Compute the minimal number of swaps to order a sequence

Question

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?

Answer 1

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.

Answer 2

All the cycle counting is very difficult to keep in your head. There is a way that is much simpler to memorize.

First, let's go through a sample case manually.

• Sequence: [7, 1, 3, 2, 4, 5, 6]
• Enumerate it: [(0, 7), (1, 1), (2, 3), (3, 2), (4, 4), (5, 5), (6, 6)]
• Sort the enumeration by value: [(1, 1), (3, 2), (2, 3), (4, 4), (5, 5), (6, 6), (0, 7)]
• Start from the beginning. While the index is different from the enumerated index keep on swapping the elements defined by index and enumerated index. Remember: swap(0,2);swap(0,3) is the same as swap(2,3);swap(0,2)
  • swap(0, 1) => [(3, 2), (1, 1), (2, 3), (4, 4), (5, 5), (6, 6), (0, 7)]
  • swap(0, 3) => [(4, 4), (1, 1), (2, 3), (3, 2), (5, 5), (6, 6), (0, 7)]
  • swap(0, 4) => [(5, 5), (1, 1), (2, 3), (3, 2), (4, 4), (6, 6), (0, 7)]
  • swap(0, 5) => [(6, 6), (1, 1), (2, 3), (3, 2), (4, 4), (5, 5), (0, 7)]
  • swap(0, 6) => [(0, 7), (1, 1), (2, 3), (3, 2), (4, 4), (5, 5), (6, 6)]

I.e. semantically you sort the elements and then figure out how to put them to the initial state via swapping through the leftmost item that is out of place.

Python algorithm is as simple as this:

def swap(arr, i, j):
    arr[i], arr[j] = arr[j], arr[i]


def minimum_swaps(arr):
    annotated = [*enumerate(arr)]
    annotated.sort(key = lambda it: it[1])

    count = 0

    i = 0
    while i < len(arr):
        if annotated[i][0] == i:
            i += 1
            continue
        swap(annotated, i, annotated[i][0])
        count += 1

    return count

Thus, you don't need to memorize visited nodes or compute some cycle length.

Answer 3

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;
}

Answer 4

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;


    }

Answer 5

@Archibald, I like your solution, and such was my initial assumptions that sorting the array would be the simplest solution, but I don't see the need to go through the effort of the reverse-traverse as I've dubbed it, ie enumerating then sorting the array and then computing the swaps for the enums.

I find it simpler to subtract 1 from each element in the array and then to compute the swaps required to sort that list

here is my tweak/solution:

def swap(arr, i, j):
    tmp = arr[i]
    arr[i] = arr[j]
    arr[j] = tmp

def minimum_swaps(arr):

    a = [x - 1 for x in arr]

    swaps = 0
    i = 0
    while i < len(a):
        if a[i] == i:
            i += 1
            continue
        swap(a, i, a[i])
        swaps += 1

    return swaps

As for proving optimality, I think @arax has a good point.

Answer 6

// Assuming that we are dealing with only sequence started with zero

function minimumSwaps(arr) {
    var len = arr.length
    var visitedarr = []
    var i, start, j, swap = 0
    for (i = 0; i < len; i++) {
        if (!visitedarr[i]) {
            start = j = i
            var cycleNode = 1
            while (arr[j] != start) {
                j = arr[j]
                visitedarr[j] = true
                cycleNode++
            }
            swap += cycleNode - 1
        }
    }
    return swap
}

Answer 7

I really liked the solution of @Ieuan Uys in Python.

What I improved on his solution;

• While loop is iterated one less to increase speed; while i < len(a) - 1
• Swap function is de-capsulated to make one, single function.
• Extensive code comments are added to increase readability.

