# Maximum Manhattan distance between a distinct pair from N coordinates

• Difficulty Level : Medium
• Last Updated : 15 Jun, 2022

Given an array arr[] consisting of N integer coordinates, the task is to find the maximum Manhattan Distance between any two distinct pairs of coordinates.

The Manhattan Distance between two points (X1, Y1) and (X2, Y2) is given by |X1 â€“ X2| + |Y1 â€“ Y2|.

Examples:

Input: arr[] = {(1, 2), (2, 3), (3, 4)}
Output: 4
Explanation:
The maximum Manhattan distance is found between (1, 2) and (3, 4) i.e., |3 – 1| + |4- 2 | = 4.

Input: arr[] = {(-1, 2), (-4, 6), (3, -4), (-2, -4)}
Output: 17
Explanation:
The maximum Manhattan distance is found between (-4, 6) and (3, -4) i.e.,  |-4 – 3| + |6 – (-4)| = 17.

Naive Approach: The simplest approach is to iterate over the array, and for each coordinate, calculate its Manhattan distance from all remaining points. Keep updating the maximum distance obtained after each calculation. Finally, print the maximum distance obtained.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to calculate the maximum` `// Manhattan distance` `void` `MaxDist(vector >& A, ``int` `N)` `{` `    ``// Stores the maximum distance` `    ``int` `maximum = INT_MIN;`   `    ``for` `(``int` `i = 0; i < N; i++) {`   `        ``int` `sum = 0;`   `        ``for` `(``int` `j = i + 1; j < N; j++) {`   `            ``// Find Manhattan distance` `            ``// using the formula` `            ``// |x1 - x2| + |y1 - y2|` `            ``sum = ``abs``(A[i].first - A[j].first)` `                  ``+ ``abs``(A[i].second - A[j].second);`   `            ``// Updating the maximum` `            ``maximum = max(maximum, sum);` `        ``}` `    ``}`   `    ``cout << maximum;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `N = 3;`   `    ``// Given Co-ordinates` `    ``vector > A` `        ``= { { 1, 2 }, { 2, 3 }, { 3, 4 } };`   `    ``// Function Call` `    ``MaxDist(A, N);`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG {`   `    ``// Pair class` `    ``public` `static` `class` `Pair {` `        ``int` `x;` `        ``int` `y;`   `        ``Pair(``int` `x, ``int` `y)` `        ``{` `            ``this``.x = x;` `            ``this``.y = y;` `        ``}` `    ``}`   `    ``// Function to calculate the maximum` `    ``// Manhattan distance` `    ``static` `void` `MaxDist(ArrayList A, ``int` `N)` `    ``{`   `        ``// Stores the maximum distance` `        ``int` `maximum = Integer.MIN_VALUE;`   `        ``for` `(``int` `i = ``0``; i < N; i++) {` `            ``int` `sum = ``0``;`   `            ``for` `(``int` `j = i + ``1``; j < N; j++) {`   `                ``// Find Manhattan distance` `                ``// using the formula` `                ``// |x1 - x2| + |y1 - y2|` `                ``sum = Math.abs(A.get(i).x - A.get(j).x)` `                      ``+ Math.abs(A.get(i).y - A.get(j).y);`   `                ``// Updating the maximum` `                ``maximum = Math.max(maximum, sum);` `            ``}` `        ``}` `        ``System.out.println(maximum);` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `n = ``3``;`   `        ``ArrayList al = ``new` `ArrayList<>();`   `        ``// Given Co-ordinates` `        ``Pair p1 = ``new` `Pair(``1``, ``2``);` `        ``al.add(p1);`   `        ``Pair p2 = ``new` `Pair(``2``, ``3``);` `        ``al.add(p2);`   `        ``Pair p3 = ``new` `Pair(``3``, ``4``);` `        ``al.add(p3);`   `        ``// Function call` `        ``MaxDist(al, n);` `    ``}` `}`   `// This code is contributed by bikram2001jha`

## Python3

 `# Python3 program for the above approach` `import` `sys`   `# Function to calculate the maximum` `# Manhattan distance`     `def` `MaxDist(A, N):`   `    ``# Stores the maximum distance` `    ``maximum ``=` `-` `sys.maxsize`   `    ``for` `i ``in` `range``(N):` `        ``sum` `=` `0`   `        ``for` `j ``in` `range``(i ``+` `1``, N):`   `            ``# Find Manhattan distance` `            ``# using the formula` `            ``# |x1 - x2| + |y1 - y2|` `            ``Sum` `=` `(``abs``(A[i][``0``] ``-` `A[j][``0``]) ``+` `                   ``abs``(A[i][``1``] ``-` `A[j][``1``]))`   `            ``# Updating the maximum` `            ``maximum ``=` `max``(maximum, ``Sum``)`   `    ``print``(maximum)`     `# Driver code` `N ``=` `3`   `# Given co-ordinates` `A ``=` `[[``1``, ``2``], [``2``, ``3``], [``3``, ``4``]]`   `# Function call` `MaxDist(A, N)`   `# This code is contributed by divyeshrabadiya07`

