Maximum value of |arr[i] – arr[j]| + |i – j|

Given a array of N positive integers. The task is to find the maximum value of |arr[i] – arr[j]| + |i – j|, where 0 <= i, j <= N – 1 and arr[i], arr[j] belong to the array.

Examples:

Input : N = 4, arr[] = { 1, 2, 3, 1 } 
Output : 4
Explanation:
Choose i = 0 and j = 2. This will result in |1-3|+|0-2| = 4 which is the maximum possible value.

Input : N = 3, arr[] = { 1, 1, 1 }
Output : 2

Method 1: The idea is to use brute force i.e iterate in two for loops.

Below is the implementation of this approach:  

C++

#include <bits/stdc++.h>
using namespace std;
#define MAX 10

// Return maximum value of |arr[i] - arr[j]| + |i - j|
int findValue(int arr[], int n)
{
    int ans = 0;

    // Iterating two for loop, one for
    // i and another for j.
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)

            // Evaluating |arr[i] - arr[j]| + |i - j|
            // and compare with previous maximum.
            ans = max(ans,
                      abs(arr[i] - arr[j]) + abs(i - j));

    return ans;
}

// Driven Program
int main()
{
    int arr[] = { 1, 2, 3, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << findValue(arr, n) << endl;

    return 0;
}

Java

// java program to find maximum value of
// |arr[i] - arr[j]| + |i - j|
class GFG {
    static final int MAX = 10;

    // Return maximum value of
    // |arr[i] - arr[j]| + |i - j|
    static int findValue(int arr[], int n)
    {
        int ans = 0;

        // Iterating two for loop,
        // one for i and another for j.
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)

                // Evaluating |arr[i] - arr[j]|
                // + |i - j| and compare with
                // previous maximum.
                ans = Math.max(ans,
                               Math.abs(arr[i] - arr[j])
                               + Math.abs(i - j));

        return ans;
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 1, 2, 3, 1 };
        int n = arr.length;

        System.out.println(findValue(arr, n));
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Python3 program to find
# maximum value of
# |arr[i] - arr[j]| + |i - j|

# Return maximum value of
# |arr[i] - arr[j]| + |i - j|
def findValue(arr, n):
    ans = 0;

    # Iterating two for loop,
    # one for i and another for j.
    for i in range(n):
        for j in range(n):

            # Evaluating |arr[i] -
            # arr[j]| + |i - j|
            # and compare with
            # previous maximum.
            ans = ans if ans>(abs(arr[i] - arr[j]) +
                              abs(i - j)) else (abs(arr[i] -
                                      arr[j]) + abs(i - j)) ;
    return ans;

# Driver Code
arr = [1, 2, 3, 1];
n = len(arr);
print(findValue(arr, n));

# This code is contributed by mits.

Output

4

Method 2 (tricky): First of all lets make four equations by removing the absolute value signs (“|”). The following 4 equations will be formed, and we need to find the maximum value of these equations and that will be our answer. 

1. arr[i] – arr[j] + i – j = (arr[i] + i) – (arr[j] + j)
2. arr[i] – arr[j] – i + j = (arr[i] – i) – (arr[j] – j)
3. -arr[i] + arr[j] + i – j = -(arr[i] – i) + (arr[j] – j)
4. -arr[i] + arr[j] – i + j = -(arr[i] + i) + (arr[j] + j)

Observe the equations (1) and (4) are identical. Similarly, equations (2) and (3) are identical.
Now the task is to find the maximum value of these equations. So the approach is to form two arrays, first_array[], it will store arr[i] + i, 0 <= i < n, second_array[], it will store arr[i] – i, 0 <= i < n. 
Now our task is easy, we just need to find the maximum difference between the two values of these two arrays.
For that, we find the maximum value and minimum value in the first_array and store their difference: 
ans1 = (maximum value in first_array – minimum value in first_array) 
Similarly, we need to find the maximum value and minimum value in the second_array and store their difference: 
ans2 = (maximum value in second_array – minimum value in second_array) 
Our answer will be a maximum of ans1 and ans2.

Below is the implementation of the above approach: 

PHP

<?php
// Efficient CPP program
// to find maximum value
// of |arr[i] - arr[j]| + |i - j|

// Return maximum |arr[i] -
// arr[j]| + |i - j|
function findValue($arr, $n)
{
    $a[] =array(); $b=array();$tmp;

    // Calculating first_array
    // and second_array
    for ($i = 0; $i < $n; $i++)
    {
        $a[$i] = ($arr[$i] + $i);
        $b[$i] = ($arr[$i] - $i);
    }

    $x = $a[0]; $y = $a[0];

