You have been given an array 'ARR' of 'N' distinct elements.
Your task is to find the minimum no. of swaps required to sort the array.
For example:
For the given input array [4, 3, 2, 1], the minimum no. of swaps required to sort the array is 2, i.e. swap index 0 with 3 and 1 with 2 to form the sorted array [1, 2, 3, 4].
The first line of input contains an integer ‘T’ representing the number of test cases. Then the test cases follow.
The first line of each test case contains an integer ‘N’ representing the size of the input array.
The second line of each test case contains the 'N' elements of the array separated by a single space.
For each test case, print a single line containing a single integer which represents the minimum no. of swaps required to sort the array.
The output for each test case is in a separate line.
You do not need to print anything; it has already been taken care of. Just implement the given function.
1 <= T <= 100
1 <= N <= 1000
0 <= ARR[i] <= 10 ^ 9
Where 'ARR[i]' is the value of the input array elements.
Time Limit: 1 sec
2
4
4 3 2 1
5
1 5 4 3 2
2
2
For the first test case, swap index 0 with 3 i.e. 4 -> 1 and 1 with 2 i.e. 3 -> 2 to form the sorted array {1, 2, 3, 4}.
For the second test case, swap index 1 with 4 i.e. 5 -> 2 and 2 with 3 i.e. 4 -> 3 to form the sorted array {1, 2, 3, 4, 5}.
2
4
1 2 3 4
6
3 5 2 4 6 8
0
3
Swap every element of the given array with its right index, if it is not at the right place.
While iterating over the array, check the current element, and if not in the correct place, replace that element with the index of the element which should have come in this place.
Below is the algorithm:
O(N ^ 2), where ‘N’ is the size of the given array.
Since we are using indexOf() function to check the right index of every element of input array, having the worst case time complexity of O(N) for each array element. So, the overall complexity is O(N ^ 2).
O(N), where ‘N’ is the size of the given array.
Since we are using a temp array to store the sorted form of a given input array, thus, the space complexity will be O(N).
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Interview problems
min swap easy soln
#include<bits/stdc++.h>
int minSwaps(vector<int> &arr)
{
int n=arr.size();
vector<pair<int,int>>v(n);
for(int i=0;i<n;i++){
v[i]={arr[i],i};
}
sort(v.begin(),v.end());
int cnt=0;
for(int i=0;i<n;i++){
if(v[i].second==i){
}
else{
swap(v[i],v[v[i].second]);
cnt++;
i--;
}
}
return cnt;
}
Interview problems
C++ solution please upvote
int minSwaps(vector<int> &arr)
{
int n = arr.size();
int count = 0;
vector<int> arr2 = arr;
sort(arr2.begin(), arr2.end());
for (int i = 0; i < n; i++) {
int minIdx = i;
for (int j = i + 1; j < n; j++) {
if (arr[j] < arr[minIdx]) {
minIdx = j;
}
}
if (minIdx != i) {
swap(arr[i], arr[minIdx]);
count++;
}
if (equal(arr.begin(), arr.end(), arr2.begin())) {
return count;
}
}
return count;
}Interview problems
EASY SOLUTION C++
int minSwaps(vector<int> &arr)
{
// using selection sort.
int n = arr.size();
int cnt = 0;
for(int i=0; i<n-1; i++){
int min = i;
for(int j=i+1; j<n; j++){
if(arr[j] < arr[min])
min = j;
}
if(min != i){
swap(arr[min], arr[i]);
cnt++;
}
}
return cnt;
}
Interview problems
Simple java Solution
import java.util.*;
public class Solution {
public static class Pair
{
int n;
int i;
Pair(int element,int index)
{
n=element;
i=index;
}
}
public static int minSwaps(int[] arr) {
// Write your code here.
