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 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
# 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# 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.
<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>
Time Complexity: O(N)
Auxiliary Space: O(N)
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