My code in python.

def minimumSwaps(arr):
    #make array values starting from zero to match index values. 
    a = [x - 1 for x in arr] 

    #initialize number of swaps and iterator.
    swaps = 0
    i = 0

    while i < len(a)-1:
        if a[i] == i:
            i += 1
            continue

        #swap.
        tmp = a[i] #create temp variable assign it to a[i]
        a[i] = a[tmp] #assign value of a[i] with a[tmp]
        a[tmp] = tmp #assign value of a[tmp] with tmp (or initial a[i])

        #calculate number of swaps.
        swaps += 1

    return swaps

Detailed explanation on what code does on an array with size n;

We check every value except last one (n-1 iterations) in the array one by one. If the value does not match with array index, then we send this value to its place where index value is equal to its value. For instance, if at a[0] = 3. Then this value should swap with a[3]. a[0] and a[3] is swapped. Value 3 will be at a[3] where it is supposed to be. One value is sent to its place. We have n-2 iteration left. I am not interested what is now a[0]. If it is not 0 at that location, it will be swapped by another value latter. Because that another value also exists in a wrong place, this will be recognized by while loop latter.

Real Example

a[4, 2, 1, 0, 3]
#iteration 0, check a[0]. 4 should be located at a[4] where the value is 3. Swap them. 
a[3, 2, 1, 0, 4] #we sent 4 to the right location now.  
#iteration 1, check a[1]. 2 should be located at a[2] where the value is 1. Swap them. 
a[3, 1, 2, 0, 4] #we sent 2 to the right location now.  
#iteration 2, check a[2]. 2 is already located at a[2]. Don't do anything, continue. 
a[3, 1, 2, 0, 4] 
#iteration 3, check a[3]. 0 should be located at a[0] where the value is 3. Swap them. 
a[0, 1, 2, 3, 4] #we sent 0 to the right location now.  
# There is no need to check final value of array. Since all swaps are done. 

Answer 8

Nicely done solution by @bekce. If using C#, the initial code of setting up the modified array ar can be succinctly expressed as:

var origIndexes = Enumerable.Range(0, n).ToArray();
Array.Sort(ar, origIndexes);

then use origIndexes instead of ar in the rest of the code.

Answer 9

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
}

Answer 10

This is the sample code in C++ that finds the minimum number of swaps to sort a permutation of the sequence of (1,2,3,4,5,.......n-2,n-1,n)

#include<bits/stdc++.h>
using namespace std;


int main()
{
    int n,i,j,k,num = 0;
    cin >> n;
    int arr[n+1];
    for(i = 1;i <= n;++i)cin >> arr[i];
    for(i = 1;i <= n;++i)
    {
        if(i != arr[i])// condition to check if an element is in a cycle r nt
        {
            j = arr[i];
            arr[i] = 0;
            while(j != 0)// Here i am traversing a cycle as mentioned in 
            {             // first answer
                k = arr[j];
                arr[j] = j;
                j = k;
                num++;// reducing cycle by one node each time
            }
            num--;
        }
    }
    for(i = 1;i <= n;++i)cout << arr[i] << " ";cout << endl;
    cout << num << endl;
    return 0;
}

Answer 11

Solution using Javascript.

First I set all the elements with their current index that need to be ordered, and then I iterate over the map to order only the elements that need to be swapped.

function minimumSwaps(arr) {
  const mapUnorderedPositions = new Map()
  for (let i = 0; i < arr.length; i++) {
    if (arr[i] !== i+1) {
      mapUnorderedPositions.set(arr[i], i)
    }
  }

  let minSwaps = 0
  while (mapUnorderedPositions.size > 1) {
    const currentElement = mapUnorderedPositions.entries().next().value
    const x = currentElement[0]
    const y = currentElement[1]

    // Skip element in map if its already ordered
    if (x-1 !== y) {
      // Update unordered position index of swapped element
      mapUnorderedPositions.set(arr[x-1], y)

      // swap in array
      arr[y] = arr[x-1]
      arr[x-1] = x

      // Increment swaps
      minSwaps++
    }

    mapUnorderedPositions.delete(x)
  }


  return minSwaps
}

If you have an input like 7 2 4 3 5 6 1, this is how the debugging will go:

Map { 7 => 0, 4 => 2, 3 => 3, 1 => 6 }
currentElement [ 7, 0 ]
swapping 1 with 7
[ 1, 2, 4, 3, 5, 6, 7 ]

currentElement [ 4, 2 ]
swapping 3 with 4
[ 1, 2, 3, 4, 5, 6, 7 ]

currentElement [ 3, 2 ]
skipped

minSwaps = 2

Answer 12

Finding the minimum number of swaps required to put a permutation of 1..N in order.