    // Finding maximum and
    // minimum value in
    // first_array
    for ($i = 0; $i < $n; $i++)
    {
        if ($a[$i] > $x)
        $x = $a[$i];

        if ($a[$i] < $y)
            $y = $a[$i];
    }

    // Storing the difference
    // between maximum and
    // minimum value in first_array
    $ans1 = ($x - $y);

    $x = $b[0];
    $y = $b[0];

    // Finding maximum and
    // minimum value in
    // second_array
    for ($i = 0; $i < $n; $i++)
    {
        if ($b[$i] > $x)
            $x = $b[$i];

        if ($b[$i] < $y)
            $y = $b[$i];
    }

    // Storing the difference
    // between maximum and
    // minimum value in
    // second_array
    $ans2 = ($x -$y);

    return max($ans1, $ans2);
}

    // Driver Code
    $arr = array(1, 2, 3, 1);
    $n = count($arr);

    echo findValue($arr, $n);

// This code is contributed by anuj_67.
?>

Javascript

<script>
    // Efficient Javascript program to find maximum
    // value of |arr[i] - arr[j]| + |i - j|

    // Return maximum |arr[i] -
    // arr[j]| + |i - j|
    function findValue(arr, n)
    {
        let a = new Array(n);
        let b = new Array(n);
        // int tmp;

        // Calculating first_array
        // and second_array
        for (let i = 0; i < n; i++)
        {
            a[i] = (arr[i] + i);
            b[i] = (arr[i] - i);
        }

        let x = a[0], y = a[0];

        // Finding maximum and
        // minimum value in
        // first_array
        for (let i = 0; i < n; i++)
        {
            if (a[i] > x)
                x = a[i];

            if (a[i] < y)
                y = a[i];
        }

        // Storing the difference
        // between maximum and
        // minimum value in first_array
        let ans1 = (x - y);

        x = b[0];
        y = b[0];

        // Finding maximum and
        // minimum value in
        // second_array
        for (let i = 0; i < n; i++)
        {
            if (b[i] > x)
                x = b[i];

            if (b[i] < y)
                y = b[i];
        }

        // Storing the difference
        // between maximum and
        // minimum value in second_array
        let ans2 = (x - y);

        return Math.max(ans1, ans2);
    }

    let arr = [ 1, 2, 3, 1 ];
    let n = arr.length;
    document.write(findValue(arr, n));

</script>

Output

4

Method – 3

This solution is space optimization on above mentioned (method2) solution.
In Method 2 solution we had used two matrix of size n which laid to O(n) space complexity 
but here we only use O(1) space instead of that two n size array

Implementation:

C++

// Optimized CPP program to find maximum value of
// |arr[i] - arr[j]| + |i - j|
#include <bits/stdc++.h>
using namespace std;

// Return maximum |arr[i] - arr[j]| + |i - j|
int findValue(int A[], int n)
{
    /*
        Let us walk through all possible cases in | A[i]-A[j] | + | i-j | ->
            1. A[i]>A[j] , i>j
               A[i]-A[j] --- positive
               i-j       --- positive
               (A[i]+i)-(A[j]+j)  ------------------- Eqn 1

            2. A[i]<A[j] , i<j
               A[i]-A[j] --- negative
               i-j       --- negative
               -((A[i]+i)-(A[j]+j))  ------------------- Eqn 2

            3. A[i]<A[j] , i>j
               A[i]-A[j] --- negative
               i-j       --- positive
               (A[j]-j)-(A[i]-i)  ------------------- Eqn 3

            4. A[i]>A[j] , i<j
               A[i]-A[j] --- positive
               i-j       --- negative
               -((A[j]-j)-(A[i]-i)) ------------------- Eqn 4

            Eqn 3 and 4 are the same only difference in sign
            Eqn 1 and 2 are the same only difference in sign,
            So we consider these two sets of equations
    */

     int mx1=INT_MIN;
     int mn1=INT_MAX;

     int mn2=INT_MAX;
     int mx2=INT_MIN;

     // mx1, mn1 represents the largest and smallest arr[i]+i
     // mx2, mn2 represents the largest and smallest arr[i]-i

     for(int i=0;i<n;i++){
          mx1=max(mx1,A[i]+i);
          mn1=min(mn1,A[i]+i);

          mx2=max(mx2,A[i]-i);
          mn2=min(mn2,A[i]-i);
     }
     int res=0;

     /* First set of equations are-
         1. (A[i]+i)-(A[j]+j)
         2. -((A[i]+i)-(A[j]+j));

         The best solution can only be achieved when ---
              res=max(res,mx1-mn1);
              res=max(res,-(mn1-mx1));

         But these two res's are the same on rearranging, so take either of them
     */
     res=max(res,mx1-mn1);

    /* Second set of equations are-
            1. (A[j]-j)-(A[i]-i)
            2. -((A[j]-j)-(A[i]-i));

       The best solution can only be achieved when ---
            res=max(res,mx2-mn2);
            res=max(res,-(mn2-mx2));

