# Sum of Manhattan distances between all pairs of points

Given **n** integer coordinates. The task is to find sum of manhattan distance between all pairs of coordinates.

Manhattan Distance between two points (x_{1}, y_{1}) and (x_{2}, y_{2}) is: **|x _{1} – x_{2}| + |y_{1} – y_{2}|**

**Examples :**

Input : n = 4 point1 = { -1, 5 } point2 = { 1, 6 } point3 = { 3, 5 } point4 = { 2, 3 } Output : 22 Distance of { 1, 6 }, { 3, 5 }, { 2, 3 } from { -1, 5 } are 3, 4, 5 respectively. Therefore, sum = 3 + 4 + 5 = 12 Distance of { 3, 5 }, { 2, 3 } from { 1, 6 } are 3, 4 respectively. Therefore, sum = 12 + 3 + 4 = 19 Distance of { 2, 3 } from { 3, 5 } is 3. Therefore, sum = 19 + 3 = 22.

**Method 1: (Brute Force)**

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**Time Complexity: O(n2)**

The idea is to run two nested loop i.e for each each point, find manhattan distance for all other points.

for (i = 1; i < n; i++) for (j = i + 1; j < n; j++) sum += ((x_{i}- x_{j}) + (y_{i}- y_{j}))

Below is the implementation of this approach:

## C++

`// CPP Program to find sum of Manhattan distance` `// between all the pairs of given points` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Return the sum of distance between all` `// the pair of points.` `int` `distancesum(` `int` `x[], ` `int` `y[], ` `int` `n)` `{` ` ` `int` `sum = 0;` ` ` `// for each point, finding distance to` ` ` `// rest of the point` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `for` `(` `int` `j = i + 1; j < n; j++)` ` ` `sum += (` `abs` `(x[i] - x[j]) +` ` ` `abs` `(y[i] - y[j]));` ` ` `return` `sum;` `}` `// Driven Program` `int` `main()` `{` ` ` `int` `x[] = { -1, 1, 3, 2 };` ` ` `int` `y[] = { 5, 6, 5, 3 };` ` ` `int` `n = ` `sizeof` `(x) / ` `sizeof` `(x[0]);` ` ` `cout << distancesum(x, y, n) << endl;` ` ` `return` `0;` `}` |

## Java

`// Java Program to find sum of Manhattan distance` `// between all the pairs of given points` `import` `java.io.*;` `class` `GFG {` ` ` ` ` `// Return the sum of distance between all` ` ` `// the pair of points.` ` ` `static` `int` `distancesum(` `int` `x[], ` `int` `y[], ` `int` `n)` ` ` `{` ` ` `int` `sum = ` `0` `;` ` ` `// for each point, finding distance to` ` ` `// rest of the point` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++)` ` ` `for` `(` `int` `j = i + ` `1` `; j < n; j++)` ` ` `sum += (Math.abs(x[i] - x[j]) +` ` ` `Math.abs(y[i] - y[j]));` ` ` `return` `sum;` ` ` `}` ` ` `// Driven Program` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `x[] = { -` `1` `, ` `1` `, ` `3` `, ` `2` `};` ` ` `int` `y[] = { ` `5` `, ` `6` `, ` `5` `, ` `3` `};` ` ` `int` `n = x.length;` ` ` ` ` `System.out.println(distancesum(x, y, n));` ` ` `}` `}` `// This code is contributed by vt_m.` |

## Python3

`# Python3 code to find sum of` `# Manhattan distance between all` `# the pairs of given points` `# Return the sum of distance` `# between all the pair of points.` `def` `distancesum (x, y, n):` ` ` `sum` `=` `0` ` ` ` ` `# for each point, finding distance` ` ` `# to rest of the point` ` ` `for` `i ` `in` `range` `(n):` ` ` `for` `j ` `in` `range` `(i` `+` `1` `,n):` ` ` `sum` `+` `=` `(` `abs` `(x[i] ` `-` `x[j]) ` `+` ` ` `abs` `(y[i] ` `-` `y[j]))` ` ` ` ` `return` `sum` `# Driven Code` `x ` `=` `[ ` `-` `1` `, ` `1` `, ` `3` `, ` `2` `]` `y ` `=` `[ ` `5` `, ` `6` `, ` `5` `, ` `3` `]` `n ` `=` `len` `(x)` `print` `(distancesum(x, y, n) )` `# This code is contributed by "Sharad_Bhardwaj".` |

