Related: Number of swaps to sort when only adjacent swapping allowed
Minimum number of swaps required to sort an array - GeeksforGeeks
Given an array of n distinct elements, find the minimum number of swaps required to sort the array.
ans = Σi = 1k(cycle_size – 1)
https://www.includehelp.com/java-programs/minimum-swaps-required-to-sort-an-array.aspx
https://www.geeksforgeeks.org/minimum-number-of-swaps-required-to-sort-an-array-set-2/
http://stackoverflow.com/questions/15152322/compute-the-minimal-number-of-swaps-to-order-a-sequence/15152602#15152602
https://www.geeksforgeeks.org/number-swaps-sort-adjacent-swapping-allowed/
Minimum number of swaps required to sort an array - GeeksforGeeks
Given an array of n distinct elements, find the minimum number of swaps required to sort the array.
This can be easily done by visualizing the problem as a graph. We will have n nodes and an edge directed from node i to node j if the element at i’th index must be present at j’th index in the sorted array.
Graph for {4, 3, 2, 1}
The graph will now contain many non-intersecting cycles. Now a cycle with 2 nodes will only require 1 swap to reach the correct ordering, similarly a cycle with 3 nodes will only require 2 swap to do so.
Graph for {4, 5, 2, 1, 5}
Hence,
where k is the number of cycles
Time Complexity: O(n*log(n))
Auxiliary Space: O(n)
Auxiliary Space: O(n)
public static int minSwaps(int[] arr) { int n = arr.length; // Create two arrays and use as pairs where first // array is element and second array // is position of first element ArrayList <Pair <Integer, Integer> > arrpos = new ArrayList <Pair <Integer, Integer> > (); for (int i = 0; i < n; i++) arrpos.add(new Pair <Integer, Integer> (arr[i], i)); // Sort the array by array element values to // get right position of every element as the // elements of second array. arrpos.sort(new Comparator<Pair<Integer, Integer>>() { @Override public int compare(Pair<Integer, Integer> o1, Pair<Integer, Integer> o2) { if (o1.getValue() > o2.getValue()) return -1; // We can change this to make it then look at the // words alphabetical order else if (o1.getValue().equals(o2.getValue())) return 0; else return 1; } }); // To keep track of visited elements. Initialize // all elements as not visited or false. Boolean[] vis = new Boolean[n]; Arrays.fill(vis, false); // Initialize result int ans = 0; // Traverse array elements for (int i = 0; i < n; i++) { // already swapped and corrected or // already present at correct pos if (vis[i] || arrpos.get(i).getValue() == i) continue; // find out the number of node in // this cycle and add in ans int cycle_size = 0; int j = i; while (!vis[j]) { vis[j] = true; // move to next node j = arrpos.get(j).getValue(); cycle_size++; } // Update answer by adding current cycle. ans += (cycle_size - 1); } // Return result return ans; }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; }
Find the Elements that are already sorted and focus on the rest of the elements for swapping.
public static int minimumSwap (int[] inputarray) { int[] sorted = new int[inputarray.length];//take new array of size of the input array for (int i = 0; i < array.length; i++) { sorted[i] = inputarray[i]; }//copy the input array into sorted array using for loop int n = sizeof(arr)/sizeof(arr[0]); //now sort the inputarray using below method. inputarray.sort(inputarray,inputarray+n); int j = 0; //now compare the sorted array with input array for (int i = 0; i < inputarray.length; i++) { if (inputarray[i] == sorted[j]) j++; } //now we have count of not moved elements.Just subtract it from array lengths, will get the minimum number of swaps required to sort. return array.length - j;}
The idea is to create a vector of pair in C++ with first element as array values and second element as array indices. The next step is to sort the vector of pair according to the first element of the pair. After that traverse the vector and check if the index mapped with the value is correct or not, if not then keep swapping until the element is placed correctly and keep counting the number of swaps.
X. http://geekyjumps.blogspot.com/2013/08/find-minimum-number-of-swap-required-to.html
http://www.geeksforgeeks.org/minimum-swap-required-convert-binary-tree-binary-search-tree/
Given an array containing sequence of bits (0 or 1), you have to sort this array in the ascending order i.e. all 0' in first part of array followed by all 1's. The constraints is that you can swap only the adjacent elements in the array. Find the minimum number of swaps required to sort the given input array.
Example: Given the array (0,0,1,0,1,0,1,1) the minimum number of swaps is 3.
int swapRequired(int []array){
//Boundary condition
if(array == null || array.length == 0)
return 0;
int swap = 0, count = 0;
//iterate over all the elements of given array
for(int i = array.length; i >= 0; i--){
if(array[i] == 0)
count++; //count the number of 0s between two 1s
else
swap += count;
}
return swap; //return the count of required swap
}
Example: Given the array (0,0,1,0,1,0,1,1) the minimum number of swaps is 3.
int swapRequired(int []array){
//Boundary condition
if(array == null || array.length == 0)
return 0;
int swap = 0, count = 0;
//iterate over all the elements of given array
for(int i = array.length; i >= 0; i--){
if(array[i] == 0)
count++; //count the number of 0s between two 1s
else
swap += count;
}
return swap; //return the count of required swap
}
http://www.geeksforgeeks.org/minimum-swap-required-convert-binary-tree-binary-search-tree/
Given the array representation of Complete Binary Tree i.e, if index i is the parent, index 2*i + 1 is the left child and index 2*i + 2 is the right child. The task is to find the minimum number of swap required to convert it into Binary Search Tree.
The idea is to use the fact that inorder traversal of Binary Search Tree is in increasing order of their value.
So, find the inorder traversal of the Binary Tree and store it in the array and try to sort the array. The minimum number of swap required to get the array sorted will be the answer. Please refer below post to find minimum number of swaps required to get the array sorted.
So, find the inorder traversal of the Binary Tree and store it in the array and try to sort the array. The minimum number of swap required to get the array sorted will be the answer. Please refer below post to find minimum number of swaps required to get the array sorted.
Exercise: Can we extend this to normal binary tree, i.e., a binary tree represented using left and right pointers, and not necessarily complete?
https://www.geeksforgeeks.org/number-swaps-sort-adjacent-swapping-allowed/
Given an array arr[] of non negative integers. We can perform a swap operation on any two adjacent elements in the array. Find the minimum number of swaps needed to sort the array in ascending order.
Examples :
Input : arr[] = {3, 2, 1}
Output : 3
We need to do following swaps
(3, 2), (3, 1) and (1, 2)
Input : arr[] = {1, 20, 6, 4, 5}
Output : 5
)
There is an interesting solution to this problem. It can be solved using the fact that number of swaps needed is equal to number of inversions. So we basically need to count inversions in array.
The fact can be established using below observations:
1) A sorted array has no inversions.
2) An adjacent swap can reduce one inversion. Doing x adjacent swaps can reduce x inversions in an array.
Read full article from Minimum number of swaps required to sort an array - GeeksforGeeks1) A sorted array has no inversions.
2) An adjacent swap can reduce one inversion. Doing x adjacent swaps can reduce x inversions in an array.
