{ Minimum Swaps 2 }

Solving the Programming Challenge

Minimum Swaps 2

Problem

You are given an unordered array consisting of consecutive integers [1, 2, 3, …, n] without any duplicates. You are allowed to swap any two elements. You need to find the minimum number of swaps required to sort the array in ascending order.

For example, given the array we perform the following steps:

i   arr                         swap (indices)
0   [7, 1, 3, 2, 4, 5, 6]   swap (0,3)
1   [2, 1, 3, 7, 4, 5, 6]   swap (0,1)
2   [1, 2, 3, 7, 4, 5, 6]   swap (3,4)
3   [1, 2, 3, 4, 7, 5, 6]   swap (4,5)
4   [1, 2, 3, 4, 5, 7, 6]   swap (5,6)
5   [1, 2, 3, 4, 5, 6, 7]
It took  swaps to sort the array.

Function Description

Complete the function minimumSwaps in the editor below. It must return an integer representing the minimum number of swaps to sort the array.

minimumSwaps has the following parameter(s):

arr: an unordered array of integers

Read more on the challenge page…


My Solution

I’m providing the solution for Python and JS, please leave on the comments if you found a better way.

Python

def minimumSwaps(arr):
    swaps = 0
    tmp = {}

    for i, val in enumerate(arr):
        tmp[val] = i

    for i in range(len(arr)):
        # because they are consecutives, I can see if the number is where it belongs
        if arr[i] != i+1:
            swaps += 1
            t = arr[i]
            arr[i] = i+1
            arr[tmp[i+1]] = t

            # Switch also the tmp array, no need to change i+1 as it's already good now
            tmp[t] = tmp[i+1]

    return swaps

Javascript

function minimumSwaps(arr) {
    let swaps = 0;

    const tmp = {};
    arr.forEach((v, i) => tmp[v] = i);;

    arr.forEach((v, i) => {
        // because they are consecutives, I can see if the number is where it belongs
        if (arr[i] !== i+1) {
            swaps += 1
            const t = arr[i]
            arr[i] = i+1
            arr[tmp[i+1]] = t

            // Switch also the tmp array, no need to change i+1 as it's already good now
            tmp[t] = tmp[i+1]
        }
    });

    return swaps
}

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