## C#

`// C# Program to find sum of Manhattan distance` `// between all the pairs of given points` `using` `System;` `class` `GFG {` ` ` ` ` `// Return the sum of distance between all` ` ` `// the pair of points.` ` ` `static` `int` `distancesum(` `int` `[]x, ` `int` `[]y, ` `int` `n)` ` ` `{` ` ` `int` `sum = 0;` ` ` `// for each point, finding distance to` ` ` `// rest of the point` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `for` `(` `int` `j = i + 1; j < n; j++)` ` ` `sum += (Math.Abs(x[i] - x[j]) +` ` ` `Math.Abs(y[i] - y[j]));` ` ` `return` `sum;` ` ` `}` ` ` `// Driven Program` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `[]x = { -1, 1, 3, 2 };` ` ` `int` `[]y = { 5, 6, 5, 3 };` ` ` `int` `n = x.Length;` ` ` ` ` `Console.WriteLine(distancesum(x, y, n));` ` ` `}` `}` `// This code is contributed by vt_m.` |

## PHP

`<?php` `// PHP Program to find sum` `// of Manhattan distance` `// between all the pairs` `// of given points` `// Return the sum of distance` `// between all the pair of points.` `function` `distancesum( ` `$x` `, ` `$y` `, ` `$n` `)` `{` ` ` `$sum` `= 0;` ` ` `// for each point, finding` ` ` `// distance to rest of` ` ` `// the point` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++)` ` ` `for` `(` `$j` `= ` `$i` `+ 1; ` `$j` `< ` `$n` `; ` `$j` `++)` ` ` `$sum` `+= (` `abs` `(` `$x` `[` `$i` `] - ` `$x` `[` `$j` `]) +` ` ` `abs` `(` `$y` `[` `$i` `] - ` `$y` `[` `$j` `]));` ` ` `return` `$sum` `;` `}` ` ` `// Driver Code` ` ` `$x` `= ` `array` `(-1, 1, 3, 2);` ` ` `$y` `= ` `array` `(5, 6, 5, 3);` ` ` `$n` `= ` `count` `(` `$x` `);` ` ` `echo` `distancesum(` `$x` `, ` `$y` `, ` `$n` `);` ` ` `// This code is contributed by anuj_67.` `?>` |

## Javascript

`<script>` `// JavaScript Program to find sum of Manhattan distance` `// between all the pairs of given points` ` ` `// Return the sum of distance between all` ` ` `// the pair of points.` ` ` `function` `distancesum(x, y, n)` ` ` `{` ` ` `let sum = 0;` ` ` ` ` `// for each point, finding distance to` ` ` `// rest of the point` ` ` `for` `(let i = 0; i < n; i++)` ` ` `for` `(let j = i + 1; j < n; j++)` ` ` `sum += (Math.abs(x[i] - x[j]) +` ` ` `Math.abs(y[i] - y[j]));` ` ` `return` `sum;` ` ` `}` `// Driver code` ` ` `let x = [ -1, 1, 3, 2 ];` ` ` `let y = [ 5, 6, 5, 3 ];` ` ` `let n = x.length;` ` ` ` ` `document.write(distancesum(x, y, n));` ` ` ` ` `// This code is contributed by sanjoy_62.` `</script>` |

**Output: **

22

**Method 2: (Efficient Approach)**

- The idea is to use Greedy Approach. First observe, the manhattan formula can be decomposed into two independent sums, one for the difference between
**x**coordinates and the second between**y**coordinates. If we know how to compute one of them we can use the same method to compute the other. So now we will stick to compute the sum of**x**coordinates distance. - Let’s assume that we have calculated the sum of distances between any two points till a point
**x**for all values of_{i-1}**x**‘s smaller than**x**, let this sum be_{i-1}**res**and now we have to calculate the distance between any two points with**x**included, where_{i}**x**is the next greater point, To calculate the distance of each point from the next greater point_{i}**x**, we can add the existing sum of differences_{i}**res**with the distance of**x**from all the points_{i}**x**which are less than_{k}**x**. Hence the sum between any two points will now be equal to_{i}**res**+**∑**(**x**_{i}**–****x**) , where_{k}**x**is the current point from which differences are being measured, and_{i}**x**are all the points less than_{k}**x**._{i} - Because for every calculation
**x**remains same, we can simplify the summation as :_{i}