We can use that the we know what the sort result would be: 1..N, which means we don't actually have to do swaps just count them.

The shuffling of 1..N is called a permutation, and is composed of disjoint cyclic permutations, for example, this permutation of 1..6:

1 2 3 4 5 6
6 4 2 3 5 1

Is composed of the cyclic permutations (1,6)(2,4,3)(5)

1->6(->1) cycle: 1 swap
2->4->3(->2) cycle: 2 swaps
5(->5) cycle: 0 swaps

So a cycle of k elements requires k-1 swaps to put in order.

Since we know where each element "belongs" (i.e. value k belongs at position k-1) we can easily traverse the cycle. Start at 0, we get 6, which belongs at 5, and there we find 1, which belongs at 0 and we're back where we started.

To avoid re-counting a cycle later, we track which elements were visited - alternatively you could perform the swaps so that the elements are in the right place when you visit them later.

The resulting code:

def minimumSwaps(arr):
    visited = [False] * len(arr)
    numswaps = 0
    for i in range(len(arr)):
        if not visited[i]:
            visited[i] = True
            j = arr[i]-1
            while not visited[j]:
                numswaps += 1
                visited[j] = True
                j = arr[j]-1
    return numswaps

Answer 13

An implementation on integers with primitive types in Java (and tests).

import java.util.Arrays;

public class MinSwaps {
  public static int computate(int[] unordered) {
    int size = unordered.length;
    int[] ordered = order(unordered);
    int[] realPositions = realPositions(ordered, unordered);
    boolean[] touchs = new boolean[size];
    Arrays.fill(touchs, false);
    int i;
    int landing;
    int swaps = 0;

    for(i = 0; i < size; i++) {
      if(!touchs[i]) {
        landing = realPositions[i];

        while(!touchs[landing]) {
          touchs[landing] = true;
          landing = realPositions[landing];

          if(!touchs[landing]) { swaps++; }
        }
      }
    }

    return swaps;
  }

  private static int[] realPositions(int[] ordered, int[] unordered) {
    int i;
    int[] positions = new int[unordered.length];

    for(i = 0; i < unordered.length; i++) {
      positions[i] = position(ordered, unordered[i]);
    }

    return positions;
  }

  private static int position(int[] ordered, int value) {
    int i;

    for(i = 0; i < ordered.length; i++) {
      if(ordered[i] == value) {
        return i;
      }
    }

    return -1;
  }

  private static int[] order(int[] unordered) {
    int[] ordered = unordered.clone();
    Arrays.sort(ordered);

    return ordered;
  }
}

Tests

import org.junit.Test;

import static org.junit.Assert.assertEquals;

public class MinimumSwapsSpec {
  @Test
  public void example() {
    // setup
    int[] unordered = new int[] { 40, 23, 1, 7, 52, 31 };

    // run
    int minSwaps = MinSwaps.computate(unordered);

    // verify
    assertEquals(5, minSwaps);
  }

  @Test
  public void example2() {
    // setup
    int[] unordered = new int[] { 4, 3, 2, 1 };

    // run
    int minSwaps = MinSwaps.computate(unordered);

    // verify
    assertEquals(2, minSwaps);
  }

  @Test
  public void example3() {
    // setup
    int[] unordered = new int[] {1, 5, 4, 3, 2};

    // run
    int minSwaps = MinSwaps.computate(unordered);

