Given a permutation of first **n** natural numbers as array and an integer k. Print the lexicographically largest permutation after at most k swaps

Input: arr[] = {4, 5, 2, 1, 3} k = 3 Output: 5 4 3 2 1 Explanation: Swap 1^{st}and 2^{nd}elements: 5 4 2 1 3 Swap 3^{rd}and 5^{th}elements: 5 4 3 1 2 Swap 4^{th}and 5^{th}elements: 5 4 3 2 1 Input: arr[] = {2, 1, 3} k = 1 Output: 3 1 2

A **Naive approach** is to one by one generate permutation in lexicographically decreasing order. Compare every generated permutation with original array and count the number of swaps required to convert. If count is less than or equal to k, print this permutation. Problem of this approach is that it would be difficult to implement and will definitely time out for large value of N.

An **Efficient** approach is to think greedily. If we visualize the problem then we will get to know that largest permutation can only be obtained if it starts with n and continues with n-1, n-2,…. So we just need to put the 1^{st}, 2^{nd}, 3^{rd}, …, k^{th} largest element to their respective position.

## C++

// Below is C++ code to print largest permutation // after atmost K swaps #include<bits/stdc++.h> using namespace std; // Function to calculate largest permutation after // atmost K swaps void KswapPermutation(int arr[], int n, int k) { // Auxiliary dictionary of storing the position // of elements int pos[n+1]; for (int i = 0; i < n; ++i) pos[arr[i]] = i; for (int i=0; i<n && k; ++i) { // If element is already i'th largest, // then no need to swap if (arr[i] == n-i) continue; // Find position of i'th largest value, n-i int temp = pos[n-i]; // Swap the elements position pos[arr[i]] = pos[n-i]; pos[n-i] = i; // Swap the ith largest value with the // current value at ith place swap(arr[temp], arr[i]); // decrement number of swaps --k; } } // Driver code int main() { int arr[] = {4, 5, 2, 1, 3}; int n = sizeof(arr)/sizeof(arr[0]); int k = 3; KswapPermutation(arr, n, k); cout << "Largest permutation after " << k << " swaps:\n"; for (int i=0; i<n; ++i) printf("%d ", arr[i]); return 0; }

## Python

# Python code to print largest permutation after K swaps def KswapPermutation(arr, n, k): # Auxiliary array of storing the position of elements pos = {} for i in range(n): pos[arr[i]] = i for i in range(n): # If K is exhausted then break the loop if k == 0: break # If element is already largest then no need to swap if (arr[i] == n-i): continue # Find position of i'th largest value, n-i temp = pos[n-i] # Swap the elements position pos[arr[i]] = pos[n-i] pos[n-i] = i # Swap the ith largest value with the value at # ith place arr[temp], arr[i] = arr[i], arr[temp] # Decrement K after swap k = k-1 # Driver code arr = [4, 5, 2, 1, 3] n = len(arr) k = 3 KswapPermutation(arr, N, K) # Print the answer print "Largest permutation after", K, "swaps: " print " ".join(map(str,arr))

Output:Largest permutation after 3 swaps: 5 4 3 2 1

**Time complexity: **O(n)

**Auxiliary space: **O(n)

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