res = res +(x_{i}– x_{1}) + (x_{i}– x_{2}) + (x_{i }– x_{3})………(x_{i }– x_{i-1})

res = res + (x, because in a sorted array, there are i elements smaller than the current index i ._{i})*i – (x_{1}+ x_{2}+ …… x_{i-1})

res = res + (x, where_{i})*i – S_{i-1}Sis the sum of all the previous points till index_{i-1}i – 1

4. For the new index i , we need to add the difference of the current index **x _{i}** from all the previous indices

**x**

_{k }**< x**

_{i} 5. If we sort all points in non-decreasing order, we can easily compute the desired sum of distances along one axis between each pair of coordinates in O(N) time, processing points from left to right and using the above method.

Also, we don’t have to concern if two points are equal coordinates, after sorting points in non-decreasing order, we say that a point **x _{i-1}** is smaller

**x**if and only if it appears earlier in the sorted array.

_{i}Below is the implementation of this approach:

## C++

`// CPP Program to find sum of Manhattan` `// distances between all the pairs of` `// given points` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Return the sum of distance of one axis.` `int` `distancesum(` `int` `arr[], ` `int` `n)` `{` ` ` `// sorting the array.` ` ` `sort(arr, arr + n);` ` ` `// for each point, finding the distance.` ` ` `int` `res = 0, sum = 0;` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `res += (arr[i] * i - sum);` ` ` `sum += arr[i];` ` ` `}` ` ` `return` `res;` `}` `int` `totaldistancesum(` `int` `x[], ` `int` `y[], ` `int` `n)` `{` ` ` `return` `distancesum(x, n) + distancesum(y, n);` `}` `// Driven Program` `int` `main()` `{` ` ` `int` `x[] = { -1, 1, 3, 2 };` ` ` `int` `y[] = { 5, 6, 5, 3 };` ` ` `int` `n = ` `sizeof` `(x) / ` `sizeof` `(x[0]);` ` ` `cout << totaldistancesum(x, y, n) << endl;` ` ` `return` `0;` `}` |

## Java

`// Java Program to find sum of Manhattan` `// distances between all the pairs of` `// given points` `import` `java.io.*;` `import` `java.util.*;` `class` `GFG {` ` ` ` ` `// Return the sum of distance of one axis.` ` ` `static` `int` `distancesum(` `int` `arr[], ` `int` `n)` ` ` `{` ` ` ` ` `// sorting the array.` ` ` `Arrays.sort(arr);` ` ` `// for each point, finding the distance.` ` ` `int` `res = ` `0` `, sum = ` `0` `;` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) {` ` ` `res += (arr[i] * i - sum);` ` ` `sum += arr[i];` ` ` `}` ` ` `return` `res;` ` ` `}` ` ` `static` `int` `totaldistancesum(` `int` `x[],` ` ` `int` `y[], ` `int` `n)` ` ` `{` ` ` `return` `distancesum(x, n) +` ` ` `distancesum(y, n);` ` ` `}` ` ` `// Driven Program` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `x[] = { -` `1` `, ` `1` `, ` `3` `, ` `2` `};` ` ` `int` `y[] = { ` `5` `, ` `6` `, ` `5` `, ` `3` `};` ` ` `int` `n = x.length;` ` ` `System.out.println(totaldistancesum(x,` ` ` `y, n));` ` ` `}` `}` `// This code is contributed by vt_m.` |

## Python3

`# Python3 code to find sum of Manhattan` `# distances between all the pairs of` `# given points` `# Return the sum of distance of one axis.` `def` `distancesum (arr, n):` ` ` ` ` `# sorting the array.` ` ` `arr.sort()` ` ` ` ` `# for each point, finding` ` ` `# the distance.` ` ` `res ` `=` `0` ` ` `sum` `=` `0` ` ` `for` `i ` `in` `range` `(n):` ` ` `res ` `+` `=` `(arr[i] ` `*` `i ` `-` `sum` `)` ` ` `sum` `+` `=` `arr[i]` ` ` ` ` `return` `res` ` ` `def` `totaldistancesum( x , y , n ):` ` ` `return` `distancesum(x, n) ` `+` `distancesum(y, n)` ` ` `# Driven Code` `x ` `=` `[ ` `-` `1` `, ` `1` `, ` `3` `, ` `2` `]` `y ` `=` `[ ` `5` `, ` `6` `, ` `5` `, ` `3` `]` `n ` `=` `len` `(x)` `print` `(totaldistancesum(x, y, n) )` `# This code is contributed by "Sharad_Bhardwaj".` |