## C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG {`   `    ``// Pair class` `    ``public` `class` `Pair {` `        ``public` `int` `x;` `        ``public` `int` `y;`   `        ``public` `Pair(``int` `x, ``int` `y)` `        ``{` `            ``this``.x = x;` `            ``this``.y = y;` `        ``}` `    ``}`   `    ``// Function to calculate the maximum` `    ``// Manhattan distance` `    ``static` `void` `MaxDist(List A, ``int` `N)` `    ``{`   `        ``// Stores the maximum distance` `        ``int` `maximum = ``int``.MinValue;`   `        ``for` `(``int` `i = 0; i < N; i++) {` `            ``int` `sum = 0;`   `            ``for` `(``int` `j = i + 1; j < N; j++) {`   `                ``// Find Manhattan distance` `                ``// using the formula` `                ``// |x1 - x2| + |y1 - y2|` `                ``sum = Math.Abs(A[i].x - A[j].x)` `                      ``+ Math.Abs(A[i].y - A[j].y);`   `                ``// Updating the maximum` `                ``maximum = Math.Max(maximum, sum);` `            ``}` `        ``}` `        ``Console.WriteLine(maximum);` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `Main(String[] args)` `    ``{` `        ``int` `n = 3;`   `        ``List al = ``new` `List();`   `        ``// Given Co-ordinates` `        ``Pair p1 = ``new` `Pair(1, 2);` `        ``al.Add(p1);`   `        ``Pair p2 = ``new` `Pair(2, 3);` `        ``al.Add(p2);`   `        ``Pair p3 = ``new` `Pair(3, 4);` `        ``al.Add(p3);`   `        ``// Function call` `        ``MaxDist(al, n);` `    ``}` `}`   `// This code is contributed by Amit Katiyar`

## Javascript

 ``

Output

`4`

Time Complexity: O(N2), where N is the size of the given array.
Auxiliary Space: O(1)

Efficient Approach: The idea is to use store sums and differences between X and Y coordinates and find the answer by sorting those differences. Below are the observations to the above problem statement:

• Manhattan Distance between any two points (Xi, Yi) and (Xj, Yj) can be written as follows:

|Xi – Xj| + |Yi – Yj| = max(Xi – Xj -Yi + Yj
-Xi + Xj + Yi – Yj
-Xi + Xj – Yi + Yj
Xi – Xj + Yi – Yj).

• The above expression can be rearranged as:

|Xi – Xj| + |Yi – Yj| = max((Xi – Yi) – (Xj – Yj),
(-Xi + Yi) – (-Xj + Yj),
(-Xi – Yi) – (-Xj – Yj),
(Xi + Yi) – (Xj + Yj))

• It can be observed from the above expression, that the answer can be found by storing the sum and differences of the coordinates.

Follow the below steps to solve the problem:

