Given an array of n positive integers such that each element of an integer is from 1 to n. Find the lexicographically permutation that can be obtained by replacing minimum number of elements in array such that every element of array occurs exactly once in the entire array. First, print the minimum number of replacements required and then print the final lexicographical array.
Examples:
Input arr[] = {2, 3, 4, 3, 2}
Output 2
1 3 4 5 2
Explanation
Replace number '2' at position 1st with number
'1' and '3' at position 4th with number '5'.
The array that we obtain is [1, 3, 4, 5, 2]
which is lexicographically smallest among all
the possible suitable.
Input arr[] = {2, 1, 2, 1, 2}
Output 3
2 1 3 4 5
Naive approach is to generate all the permutation from 1 to n and pick the smallest one which renders the minimum replacements. Time complexity of this approach is O(n!) which will definitely time out for a large value of n.
Efficient approach is to pick desired elements greedily. Firstly initialize the cnt[] array which will contain the frequency of elements occurring in the array. For each element of array(ai), occurred more than once in an array, add the numbers in ascending order because of getting lexicographically minimal permutation. For instance,
Iterate the array over all elements. Let the current number of array is ai. If count of ai is equaled to 1 then move to the next number of array. If count of ai is greater than 1 then replace the number ai with element ele(the smallest number which does not occur in array) only if ele < ai. Meanwhile, decrease the count of ai in cnt[] array.
If ele > ai then mark the number ai so that we can replace it in the next iteration. This step this need because we need to make smallest lexicographically permutation.
For example, let's suppose the arr[] = {1, 5, 4, 5, 3, 7, 3}
In first iteration '5' occurs two times in array(indexing 1),
therefore we have to replace '5' at position '2' with '2'(2 < 5).
Now the updated array = {1, 2, 4, 5, 3, 7, 3}
In next iteration, '3' would be consider as it occurs two times
in array. But this time the next element of replacement would
be equals to 6 which is greater than 3. Therefore visit element
3 in boolean array vis[] and iterate over other elements.
Now again '3' occurred at position 7th, this time replace it with
number '6'.
Final array is arr[] = {1, 2, 4, 5, 3, 7, 6}
Implementation:
C++
// C++ program to print lexicographically // permutation array by replacing minimum // element of array #include <bits/stdc++.h> using namespace std; // Function to calculate lexicographically permutation // in array void lexicoSmallestPermuatation(int arr[], int n) { // Calculate frequency of array elements int cnt[n + 1]; memset(cnt, 0, sizeof(cnt)); for (int i = 0; i < n; ++i) ++cnt[arr[i]]; int ele = 1, replacement = 0; bool vis[n + 1]; memset(vis, 0, sizeof(vis)); for (int i = 0; i < n; ++i) { // If count of element is 1, no // need to replace if (cnt[arr[i]] == 1) continue; // Find the element that has not // occurred in array while (cnt[ele]) ++ele; // If replacement element is greater // than current arr[i] then visit // that element for next iteration if (ele > arr[i] && !vis[arr[i]]) vis[arr[i]] = 1; else { // Decrement count and assign the element // to array --cnt[arr[i]]; arr[i] = ele; // Increment the replacement count ++replacement; // Increment element after assigning // to the array ++ele; } } cout << replacement << "\n"; for (int i = 0; i < n; ++i) cout << arr[i] << " "; } // Driver code int main() { int arr[] = { 2, 3, 4, 3, 2 }; int sz = sizeof(arr) / sizeof(arr[0]); lexicoSmallestPermuatation(arr, sz); return 0; } |
Java
// Java program to print lexicographically // permutation array by replacing minimum // element of array class GFG { // Function to calculate lexicographically permutation // in array static void lexicoSmallestPermuatation(int arr[], int n) { // Calculate frequency of array elements int cnt[] = new int[n + 1]; for (int i = 0; i < n; ++i) { ++cnt[arr[i]]; } int ele = 1, replacement = 0; boolean vis[] = new boolean[n + 1]; for (int i = 0; i < n; ++i) { // If count of element is 1, no // need to replace if (cnt[arr[i]] == 1) { continue; } // Find the element that has not // occurred in array while (cnt[ele]>0) { ++ele; } // If replacement element is greater // than current arr[i] then visit // that element for next iteration if (ele > arr[i] && !vis[arr[i]]) { vis[arr[i]] = true; } else { // Decrement count and assign the element // to array --cnt[arr[i]]; arr[i] = ele; // Increment the replacement count ++replacement; // Increment element after assigning // to the array ++ele; } } System.out.print(replacement + "\n"); for (int i = 0; i < n; ++i) { System.out.print(arr[i] + " "); } } // Driver code public static void main(String[] args) { int arr[] = {2, 3, 4, 3, 2}; int sz = arr.length; lexicoSmallestPermuatation(arr, sz); } } // This code is contributed by 29AjayKumar |