## C#

`// C# Program to find sum of Manhattan` `// distances between all the pairs of` `// given points` `using` `System;` `class` `GFG {` ` ` ` ` `// Return the sum of distance of one axis.` ` ` `static` `int` `distancesum(` `int` `[]arr, ` `int` `n)` ` ` `{` ` ` ` ` `// sorting the array.` ` ` `Array.Sort(arr);` ` ` `// for each point, finding the distance.` ` ` `int` `res = 0, sum = 0;` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `res += (arr[i] * i - sum);` ` ` `sum += arr[i];` ` ` `}` ` ` `return` `res;` ` ` `}` ` ` `static` `int` `totaldistancesum(` `int` `[]x,` ` ` `int` `[]y, ` `int` `n)` ` ` `{` ` ` `return` `distancesum(x, n) +` ` ` `distancesum(y, n);` ` ` `}` ` ` `// Driven Program` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `[]x = { -1, 1, 3, 2 };` ` ` `int` `[]y = { 5, 6, 5, 3 };` ` ` `int` `n = x.Length;` ` ` `Console.WriteLine(totaldistancesum(x,` ` ` `y, n));` ` ` `}` `}` `// This code is contributed by vt_m.` |

## PHP

`<?php` `// PHP Program to find sum of` `// Manhattan distances between` `// all the pairs of given points` `// Return the sum of` `// distance of one axis.` `function` `distancesum(` `$arr` `, ` `$n` `)` `{` ` ` `// sorting the array.` ` ` `sort(` `$arr` `);` ` ` `// for each point,` ` ` `// finding the distance.` ` ` `$res` `= 0; ` `$sum` `= 0;` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++)` ` ` `{` ` ` `$res` `+= (` `$arr` `[` `$i` `] * ` `$i` `- ` `$sum` `);` ` ` `$sum` `+= ` `$arr` `[` `$i` `];` ` ` `}` ` ` `return` `$res` `;` `}` `function` `totaldistancesum(` `$x` `, ` `$y` `, ` `$n` `)` `{` ` ` `return` `distancesum(` `$x` `, ` `$n` `) +` ` ` `distancesum(` `$y` `, ` `$n` `);` `}` `// Driver Code` `$x` `= ` `array` `(-1, 1, 3, 2);` `$y` `= ` `array` `(5, 6, 5, 3);` `$n` `= sizeof(` `$x` `);` `echo` `totaldistancesum(` `$x` `, ` `$y` `, ` `$n` `), ` `"\n"` `;` `// This code is contributed by m_kit` `?>` |

## Javascript

`<script>` `// JavaScript Program to find sum of Manhattan` `// distances between all the pairs of` `// given points` ` ` `// Return the sum of distance of one axis.` ` ` `function` `distancesum(arr, n)` ` ` `{` ` ` ` ` `// sorting the array.` ` ` `arr.sort();` ` ` ` ` `// for each point, finding the distance.` ` ` `let res = 0, sum = 0;` ` ` `for` `(let i = 0; i < n; i++) {` ` ` `res += (arr[i] * i - sum);` ` ` `sum += arr[i];` ` ` `}` ` ` ` ` `return` `res;` ` ` `}` ` ` ` ` `function` `totaldistancesum(x,` ` ` `y, n)` ` ` `{` ` ` `return` `distancesum(x, n) +` ` ` `distancesum(y, n);` ` ` `}` ` ` `// Driver code` ` ` ` ` `let x = [ -1, 1, 3, 2 ];` ` ` `let y = [ 5, 6, 5, 3 ];` ` ` `let n = x.length;` ` ` `document.write(totaldistancesum(x,` ` ` `y, n));` ` ` `</script>` |

**Output :**

22

**Time Complexity : **O(nlogn)