# Find the original coordinates whose Manhattan distances are given

Given the Manhattan distances of three coordinates on a 2-D plane, the task is to find the original coordinates. Print any solution if multiple solutions are possible else print -1.

Input: d1 = 3, d2 = 4, d3 = 5
Output: (0, 0), (3, 0) and (1, 3)
Manhattan distance between (0, 0) to (3, 0) is 3,
(3, 0) to (1, 3) is 5 and (0, 0) to (1, 3) is 4

Input: d1 = 5, d2 = 10, d3 = 12
Output: -1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Let’s analyze when no solution exists. First the triangle inequality must hold true i.e. the largest distance should not exceed the sum of other two. Second, sum of all Manhattan distances should be even.
Here’s why, if we have three points and their x-coordinates are x1, x2 and x3 such that x1 < x2 < x3. They will contribute to the sum (x2 – x1) + (x3 – x1) + (x3 – x2) = 2 * (x3 – x1). Same logic applied for y-coordinates.
In all the other cases, we have a solution. Let d1, d2 and d3 be the given Manhattan distances. Fix two points as (0, 0) and (d1, 0). Now since two points are fixed, we can easily find the third point as x3 = (d1 + d2 – d3) / 2 and y3 = (d2 – x3).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the original coordinated ` `void` `solve(``int` `d1, ``int` `d2, ``int` `d3) ` `{ ` ` `  `    ``// Maximum of the given distances ` `    ``int` `maxx = max(d1, max(d2, d3)); ` ` `  `    ``// Sum of the given distances ` `    ``int` `sum = (d1 + d2 + d3); ` ` `  `    ``// Conditions when the ` `    ``// solution doesn't exist ` `    ``if` `(2 * maxx > sum or sum % 2 == 1) { ` `        ``cout << ``"-1"``; ` `        ``return``; ` `    ``} ` ` `  `    ``// First coordinate ` `    ``int` `x1 = 0, y1 = 0; ` ` `  `    ``// Second coordinate ` `    ``int` `x2 = d1, y2 = 0; ` ` `  `    ``// Third coordinate ` `    ``int` `x3 = (d1 + d2 - d3) / 2; ` `    ``int` `y3 = (d2 + d3 - d1) / 2; ` `    ``cout << ``"("` `<< x1 << ``", "` `<< y1 << ``"), ("` `         ``<< x2 << ``", "` `<< y2 << ``") and ("` `         ``<< x3 << ``", "` `<< y3 << ``")"``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `d1 = 3, d2 = 4, d3 = 5; ` `    ``solve(d1, d2, d3); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java .io.*; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to find the original coordinated ` `static` `void` `solve(``int` `d1, ``int` `d2, ``int` `d3) ` `{ ` ` `  `    ``// Maximum of the given distances ` `    ``int` `maxx = Math.max(d1, Math.max(d2, d3)); ` ` `  `    ``// Sum of the given distances ` `    ``int` `sum = (d1 + d2 + d3); ` ` `  `    ``// Conditions when the ` `    ``// solution doesn't exist ` `    ``if` `(``2` `* maxx > sum || sum % ``2` `== ``1``)  ` `    ``{ ` `        ``System.out.print(``"-1"``); ` `        ``return``; ` `    ``} ` ` `  `    ``// First coordinate ` `    ``int` `x1 = ``0``, y1 = ``0``; ` ` `  `    ``// Second coordinate ` `    ``int` `x2 = d1, y2 = ``0``; ` ` `  `    ``// Third coordinate ` `    ``int` `x3 = (d1 + d2 - d3) / ``2``; ` `    ``int` `y3 = (d2 + d3 - d1) / ``2``; ` `    ``System.out.print(``"("``+x1+``", "``+y1+``"), ("``+x2+``", "``+y2+``") and ("``+x3+``", "``+y3+``")"``); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `d1 = ``3``, d2 = ``4``, d3 = ``5``; ` `    ``solve(d1, d2, d3); ` ` `  `} ` `} ` ` `  `// This code is contributed by anuj_67.. `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to find the original coordinated ` `def` `solve(d1, d2, d3) : ` ` `  `     `  `    ``# Maximum of the given distances ` `    ``maxx ``=` `max``(d1, ``max``(d2, d3)) ` `     `  `    ``# Sum of the given distances ` `    ``sum` `=` `(d1 ``+` `d2 ``+` `d3) ` `     `  `    ``# Conditions when the ` `    ``# solution doesn't exist ` `    ``if` `(``2` `*` `maxx > ``sum` `or` `sum` `%` `2` `=``=` `1``) : ` `        ``print``(``"-1"``) ` `        ``return` `     `  `     `  `    ``# First coordinate ` `    ``x1 ``=` `0` `    ``y1 ``=` `0` `     `  `    ``# Second coordinate ` `    ``x2 ``=` `d1  ` `    ``y2 ``=` `0` `     `  `    ``# Third coordinate ` `    ``x3 ``=` `(d1 ``+` `d2 ``-` `d3) ``/``/` `2` `    ``y3 ``=` `(d2 ``+` `d3 ``-` `d1) ``/``/` `2` `    ``print``(``"("` `, x1 , ``","` `, y1 , ``"), ("` `        ``, x2 , ``","` `, y2 , ``") and ("` `        ``, x3 , ``","` `, y3 , ``")"``) ` ` `  ` `  `# Driver code ` `d1 ``=` `3` `d2 ``=` `4` `d3 ``=` `5` `solve(d1, d2, d3) ` ` `  `# This code is contributed by ihritik `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to find the original coordinated ` `static` `void` `solve(``int` `d1, ``int` `d2, ``int` `d3) ` `{ ` ` `  `    ``// Maximum of the given distances ` `    ``int` `maxx = Math.Max(d1, Math.Max(d2, d3)); ` ` `  `    ``// Sum of the given distances ` `    ``int` `sum = (d1 + d2 + d3); ` ` `  `    ``// Conditions when the ` `    ``// solution doesn't exist ` `    ``if` `(2 * maxx > sum || sum % 2 == 1)  ` `    ``{ ` `        ``Console.WriteLine(``"-1"``); ` `        ``return``; ` `    ``} ` ` `  `    ``// First coordinate ` `    ``int` `x1 = 0, y1 = 0; ` ` `  `    ``// Second coordinate ` `    ``int` `x2 = d1, y2 = 0; ` ` `  `    ``// Third coordinate ` `    ``int` `x3 = (d1 + d2 - d3) / 2; ` `    ``int` `y3 = (d2 + d3 - d1) / 2; ` `    ``Console.WriteLine(``"("``+x1+``", "``+y1+``"), ("``+x2+``", "``+y2+``") and ("``+x3+``", "``+y3+``")"``); ` `} ` ` `  `// Driver code ` `static` `void` `Main() ` `{ ` `    ``int` `d1 = 3, d2 = 4, d3 = 5; ` `    ``solve(d1, d2, d3); ` ` `  `} ` `} ` ` `  `// This code is contributed by mits `

Output:

```(0, 0), (3, 0) and (1, 3)
```

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Improved By : Mithun Kumar, vt_m, ihritik