Given an array arr[] of length N and another array P[] containing {a1, a2, … ak} which represents the positions of the given array arr[], the task is to check if the array can be sorted by only swapping the elements arr[ai], arr[ai+1] where ‘i’ is some element in the array P[].
Examples:
Input: arr[] = {3, 2, 1}, P[] = {1, 2}
Output: Yes
Explanation:
Initially, i = 1 (i.e.) first element and second element are swapped. Therefore, arr[0] <=> arr[1]. arr[] = {2, 3, 1}.
Similarly, i = 2 (i.e.) second element and third element are swapped. arr[] = {2, 1, 3}.
Finally, i = 1 (i.e.) first element and second element are swapped. arr[] = {1, 2, 3}.
Since this array is sorted, therefore, the given array can be sorted.Input: arr[] = {5, 3, -4, 1, 12}, P[] = {2, 4, 3}
Output: No
Approach: The idea is to use two pointer approach to check if the array can be sorted or not.
- Initially, we create a position array pos[] of size N. This array will be used to mark the given positions in the array P[]. That is:
if j = ai (1 ≤ i ≤ K) then the element pos[ai-1] will be 1 else 0
- Now, iterate through the array and check if pos[i] = 1 or not.
- If we encounter the pos[i]=1, we store the iterator in a temporary variable, and then we increment the iterator with value 1, till we have pos[i]=1 continuously, i.e.,
j = i while (j < N and pos[j]) j=j+1 - After this increment, we sort this segment that we obtained from i to j+1 and finally, check after the position j, in the vector that we have to check, because we have sorted till this segment.
Sort(arr[i] to arr[j+1]) i=j
- Finally, after this loop completion, we have to check if the array has been sorted or not.
Below is the implementation of the above approach:
C++
// C++ program to check if the array // can be sorted only if the elements // on the given positions can be swapped #include <bits/stdc++.h> using namespace std; // Function to check if the array can // be sorted only if the elements on // the given positions can be swapped void check_vector(vector<int> A, int n, vector<int> p) { // Creating an array for marking // the positions vector<int> pos(A.size()); // Iterating through the array and // mark the positions for (int i = 0; i < p.size(); i++) { pos[p[i] - 1] = 1; } int flag = 1; // Iterating through the given array for (int i = 0; i < n; i++) { if (pos[i] == 0) continue; int j = i; // If pos[i] is 1, then incrementing // till 1 is continuously present in pos while (j < n && pos[j]) ++j; // Sorting the required segment sort(A.begin() + i, A.begin() + j + 1); i = j; } // Checking if the vector is sorted or not for (int i = 0; i < n - 1; i++) { if (A[i] > A[i + 1]) { flag = 0; break; } } // Print yes if it is sorted if (flag == 1) cout << "Yes"; else cout << "No"; } // Driver code int main() { vector<int> A{ 3, 2, 1 }; vector<int> p{ 1, 2 }; check_vector(A, A.size(), p); return 0; } |
Python3
# Python3 program to check if the array # can be sorted only if the elements # on the given positions can be swapped # Function to check if the array can # be sorted only if the elements on # the given positions can be swapped def check_vector(A, n, p): # Creating an array for marking # the positions pos = [0 for i in range(len(A))] # Iterating through the array and # mark the positions for i in range(len(p)): pos[p[i] - 1] = 1 flag = 1 # Iterating through the given array for i in range(n): if (pos[i] == 0): continue j = i # If pos[i] is 1, then incrementing # till 1 is continuously present in pos while (j < n and pos[j]): j += 1 # Sorting the required segment p = A[: i] q = A[i : i + j + 1] r = A[i + j + 1 : len(A)] q.sort(reverse = False) A = p + q + r i = j # Checking if the vector is sorted or not for i in range(n - 1): if (A[i] > A[i + 1]): flag = 0 break # Print yes if it is sorted if (flag == 1): print("Yes") else: print("No"); # Driver code if __name__ == '__main__': A = [ 3, 2, 1 ] p = [ 1, 2 ] check_vector(A,len(A), p) # This code is contributed by Samarth |
Yes
Time Complexity: O(N * log(N)), where N is the size of the array.
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