    // verify
    assertEquals(2, minSwaps);
  }
}

Answer 14

Swift 4.2:

func minimumSwaps(arr: [Int]) -> Int {
    let sortedValueIdx = arr.sorted().enumerated()
        .reduce(into: [Int: Int](), { $0[$1.element] = $1.offset })

    var checked = Array(repeating: false, count: arr.count)
    var swaps = 0

    for idx in 0 ..< arr.count {
        if checked[idx] { continue }

        var edges = 1
        var cursorIdx = idx
        while true {
            let cursorEl = arr[cursorIdx]
            let targetIdx = sortedValueIdx[cursorEl]!
            if targetIdx == idx {
                break
            } else {
                cursorIdx = targetIdx
                edges += 1
            }
            checked[targetIdx] = true
        }
        swaps += edges - 1
    }

    return swaps
}

Answer 15

Python code

A = [4,3,2,1]
count = 0
for i in range (len(A)):
    min_idx = i
    for j in range (i+1,len(A)):
        if A[min_idx] > A[j]:
            min_idx = j
    if min_idx > i:
        A[i],A[min_idx] = A[min_idx],A[i]
        count = count + 1
print "Swap required : %d" %count

Answer 16

In Javascript

If the count of the array starts with 1

function minimumSwaps(arr) {
   var len = arr.length
    var visitedarr = []
    var i, start, j, swap = 0
    for (i = 0; i < len; i++) {
        if (!visitedarr[i]) {
            start = j = i
            var cycleNode = 1
            while (arr[j] != start + 1) {
                j = arr[j] - 1
                visitedarr[j] = true
                cycleNode++
            }
            swap += cycleNode - 1
        }
    }
    return swap
}

else for input starting with 0

function minimumSwaps(arr) {
    var len = arr.length
    var visitedarr = []
    var i, start, j, swap = 0
    for (i = 0; i < len; i++) {
        if (!visitedarr[i]) {
            start = j = i
            var cycleNode = 1
            while (arr[j] != start) {
                j = arr[j]
                visitedarr[j] = true
                cycleNode++
            }
            swap += cycleNode - 1
        }
    }
    return swap
}

Just extending Darshan Puttaswamy code for current HackerEarth inputs

Answer 17

def swap_sort(arr)
  changes = 0
  loop do
    # Find a number that is out-of-place
    _, i = arr.each_with_index.find { |val, index| val != (index + 1) }
    if i != nil
      # If such a number is found, then `j` is the position that the out-of-place number points to.
      j = arr[i] - 1

      # Swap the out-of-place number with number from position `j`.
      arr[i], arr[j] = arr[j], arr[i]

      # Increase swap counter.
      changes += 1
    else
      # If there are no out-of-place number, it means the array is sorted, and we're done.
      return changes
    end
  end
end                                                                                                

Answer 18

Apple Swift version 5.2.4

func minimumSwaps(arr: [Int]) -> Int {
    var swapCount = 0
    var arrayPositionValue = [(Int, Int)]()
    var visitedDictionary = [Int: Bool]()
    
    for (index, number) in arr.enumerated() {
        arrayPositionValue.append((index, number))
        visitedDictionary[index] = false
    }
    
    arrayPositionValue = arrayPositionValue.sorted{ $0.1 < $1.1 }

    for i in 0..<arr.count {
        var cycleSize = 0
        var visitedIndex = i

        while !visitedDictionary[visitedIndex]! {
            visitedDictionary[visitedIndex] = true
            visitedIndex = arrayPositionValue[visitedIndex].0
            cycleSize += 1
        }

        if cycleSize > 0 {
            swapCount += cycleSize - 1
        }
    }

    return swapCount
}

Answer 19

Go version 1.17:

func minimumSwaps(arr []int32) int32 {

var swap int32
for i := 0; i < len(arr) - 1; i++{
    for j := 0; j < len(arr); j++ {
        if arr[j] > arr[i] {
            arr[i], arr[j] = arr[j], arr[i]
            swap++
        }else {
        continue
    }
    }
} 
return swap
}