximum sum
    for i in range(N-1):
        if (A[i] < SuffMaxArr[i]):
            MaximumSum = max(MaximumSum, A[i] + SuffMaxArr[i])
 
    # Stores the maximum sum of any pair
    MinimumSum = sys.maxsize
 
    # Stores the minimum of suffixes
    # from the given array
    SuffMinArr = [0 for i in range(N)]
 
    # Set the last element
    SuffMinArr[N - 1] = sys.maxsize
 
    # Traverse the remaining array
    i = N-2
    while(i >= 0):
       
        # Update the maximum from suffix
        # for the remaining indices
        SuffMinArr[i] = min(SuffMinArr[i + 1], A[i + 1])
        i -= 1
 
    # Calculate the minimum sum
    for i in range(N - 1):
        if (A[i] < SuffMinArr[i]):
            MinimumSum = min(MinimumSum,A[i] + SuffMinArr[i])
 
    # Return the resultant difference
    return abs(MaximumSum - MinimumSum)
 
# Driver Code
if __name__ == '__main__':
    arr = [2, 4, 1, 3, 7, 5, 6]
    N = len(arr)
     
    # Function Call
    print(GetDiff(arr, N))
 
    # This code is contributed by SURENDRA_GANGWAR.

C#




// C# Program to implement
// the above approach
using System;
 
class GFG
{
 
// Function to find the difference between
// the maximum and minimum sum of a pair
// (arr[i], arr[j]) from the array such
// that i < j and arr[i] < arr[j]
static int GetDiff(int[] A, int N)
{
     
    // Stores the maximum from
    // the suffix of the array
    int[] SuffMaxArr = new int[N];
 
    // Set the last element
    SuffMaxArr[N - 1] = A[N - 1];
 
    // Traverse the remaining array
    for(int i = N - 2; i >= 0; --i)
    {
         
        // Update the maximum from suffix
        // for the remaining indices
        SuffMaxArr[i] = Math.Max(SuffMaxArr[i + 1],
                                          A[i + 1]);
    }
 
    // Stores the maximum sum of any pair
    int MaximumSum = Int32.MinValue;
 
    // Calculate the maximum sum
    for(int i = 0; i < N - 1; i++)
    {
        if (A[i] < SuffMaxArr[i])
            MaximumSum = Math.Max(MaximumSum,
                                  A[i] + SuffMaxArr[i]);
    }
 
    // Stores the maximum sum of any pair
    int MinimumSum = Int32.MaxValue;
 
    // Stores the minimum of suffixes
    // from the given array
    int[] SuffMinArr = new int[N];
 
    // Set the last element
    SuffMinArr[N - 1] = Int32.MaxValue;
 
    // Traverse the remaining array
    for(int i = N - 2; i >= 0; --i)
    {
         
        // Update the maximum from suffix
        // for the remaining indices
        SuffMinArr[i] = Math.Min(SuffMinArr[i + 1],
                                          A[i + 1]);
    }
 
    // Calculate the minimum sum
    for(int i = 0; i < N - 1; i++)
    {
        if (A[i] < SuffMinArr[i])
        {
            MinimumSum = Math.Min(MinimumSum,
                                  A[i] + SuffMinArr[i]);
        }
    }
 
    // Return the resultant difference
    return Math.Abs(MaximumSum - MinimumSum);
}
 
// Driver Code
public static void Main(String[] args)
{
    int[] arr = { 2, 4, 1, 3, 7, 5, 6 };
    int N = arr.Length;
 
    // Function Call
    Console.WriteLine(GetDiff(arr, N));
}
}
 
// This code is contributed by code_hunt.

Javascript




<script>
// JavaScript program for the above approach
 
// Function to find the difference between
// the maximum and minimum sum of a pair
// (arr[i], arr[j]) from the array such
// that i < j and arr[i] < arr[j]
function GetDiff(A, N)
{
      
    // Stores the maximum from
    // the suffix of the array
    let SuffMaxArr = Array(N).fill(0);
  
    // Set the last element
    SuffMaxArr[N - 1] = A[N - 1];
  
    // Traverse the remaining array
    for(let i = N - 2; i >= 0; --i)
    {
          
        // Update the maximum from suffix
        // for the remaining indices
        SuffMaxArr[i] = Math.max(SuffMaxArr[i + 1],
                                          A[i + 1]);
    }
  
    // Stores the maximum sum of any pair
    let MaximumSum = Number.MIN_VALUE;
  
    // Calculate the maximum sum
    for(let i = 0; i < N - 1; i++)
    {
        if (A[i] < SuffMaxArr[i])
            MaximumSum = Math.max(MaximumSum,
                                  A[i] + SuffMaxArr[i]);
    }
  
    // Stores the maximum sum of any pair
    let MinimumSum = Number.MAX_VALUE;
  
    // Stores the minimum of suffixes
    // from the given array
    let SuffMinArr = Array(N).fill(0);
  
    // Set the last element
    SuffMinArr[N - 1] = Number.MAX_VALUE;
  
    // Traverse the remaining array
    for(let i = N - 2; i >= 0; --i)
    {
          
        // Update the maximum from suffix
        // for the remaining indices
        SuffMinArr[i] = Math.min(SuffMinArr[i + 1],
                                          A[i + 1]);
    }
  
    // Calculate the minimum sum
    for(let i = 0; i < N - 1; i++)
    {
        if (A[i] < SuffMinArr[i])
        {
            MinimumSum = Math.min(MinimumSum,
                                  A[i] + SuffMinArr[i]);
        }
    }
  
    // Return the resultant difference
    return Math.abs(MaximumSum - MinimumSum);
}
 
// Driver Code
 
    let arr = [ 2, 4, 1, 3, 7, 5, 6 ];
    let N = arr.length;
  
    // Function Call
    document.write(GetDiff(arr, N));
     
</script>
Output: 
7

 

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

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