## Compute the minimal number of swaps to order a sequence

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I'm working on sorting an integer sequence with no identical numbers (without loss of generality, let's assume the sequence is a permutation of `1,2,...,n`

) into its natural increasing order (i.e. `1,2,...,n`

). I was thinking about directly swapping the elements (regardless of the positions of elements; in other words, a swap is valid for any two elements) with minimal number of swaps (the following may be a feasible solution):

Swap two elements with the constraint that either one or both of them should be swapped into the correct position(s). Until every element is put in its correct position.

But I don't know how to mathematically prove if the above solution is optimal. Anyone can help?

I was able to prove this with graph-theory. Might want to add that tag in :)

Create a graph with `n`

vertices. Create an edge from node `n_i`

to `n_j`

if the element in position `i`

should be in position `j`

in the correct ordering. You will now have a graph consisting of several non-intersecting cycles. I argue that the minimum number of swaps needed to order the graph correctly is

M = sum (c in cycles) size(c) - 1

Take a second to convince yourself of that...if two items are in a cycle, one swap can just take care of them. If three items are in a cycle, you can swap a pair to put one in the right spot, and a two-cycle remains, etc. If `n`

items are in a cycle, you need `n-1`

swaps. (This is always true even if you don't swap with immediate neighbors.)

Given that, you may now be able to see why your algorithm is optimal. If you do a swap and at least one item is in the right position, then it will always reduce the value of `M`

by 1. For any cycle of length `n`

, consider swapping an element into the correct spot, occupied by its neighbor. You now have a correctly ordered element, and a cycle of length `n-1`

.

Since `M`

is the minimum number of swaps, and your algorithm always reduces `M`

by 1 for each swap, it must be optimal.

**Minimum number of swaps required to sort an array,** The problem is to find the minimum swaps in which the array can be sorted. Examples: Input : arr[] = {4, 3, 2, 1} Output Given an array of n distinct elements, find the minimum number of swaps required to sort the array. Examples: Input : {4, 3, 2, 1} Output : 2 Explanation : Swap index 0 with 3 and 1 with 2 to form the sorted array {1, 2, 3, 4}.

**Compute the minimal number of swaps to order a sequence,** I was able to prove this with graph-theory. Might want to add that tag in :) Create a graph with n vertices. Create an edge from node n_i to n_j if It takes n − 1 swaps to implement a cycle of length n. So if the sorted permutation of a list can be decomposed into cycles with lengths a, b, c, …, z then the minimum number of swaps needed to sort the list is (a − 1) + (b − 1) + (c − 1) + ⋯ + (z − 1).

For your reference, here is an algorithm that I wrote, to generate the minimum number of swaps needed to sort the array. It finds the cycles as described by @Andrew Mao.

/** * Finds the minimum number of swaps to sort given array in increasing order. * @param ar array of <strong>non-negative distinct</strong> integers. * input array will be overwritten during the call! * @return min no of swaps */ public int findMinSwapsToSort(int[] ar) { int n = ar.length; Map<Integer, Integer> m = new HashMap<>(); for (int i = 0; i < n; i++) { m.put(ar[i], i); } Arrays.sort(ar); for (int i = 0; i < n; i++) { ar[i] = m.get(ar[i]); } m = null; int swaps = 0; for (int i = 0; i < n; i++) { int val = ar[i]; if (val < 0) continue; while (val != i) { int new_val = ar[val]; ar[val] = -1; val = new_val; swaps++; } ar[i] = -1; } return swaps; }

**Minimum number of swaps needed to sort array,** Compute the minimal number of swaps to order a sequence – Andrew Morton Apr 15 at 19:13 noofswaps = 0 for i in range(len(A)): while A[i] != i + 1: temp = A[i] A[i] i=0;i<arr.length;i++){ int j=i,count=0; while(!newarr[j]){ newarr[j]=true; Hackerrank Python code for minimum swaps 2 using hashmaps 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.

