Sort an array by swapping adjacent elements from indices that contains ‘1’ in a given string

Sorting an array means ordering all the elements of the array in increasing order. Sorting an array by swapping adjacent elements means we can only swap elements that are adjacent to each other but we can swap adjacent elements any number of times. We will be given a binary string that will contain only two types of characters ‘0’ and ‘1’. If any character is ‘0’ in the given string then we cannot swap the element present at that index of the array with adjacent elements.

Sample Examples

Input 1

Given array: {1, 4, 3, 2, 5, 7, 6}

Given string: 0111011

Output: Yes

Explanation − In the given array 1 is at its position and we can swap elements from the index1 to index3. We can swap the index1 element with the next index and then swap index2 with index3 and then again index1 with index2 making 2,3,4 in the sorted form. For the last two elements, we can swap them and get the required sorted array.

Input 2

Given array: {1, 2, 4, 3, 5, 6}

Given string: 001011

Output: No

Explanation − In the given array only 3 and 4 are not at the place where they are sorted and index3 cannot be swapped which means it’s not possible to sort the array by swapping.

Naive Approach

In this approach, we will traverse over the array and find the sections which are sortable and will sort them. After sorting them will check for the complete array whether it is sorted or not. If sorted then we will print yes otherwise no.

Example

#include <bits/stdc++.h>
using namespace std;
// function to check if the given array is sorted or not 
bool checkSorted(int arr[], int n){
   // traversing over the array 
   for(int i=1; i<n; i++){
      if(arr[i] < arr[i-1]){
         // means the current element is smaller as compared to the previous 
         return false;
      }
   }
   return true;
}
// function to sort the array 
bool isSorted(int arr[], int n, string str){
   // traversing over the given array 
   for(int i=0; i<n; i++){
      if(str[i] == '0'){
         // current index is not allowed to swap
         continue;
      }
      else if(arr[i] == i+1){
         // if the current index element is at the required position 
         // move to the next element 
         continue;
      }
      else{
         int j = i+1; // variable to traverse over the string
         while(j < n && str[j] == '1'){
            j++;
         }
         // sorting the sortable elements
         sort(arr+i,arr+j);
         i = j-1; // updating the value of i
      }
   }
   return  checkSorted(arr,n);
}
int main(){
   int arr[] = {1, 4, 3, 2, 5, 7, 6}; // given array 
   int n = 7; // size of array 
   string s = "0111011";    
   // calling the function 
   if(isSorted(arr,n,s)){
      cout<<"Yes, it is possible to sort the given array by swapping the allowed indexes";
   }
   else{
      cout<<"No, it is not possible to sort the given array by swapping the allowed indexes";
   }
   return 0;
}

Output

Yes, it is possible to sort the given array by swapping the allowed indexes

Time and Space Complexity

The time complexity of the above code is O(N*log(N)), where N is the size of the array. Here log factor is due to the sorting of the sections using the sort function.

The space complexity of the above code is O(1), as we are not using any extra space here.

Efficient Approach

In this approach, we will only check if all the elements present in the sortable range are of the current range only rather than performing actual sorting. With the help of this approach, we can save time, and time complexity will become linear.

Example

#include <bits/stdc++.h>
using namespace std;
// function to find the solution  
bool isSorted(int arr[], int n, string str){
   // traversing over the given array 
   for(int i=0; i<n; i++){
      if(arr[i] == i+1){
         // if the current index element is at the required position 
         // move to the next element 
         continue;
      }
      else if(str[i] == '0'){
         // the current index element is not at the required position 
         // its not moveable also so return false 
         return false;
      }
      else{
         int j = i+1; // variable to traverse over the string
         while(j < n && str[j] == '1'){
            j++;
         }
         // we have limit now check if all the elements in the sortable 
         // limit is in the required final range             
         for(int k = i; k<j; k++){
            if(arr[k] <= i || arr[k] > j){
               return false;
            }
         }
         i = j-1; // updating the value of i
      }
   }
   return  true;
}
int main(){
   int arr[] = {1, 4, 3, 2, 5, 7, 6}; // given array 
   int n = 7; // size of array 
   string s = "0111011";    
   // calling the function 
   if(isSorted(arr,n,s)){
      cout<<"Yes, it is possible to sort the given array by swapping the allowed indexes";
   }
   else{
      cout<<"No, it is not possible to sort the given array by swapping the allowed indexes";
   }
   return 0;
}

Output

Yes, it is possible to sort the given array by swapping the allowed indexes

Time and Space Complexity

The time complexity of the above code is O(N), where N is the size of the array.

The space complexity of the above code is O(1), as we are not using any extra space here.

Conclusion

In this tutorial, we have implemented a program to check if the given current array can be sorted or not if the sorting is done by swapping the adjacent elements only. Also, a binary string is given which indicates which index element can be sorted and which is not possible to sort. We have implemented two approaches one with the O(N*log(N)) time complexity and another with linear time complexity.