List<Pair> lst=new ArrayList<>();
for(int i=0;i<arr.length;i++)
{
lst.add(new Pair(arr[i],i));
}
Comparator<Pair> cmp=(a,b)->(a.n-b.n);
Collections.sort(lst,cmp);
int swap=0;
int k=0;
int n=arr.length;
while(k<n)
{
Pair p=lst.get(k);
if(p.i!=k)
{
swap++;
Pair temp=lst.get(k);
lst.set(k,lst.get(p.i));
lst.set(p.i,temp);
}
else
{
k++;
}
}
return swap;
}
}
Interview problems
easy c++ code || Upvote
int minSwaps(vector<int> &arr)
{
// Write your code here.
vector<pair<int,int>>v1;
for(int i=0; i<arr.size(); ++i)
{
v1.push_back({arr[i],i});
}
sort(v1.begin(),v1.end());
int count=0;
for(int i=0; i<arr.size(); ++i)
{
pair<int,int>p = v1[i];
int value = p.first;
int index = p.second;
if(index==i)
{
continue;
}
else
{
swap(v1[i],v1[index]);
count++;
i--;
}
}
return count;
}
Interview problems
Let make it Easy !!!
#include<bits/stdc++.h>
int minSwaps(vector<int> &arr)
{
vector<pair<int,int>>map(arr.size());
for(int i=0;i<arr.size();i++){
map[i] = {arr[i],i};
}
sort(map.begin(),map.end());
int swaps =0;
for(int i=0;i<arr.size();i++){
if(map[i].second == i){
continue;
}
swap(map[i], map[map[i].second]);
swaps++;
i--;
}
return swaps;
}Interview problems
Easy and short
#include <climits> #include <iostream> #include <vector> using namespace std;
int arraySortedOrNot(vector<int> &arr, int n) { if (n == 1 || n == 0) return 1; if (arr[n - 1] < arr[n - 2]) return 0; return arraySortedOrNot(arr, n - 1); }
int minSwaps(vector<int> &arr) { int N = arr.size(); int count = 0;
for (int i = 0; i < N - 1; ++i) { int minIndex = i; for (int j = i + 1; j < N; ++j) { if (arr[j] < arr[minIndex]) { minIndex = j; } }
if (minIndex != i) { swap(arr[i], arr[minIndex]); count++; }
if (arraySortedOrNot(arr, N)) { break; } }
return count; }
Interview problems
easy cpp
#include <bits/stdc++.h>
using namespace std;
int arraySortedOrNot(vector<int> &arr, int n) {
if (n == 1 || n == 0)
return 1;
if (arr[n - 1] < arr[n - 2])
return 0;
return arraySortedOrNot(arr, n - 1);
}
int minSwaps(vector<int> &arr) {
int N = arr.size();
int count = 0;
if (arraySortedOrNot(arr, N)) {
return count;
} else {
int min;
int k;
for (int i = 0; i < N; i++) {
min=INT_MAX;
for (int j = i; j < N; j++) {
if (arr[j] < min) {
min = arr[j];
k = j;
}
}
if (k!= i) {
int temp = arr[i];
arr[i] = arr[k];
arr[k] = temp;
count++;
}
if (arraySortedOrNot(arr, N)) {
return count;
}
}
}
}
Interview problems
easy c++solution
int minSwaps(vector<int> &arr)
{
// Write your code here.
int n=arr.size();
pair<int ,int>arrpos[n];
for(int i=0;i<n;i++){
arrpos[i].first=arr[i];
arrpos[i].second=i;
}
sort(arrpos,arrpos+n);
vector<bool>vis(n,false);
int ans=0;
for(int i=0;i<n;i++){
if(vis[i]||arrpos[i].second==i){
continue;
}
int cycle_size=0;
int j=i;
while(!vis[j]){
vis[j]=1;
j=arrpos[j].second;
cycle_size++;
}
if(cycle_size>0){
ans +=(cycle_size-1);
}
}
return ans;
}
Interview problems
Best Optimized solution ever with explanation
In this solution, we perform selection sort while keeping track of the number of swaps performed. The selection sort algorithm selects the minimum element from the unsorted portion of the array and places it at the correct position in the sorted portion. By counting the number of swaps, we can determine the minimum number of swaps required to sort the array
public class Solution {
public static int minSwaps(int[] arr) {
// Write your code here.
int n = arr.length;
int swaps = 0;
for (int i = 0; i < n; i++) {
// Find the index of the minimum element from i to n-1
int minIndex = i;
for (int j = i + 1; j < n; j++) {
if (arr[j] < arr[minIndex]) {
minIndex = j;
}
}
// Swap the minimum element with the current element at index i
if (minIndex != i) {
int temp = arr[i];
arr[i] = arr[minIndex];
arr[minIndex] = temp;
swaps++;
}
}
return swaps;
}
}