We do not need to swap the actual elements, just find how many elements are not in the right index (Cycle). The min swaps will be Cycle - 1; Here is the code...

static int minimumSwaps(int[] arr) { int swap=0; boolean visited[]=new boolean[arr.length]; for(int i=0;i<arr.length;i++){ int j=i,cycle=0; while(!visited[j]){ visited[j]=true; j=arr[j]-1; cycle++; } if(cycle!=0) swap+=cycle-1; } return swap; }

**Minimum number of swaps required to sort the array in ascending ,** Analyzing the problem, finding clues. I used selection sort approach as selection sort makes minimum swap operation. Really? (I don't know.) Minimum number of swaps required for arranging pairs adjacent to each other There are n-pairs and therefore 2n people. everyone has one unique number ranging from 1 to 2n. All these 2n persons are arranged in random fashion in an Array of size 2n.

Swift 4 version:

func minimumSwaps(arr: [Int]) -> Int { struct Pair { let index: Int let value: Int } var positions = arr.enumerated().map { Pair(index: $0, value: $1) } positions.sort { $0.value < $1.value } var indexes = positions.map { $0.index } var swaps = 0 for i in 0 ..< indexes.count { var val = indexes[i] if val < 0 { continue // Already visited. } while val != i { let new_val = indexes[val] indexes[val] = -1 val = new_val swaps += 1 } indexes[i] = -1 } return swaps }

**Arrays Minimum Number of Swaps,** for determining the minimum number of swaps necessary to put and unsorted array of Duration: 11:07
Posted: Aug 24, 2018 This is a bit of a trick question. The absolutely upper-bound on the number of swaps needed will always be N, with SelectionSort. It makes the least number of array writes, compared to QuickSort or other NlogN sorts.

**Minimum Swaps Required to Sort an Array,** An efficient algorithm to find the minimum number of swaps required to sort the array in Duration: 7:57
Posted: Dec 12, 2019 Swap items in the list to get to the correct order, but only swap items within the same subgraph. So, I hope you can see it's not impossible to do the minimal number of swaps to get to sorted order, but it's not worth it, because it requires a ridiculous number of comparisons. Just use heapsort.

**Minimum Swaps 2,** 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 In 1985 I published an algorithm I called UnShuffle. It performs no swaps (0) of data elements, so its performance is unaffected by large data elements. It operates on linked lists and does not perform swaps even on the list pointers.

**How to find the minimum number of swaps required to sort an array ,** This is a bit of a trick question. The absolutely upper-bound on the number of swaps needed will always be N, with SelectionSort. It makes the least number of Minimum insertions to sort an array Given an array of integer numbers, we need to sort this array in a minimum number of steps where in one step we can insert any array element from its position to any other position.

##### Comments

- Highly related / duplicate: Minimum number of swaps needed to change Array 1 to Array 2?
- what will be time complexity of this?
- Time complexity : O(n*logn) Space complexity : O(n) @puneet
- But how is that a proof of
*minimality*? "I argue that the minimum number of swaps...", "Take a second to convince yourself of that..." Sorry, "arguing" and "convincing yourself" is not enough. You have to actually prove that the above`M`

is minimal. - @AnT, I agree. Specifically, I can conceive of an algorithm that involves swaps where neither item ends it's intended positions, but achieves the same number of moves. Specifically, one can make swaps to reduce any cycle to a number of two cycles (possibly ending with a single one cycle if
`n`

is odd), and then swap all of the two cycles into the correct positions. This also involves`n-1`

moves. Although this is not faster than the algorithm provided, it at least shows that the optimality of the provided algorithm is far from obvious. - @AnT Minimality comes from the fact that any
`n`

-element cycle is minimally the product of`n-1`

transpositions - Can you explain what is happening in last while loop
- Can anyone help with understanding the code? I can't seem to grasp the logic behind what is happening
- @GURMEETSINGH did you figure out the algorithm?
- @Spindoctor yes I figured it out
- @Spindoctor in first for loop it is keeping the actual value as key and the position in the original array as value. Then the array is sorted using Collections.sort(). in second for loop we are getting index of array prior to sorting. in the last for loop we are making elements of cycle as -1
- I am not able to relate how the while loops works to find the number of cycles. Specifically, the 2nd statement in the while loop.
`j=arr[j]-1;`

Why the value of j getting derived by subtracting 1 whereas we are setting it to i at the start.