1. Initialize two arrays sum[] and diff[].
2. Store the sum of X and Y coordinates i.e., Xi + Yi in sum[] and their difference i.e., Xi – Yi in diff[].
3. Sort the sum[] and diff[] in ascending order.
4. The maximum of the values (sum[N-1] – sum[0]) and (diff[N-1] – diff[0]) is the required result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to calculate the maximum` `// Manhattan distance` `void` `MaxDist(vector >& A, ``int` `N)` `{` `    ``// Vectors to store maximum and` `    ``// minimum of all the four forms` `    ``vector<``int``> V(N), V1(N);`   `    ``for` `(``int` `i = 0; i < N; i++) {` `        ``V[i] = A[i].first + A[i].second;` `        ``V1[i] = A[i].first - A[i].second;` `    ``}`   `    ``// Sorting both the vectors` `    ``sort(V.begin(), V.end());` `    ``sort(V1.begin(), V1.end());`   `    ``int` `maximum` `        ``= max(V.back() - V.front(), V1.back() - V1.front());`   `    ``cout << maximum << endl;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `N = 3;`   `    ``// Given Co-ordinates` `    ``vector > A` `        ``= { { 1, 2 }, { 2, 3 }, { 3, 4 } };`   `    ``// Function Call` `    ``MaxDist(A, N);`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG {`   `    ``// Pair class` `    ``public` `static` `class` `Pair {` `        ``int` `x;` `        ``int` `y;`   `        ``Pair(``int` `x, ``int` `y)` `        ``{` `            ``this``.x = x;` `            ``this``.y = y;` `        ``}` `    ``}`   `    ``// Function to calculate the maximum` `    ``// Manhattan distance` `    ``static` `void` `MaxDist(ArrayList A, ``int` `N)` `    ``{`   `        ``// ArrayLists to store maximum and` `        ``// minimum of all the four forms` `        ``ArrayList V = ``new` `ArrayList<>();` `        ``ArrayList V1 = ``new` `ArrayList<>();`   `        ``for` `(``int` `i = ``0``; i < N; i++) {` `            ``V.add(A.get(i).x + A.get(i).y);` `            ``V1.add(A.get(i).x - A.get(i).y);` `        ``}`   `        ``// Sorting both the ArrayLists` `        ``Collections.sort(V);` `        ``Collections.sort(V1);`   `        ``int` `maximum` `            ``= Math.max((V.get(V.size() - ``1``) - V.get(``0``)),` `                       ``(V1.get(V1.size() - ``1``) - V1.get(``0``)));`   `        ``System.out.println(maximum);` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `n = ``3``;`   `        ``ArrayList al = ``new` `ArrayList<>();`   `        ``// Given Co-ordinates` `        ``Pair p1 = ``new` `Pair(``1``, ``2``);` `        ``al.add(p1);` `        ``Pair p2 = ``new` `Pair(``2``, ``3``);` `        ``al.add(p2);` `        ``Pair p3 = ``new` `Pair(``3``, ``4``);` `        ``al.add(p3);`   `        ``// Function call` `        ``MaxDist(al, n);` `    ``}` `}`   `// This code is contributed by bikram2001jha`

## Python3

 `# Python3 program for the above approach`   `# Function to calculate the maximum` `# Manhattan distance`     `def` `MaxDist(A, N):`   `    ``# List to store maximum and` `    ``# minimum of all the four forms` `    ``V ``=` `[``0` `for` `i ``in` `range``(N)]` `    ``V1 ``=` `[``0` `for` `i ``in` `range``(N)]`   `    ``for` `i ``in` `range``(N):` `        ``V[i] ``=` `A[i][``0``] ``+` `A[i][``1``]` `        ``V1[i] ``=` `A[i][``0``] ``-` `A[i][``1``]`   `    ``# Sorting both the vectors` `    ``V.sort()` `    ``V1.sort()`   `    ``maximum ``=` `max``(V[``-``1``] ``-` `V[``0``],` `                  ``V1[``-``1``] ``-` `V1[``0``])`   `    ``print``(maximum)`     `# Driver code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``N ``=` `3`   `    ``# Given Co-ordinates` `    ``A ``=` `[[``1``, ``2``],` `         ``[``2``, ``3``],` `         ``[``3``, ``4``]]`   `    ``# Function call` `    ``MaxDist(A, N)`   `# This code is contributed by rutvik_56`

## C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG {`   `    ``// Pair class` `    ``class` `Pair {` `        ``public` `int` `x;` `        ``public` `int` `y;`   `        ``public` `Pair(``int` `x, ``int` `y)` `        ``{` `            ``this``.x = x;` `            ``this``.y = y;` `        ``}` `    ``}`   `    ``// Function to calculate the maximum` `    ``// Manhattan distance` `    ``static` `void` `MaxDist(List A, ``int` `N)` `    ``{`   `        ``// Lists to store maximum and` `        ``// minimum of all the four forms` `        ``List<``int``> V = ``new` `List<``int``>();` `        ``List<``int``> V1 = ``new` `List<``int``>();`   `        ``for` `(``int` `i = 0; i < N; i++) {` `            ``V.Add(A[i].x + A[i].y);` `            ``V1.Add(A[i].x - A[i].y);` `        ``}`   `        ``// Sorting both the Lists` `        ``V.Sort();` `        ``V1.Sort();`   `        ``int` `maximum = Math.Max((V[V.Count - 1] - V[0]),` `                               ``(V1[V1.Count - 1] - V1[0]));`   `        ``Console.WriteLine(maximum);` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `Main(String[] args)` `    ``{` `        ``int` `n = 3;`   `        ``List al = ``new` `List();`   `        ``// Given Co-ordinates` `        ``Pair p1 = ``new` `Pair(1, 2);` `        ``al.Add(p1);` `        ``Pair p2 = ``new` `Pair(2, 3);` `        ``al.Add(p2);` `        ``Pair p3 = ``new` `Pair(3, 4);` `        ``al.Add(p3);`   `        ``// Function call` `        ``MaxDist(al, n);` `    ``}` `}`   `// This code is contributed by Amit Katiyar`

## Javascript

 ``

Output

`4`

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

Improving the Efficient Approach: Instead of storing the sums and differences in an auxiliary array, then sorting the arrays to determine the minimum and maximums, it is possible to keep a running total of the extreme sums and differences.

