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Find pair i, j such that |A[i]−A[j]| is same as sum of difference of them with any Array element

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Given an array A[] having N non-negative integers, find a pair of indices i and j such that the absolute difference between them is same as the sum of differences of those with any other array element i.e., | A[i] − A[k] | + | A[k]− A[j] | = | A[i] − A[j] |, where k can be any index.

Examples:

Input: N = 3, A[] = {2, 7, 5}
Output: 0 1
Explanation:
For k = 0: 
|A[0] – A[0]| + |A[0] – A[1]|
= |2 − 2| + |2 – 7| = 0 + 5 = 5
= |A[0] – A[1]|
For k = 1:
|A[0] – A[1]| + |A[1] – A[1]|
= |2 − 7| + |7 – 7| = 5 + 0 = 5
= |A[0] – A[1]|
For k = 2:
|A[0] – A[2]| + |A[2] – A[1]|
= |2 − 5| + |5 – 7| = 3 + 2 = 5
= |A[0] – A[1]|

Input: N = 4, arr[] = {5, 9, 1, 3}
Output: 1 2
Explanation: 
For k = 0: 
|A[1] – A[0]| + |A[0] – A[2]|
= |9 − 5| + |5 – 1| = 4 + 4 = 8
= |A[1] – A[2]|
For k = 1:
|A[1] – A[1]| + |A[1] – A[2]|
= |9 − 9| + |9 – 1| = 0 + 8 = 8
= |A[1] – A[2]|
For k = 2:
|A[1] – A[2]| + |A[2] – A[2]|
= |9 − 1| + |1 – 1| = 8 + 0 = 8
= |A[1] – A[2]|
For k = 3:
|A[1] – A[3]| + |A[3] – A[2]|
= |9 − 3| + |3 – 1| = 6 + 2 = 8
= |A[1] – A[2]|

 

Approach: The problem can be solved with the below mathematical observation:

Depending on the realtion between A[i], A[j] and A[k], the inequality can be writtten in the following 4 ways:

When A[i] ≥ A[k] ≥ A[j]:
A[i] − A[k]  + A[k]− A[j] = A[i] − A[j]
=> A[i] – A[j] =  A[i] − A[j] 

When A[k] ≥ A[i], A[k] ≥ A[j]:
A[k] − A[i]  + A[k]− A[j] = |A[i] − A[j]|
=> 2*A[k] – A[i] – A[j] =  |A[i] − A[j]| 

When A[i] ≥ A[k], A[j] ≥ A[k]:
A[i] − A[k]  – A[k]+ A[j] = |A[i] − A[j]|
=> A[i] + A[j] – 2*A[k] =  |A[i] − A[j]| 

When A[j] ≥ A[k] ≥ A[i]:
– A[i] + A[k]  – A[k] + A[j] = – A[i] + A[j]
=> A[j] – A[i] =  A[j] − A[i].

From the above equations, it is clear that if value of A[i] and A[j] are not the extreme values of the array then the probability of the equation being satisfied depends on the value of A[k] and will not hold true when A[k] lies outside the range of [A[i], A[j]].

Based on the above observation it is clear that the value of A[i] and A[j] should be the maximum and the minimum among the array elements. Follow the below steps to solve the problem:

  • Traverse the array from k = 0 to N-1:
    • Update the index of the maximum element (say i) if A[k] is greater than A[i].
    • Update the index of the minimum element (say j) if A[k] is less than A[j].
  • Return the pair (i, j) as the answer.

Below is the implementation of the above approach.

C++

  

#include <bits/stdc++.h>

using namespace std;

  

pair<int, int> findPair(int N,

                        vector<int> vec)

{

    

    int maxi = *max_element(vec.begin(),

                            vec.end());

  

    

    int mini = *min_element(vec.begin(),

                            vec.end());

  

    int idx1 = 0;

    int idx2 = 0;

  

    

    for (int i = 0; i < N; i++) {

        if (vec[i] == maxi)

            idx1 = i;

    }

  

    

    for (int i = 0; i < N; i++) {

        if (vec[i] == mini)

            idx2 = i;

    }

  

    return { idx2, idx1 };

}

  

int main()

{

    int N = 3;

    vector<int> vec{ 2, 7, 5 };

  

    

    pair<int, int> ans = findPair(N, vec);

    cout << ans.first << " " << ans.second;

    return 0;

}

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

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