       But these two res's are the same on rearranging, so take either of them
    */
    res=max(res,mx2-mn2);
    return res;
}

// Driven Code
int main()
{
      int arr[] = { 1, 2, 3, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << findValue(arr, n) << endl;

    return 0;
}

// code by RainX (Abhijit Roy)

Java

// Optimized JAVA program to find maximum value of
// |arr[i] - arr[j]| + |i - j|
import java.util.*;
class GFG
{

// Return maximum |arr[i] - arr[j]| + |i - j|
static int findValue(int arr[], int n)
{
    int temp1, temp2;
    int max1 = Integer.MIN_VALUE, max2 = Integer.MIN_VALUE;
    int min1 = Integer.MAX_VALUE, min2 = Integer.MAX_VALUE;

    // Calculating max1 , min1 and max2, min2
    for (int i = 0; i < n; i++) {
        temp1 = arr[i] + i;
        temp2 = arr[i] - i;
        max1 = Math.max(max1, temp1);
        min1 = Math.min(min1, temp1);
        max2 = Math.max(max2, temp2);
        min2 = Math.min(min2, temp2);
    }

    // required maximum ans is max of (max1-min1) and
    // (max2-min2)
    return Math.max((max1 - min1), (max2 - min2));
}

// Driven Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 1 };
    int n = arr.length;

    System.out.print(findValue(arr, n) +"\n");
}
}

// This code is contributed by gauravrajput1.

Python3

# Optimized Python program to find maximum value of
# |arr[i] - arr[j]| + |i - j|
import sys

# Return maximum |arr[i] - arr[j]| + |i - j|
def findValue(arr, n):
    temp1=0;
    temp2=0;
    max1 = -sys.maxsize;
    max2 = -sys.maxsize;
    min1 = sys.maxsize;
    min2 = sys.maxsize;

    # Calculating max1 , min1 and max2, min2
    for i in range(n):
        temp1 = arr[i] + i;
        temp2 = arr[i] - i;
        max1 = max(max1, temp1);
        min1 = min(min1, temp1);
        max2 = max(max2, temp2);
        min2 = min(min2, temp2);

    # required maximum ans is max of (max1-min1) and
    # (max2-min2)
    return max((max1 - min1), (max2 - min2));

# Driven Code
if __name__ == '__main__':
    arr = [ 1, 2, 3, 1 ];
    n = len(arr);

    print(findValue(arr, n));

# This code is contributed by umadevi9616

C#

// Optimized JAVA program to find maximum value of
// |arr[i] - arr[j]| + |i - j|
using System;

public class GFG {

    // Return maximum |arr[i] - arr[j]| + |i - j|
    static int findValue(int[] arr, int n)
    {
        int temp1, temp2;
        int max1 = int.MinValue, max2 = int.MinValue;
        int min1 = int.MaxValue, min2 = int.MaxValue;

        // Calculating max1 , min1 and max2, min2
        for (int i = 0; i < n; i++) {
            temp1 = arr[i] + i;
            temp2 = arr[i] - i;
            max1 = Math.Max(max1, temp1);
            min1 = Math.Min(min1, temp1);
            max2 = Math.Max(max2, temp2);
            min2 = Math.Min(min2, temp2);
        }

        // required maximum ans is max of (max1-min1) and
        // (max2-min2)
        return Math.Max((max1 - min1), (max2 - min2));
    }

    // Driven Code
    public static void Main(String[] args)
    {
        int[] arr = { 1, 2, 3, 1 };
        int n = arr.Length;

        Console.Write(findValue(arr, n) + "\n");
    }
}

// This code is contributed by gauravrajput1

Javascript

<script>
// Optimized javascript program to find maximum value of
// |arr[i] - arr[j]| + |i - j|

    // Return maximum |arr[i] - arr[j]| + |i - j|
    function findValue(arr , n) {
        var temp1, temp2;
        var max1 = Number.MIN_VALUE, max2 = Number.MIN_VALUE;
        var min1 = Number.MAX_VALUE, min2 = Number.MAX_VALUE;

        // Calculating max1 , min1 and max2, min2
        for (i = 0; i < n; i++) {
            temp1 = arr[i] + i;
            temp2 = arr[i] - i;
            max1 = Math.max(max1, temp1);
            min1 = Math.min(min1, temp1);
            max2 = Math.max(max2, temp2);
            min2 = Math.min(min2, temp2);
        }

        // required maximum ans is max of (max1-min1) and
        // (max2-min2)
        return Math.max((max1 - min1), (max2 - min2));
    }

    // Driven Code
        var arr = [ 1, 2, 3, 1 ];
        var n = arr.length;

        document.write(findValue(arr, n) + "\n");

// This code is contributed by gauravrajput1
</script>

Output

4

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