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to calculate the maximum` `// Manhattan distance` `void` `MaxDist(vector >& A, ``int` `N)` `{` `    ``// Variables to track running extrema` `    ``int` `minsum, maxsum, mindiff, maxdiff;`   `    ``minsum = maxsum = A[0].first + A[0].second;` `    ``mindiff = maxdiff = A[0].first - A[0].second;` `    ``for` `(``int` `i = 1; i < N; i++) {` `        ``int` `sum = A[i].first + A[i].second;` `        ``int` `diff = A[i].first - A[i].second;` `        ``if` `(sum < minsum)` `            ``minsum = sum;` `        ``else` `if` `(sum > maxsum)` `            ``maxsum = sum;` `        ``if` `(diff < mindiff)` `            ``mindiff = diff;` `        ``else` `if` `(diff > maxdiff)` `            ``maxdiff = diff;` `    ``}`   `    ``int` `maximum = max(maxsum - minsum, maxdiff - mindiff);`   `    ``cout << maximum << endl;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `N = 3;`   `    ``// Given Co-ordinates` `    ``vector > A` `        ``= { { 1, 2 }, { 2, 3 }, { 3, 4 } };`   `    ``// Function Call` `    ``MaxDist(A, N);`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `public` `class` `GFG {`   `    ``// Function to calculate the maximum` `    ``// Manhattan distance` `    ``static` `void` `MaxDist(``int``[][] A, ``int` `N)` `    ``{` `        ``// Variables to track running extrema` `        ``int` `minsum, maxsum, mindiff, maxdiff;`   `        ``minsum = maxsum = A[``0``][``0``] + A[``0``][``1``];` `        ``mindiff = maxdiff = A[``0``][``0``] - A[``0``][``1``];` `        ``for` `(``int` `i = ``1``; i < N; i++) {` `            ``int` `sum = A[i][``0``] + A[i][``1``];` `            ``int` `diff = A[i][``0``] - A[i][``1``];` `            ``if` `(sum < minsum)` `                ``minsum = sum;` `            ``else` `if` `(sum > maxsum)` `                ``maxsum = sum;` `            ``if` `(diff < mindiff)` `                ``mindiff = diff;` `            ``else` `if` `(diff > maxdiff)` `                ``maxdiff = diff;` `        ``}`   `        ``int` `maximum` `            ``= Math.max(maxsum - minsum, maxdiff - mindiff);`   `        ``System.out.println(maximum);` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `N = ``3``;`   `        ``// Given Co-ordinates` `        ``int``[][] A = { { ``1``, ``2` `}, { ``2``, ``3` `}, { ``3``, ``4` `} };`   `        ``// Function Call` `        ``MaxDist(A, N);` `    ``}` `}`   `// The code is contributed by Gautam goel (gautamgoel962)`

## Python3

 `# Python program for the above approach`   `# Function to calculate the maximum` `# Manhattan distance` `def` `MaxDist(A, N):`   `    ``# Variables to track running extrema` `    ``minsum ``=` `maxsum ``=` `A[``0``][``0``] ``+` `A[``0``][``1``]` `    ``mindiff ``=` `maxdiff ``=` `A[``0``][``0``] ``-` `A[``0``][``1``]`   `    ``for` `i ``in` `range``(``1``,N):` `        ``sum` `=` `A[i][``0``] ``+` `A[i][``1``]` `        ``diff ``=` `A[i][``0``] ``-` `A[i][``1``]` `        ``if` `(``sum` `< minsum):` `            ``minsum ``=` `sum` `        ``elif` `(``sum` `> maxsum):` `            ``maxsum ``=` `sum` `        ``if` `(diff < mindiff):` `            ``mindiff ``=` `diff` `        ``elif` `(diff > maxdiff):` `            ``maxdiff ``=` `diff`   `    ``maximum ``=` `max``(maxsum ``-` `minsum, maxdiff ``-` `mindiff)`   `    ``print``(maximum)`   `# Driver Code` `N ``=` `3`   `# Given Co-ordinates` `A ``=` `[ [ ``1``, ``2` `], [ ``2``, ``3` `], [ ``3``, ``4` `] ]`   `# Function Call` `MaxDist(A, N)`   `# This code is contributed by shinjanpatra`

## Javascript

 ``

Output

`4`

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

The ideas presented here are in two dimensions, but can be extended to further dimensions.  Each additional dimension requires double the amount of computations on each point.  For example, in 3-D space, the result is the maximum difference between the four extrema pairs computed from x+y+z, x+y-z, x-y+z, and x-y-z.

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