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
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:
(0, 0), (3, 0) and (1, 3)
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