Given a permutation {P1, P2, P3, ….. PN) of first N natural numbers. The task is to check if it is possible to make the permutation increasing by swapping any two numbers. If it is already in increasing order, do nothing.
Examples:
Input: a[] = {5, 2, 3, 4, 1}
Output: Yes
Swap 1 and 5
Input: a[] = {1, 2, 3, 4, 5}
Output: Yes
Already in increasing order
Input: a[] = {5, 2, 1, 4, 3}
Output: No
Approach: Let K be the number of positions i at which P1 ? i (1 based indexing). If K = 0, the answer is Yes as the permutation can be left as is. If K = 2, the answer is also Yes: swap the two misplaced elements. (Notice K = 1 is never possible as if any element is put in the wrong position, the element that was meant to be in that position must also be misplaced.). If K > 2 then the answer is No: a single swap can only affect two elements and can thus only correct at most two misplacements.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that returns true if it is// possible to make the permutation// increasing by swapping any two numbersbool isPossible(int a[], int n){ // To count misplaced elements int k = 0; // Count all misplaced elements for (int i = 0; i < n; i++) { if (a[i] != i + 1) k++; } // If possible if (k <= 2) return true; return false;}// Driver codeint main(){ int a[] = { 5, 2, 3, 4, 1 }; int n = sizeof(a) / sizeof(a[0]); if (isPossible(a, n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation of the approachclass GFG{ // Function that returns true if it is// possible to make the permutation// increasing by swapping any two numbersstatic boolean isPossible(int a[], int n){ // To count misplaced elements int k = 0; // Count all misplaced elements for (int i = 0; i < n; i++) { if (a[i] != i + 1) k++; } // If possible if (k <= 2) return true; return false;}// Driver codepublic static void main(String[] args){ int a[] = { 5, 2, 3, 4, 1 }; int n = a.length; if (isPossible(a, n)) System.out.println("Yes"); else System.out.println("No");}}// This code is contributed by Code_Mech |
Python3
# Python3 implementation of the approach # Function that returns true if it is # possible to make the permutation # increasing by swapping any two numbers def isPossible(a, n) : # To count misplaced elements k = 0; # Count all misplaced elements for i in range(n) : if (a[i] != i + 1) : k += 1; # If possible if (k <= 2) : return True; return False; # Driver code if __name__ == "__main__" : a = [5, 2, 3, 4, 1 ]; n = len(a); if (isPossible(a, n)) : print("Yes"); else : print("No"); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System; class GFG{ // Function that returns true if it is// possible to make the permutation// increasing by swapping any two numbersstatic Boolean isPossible(int []a, int n){ // To count misplaced elements int k = 0; // Count all misplaced elements for (int i = 0; i < n; i++) { if (a[i] != i + 1) k++; } // If possible if (k <= 2) return true; return false;}// Driver codepublic static void Main(String[] args){ int []a = { 5, 2, 3, 4, 1 }; int n = a.Length; if (isPossible(a, n)) Console.WriteLine("Yes"); else Console.WriteLine("No");}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the approach// Function that returns true if it is// possible to make the permutation// increasing by swapping any two numbersfunction isPossible(a, n){ // To count misplaced elements let k = 0; // Count all misplaced elements for (let i = 0; i < n; i++) { if (a[i] != i + 1) k++; } // If possible if (k <= 2) return true; return false;}// Driver code let a = [ 5, 2, 3, 4, 1 ]; let n = a.length; if (isPossible(a, n)) document.write("Yes"); else document.write("No");// This code is contributed by Manoj</script> |
Yes
Time Complexity: O(n)
Auxiliary Space: O(1)
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