Python3
# Python 3 program to print lexicographically # permutation array by replacing minimum # element of array # Function to calculate lexicographically # permutation in array def lexicoSmallestPermuatation(arr, n): # Calculate frequency of array elements cnt = [0 for i in range(n + 1)] for i in range(n): cnt[arr[i]] += 1 ele = 1 replacement = 0 vis = [0 for i in range(n + 1)] for i in range(n): # If count of element is 1, no # need to replace if (cnt[arr[i]] == 1): continue # Find the element that has not # occurred in array while (cnt[ele]): ele += 1 # If replacement element is greater # than current arr[i] then visit # that element for next iteration if (ele > arr[i] and vis[arr[i]] == 0): vis[arr[i]] = 1; else: # Decrement count and assign # the element to array cnt[arr[i]] -= 1 arr[i] = ele # Increment the replacement count replacement += 1 # Increment element after assigning # to the array ele += 1 print(replacement) for i in range(n): print(arr[i], end = " ") # Driver code if __name__ == '__main__': arr = [2, 3, 4, 3, 2] sz = len(arr) lexicoSmallestPermuatation(arr, sz) # This code is contributed by # Shashank_Sharma |
C#
// C# program to print lexicographically // permutation array by replacing minimum // element of array using System; public class GFG { // Function to calculate lexicographically permutation // in array static void lexicoSmallestPermuatation(int []arr, int n) { // Calculate frequency of array elements int []cnt= new int[n + 1]; for (int i = 0; i < n; ++i) { ++cnt[arr[i]]; } int ele = 1, replacement = 0; bool []vis = new bool[n + 1]; for (int i = 0; i < n; ++i) { // If count of element is 1, no // need to replace if (cnt[arr[i]] == 1) { continue; } // Find the element that has not // occurred in array while (cnt[ele]>0) { ++ele; } // If replacement element is greater // than current arr[i] then visit // that element for next iteration if (ele > arr[i] && !vis[arr[i]]) { vis[arr[i]] = true; } else { // Decrement count and assign the element // to array --cnt[arr[i]]; arr[i] = ele; // Increment the replacement count ++replacement; // Increment element after assigning // to the array ++ele; } } Console.Write(replacement + "\n"); for (int i = 0; i < n; ++i) { Console.Write(arr[i] + " "); } } // Driver code public static void Main() { int []arr = {2, 3, 4, 3, 2}; int sz = arr.Length; lexicoSmallestPermuatation(arr, sz); } } // This code is contributed by Rajput-Ji// |
PHP
<?php // PHP program to print lexicographically // permutation array by replacing minimum // element of array // Function to calculate lexicographically // permutation in array function lexicoSmallestPermuatation(&$arr, $n) { // Calculate frequency of array elements $cnt = array_fill(0, $n + 1, NULL); for ($i = 0; $i < $n; ++$i) ++$cnt[$arr[$i]]; $ele = 1; $replacement = 0; $vis = array_fill(0, $n + 1, NULL); for ($i = 0; $i < $n; ++$i) { // If count of element is 1, no // need to replace if ($cnt[$arr[$i]] == 1) continue; // Find the element that has not // occurred in array while ($cnt[$ele]) ++$ele; // If replacement element is greater // than current arr[i] then visit // that element for next iteration if ($ele > $arr[$i] && !$vis[$arr[$i]]) $vis[$arr[$i]] = 1; else { // Decrement count and assign the // element to array --$cnt[$arr[$i]]; $arr[$i] = $ele; // Increment the replacement count ++$replacement; // Increment element after assigning // to the array ++$ele; } } echo $replacement. "\n"; for ($i = 0; $i < $n; ++$i) echo $arr[$i] . " "; } // Driver code $arr = array(2, 3, 4, 3, 2 ); $sz = sizeof($arr); lexicoSmallestPermuatation($arr, $sz); // This code is contributed by ita_c ?> |
Javascript
<script> // Javascript program to // print lexicographically // permutation array by // replacing minimum // element of array // Function to calculate // lexicographically permutation // in array function lexicoSmallestPermuatation(arr, n) { // Calculate frequency of // array elements let cnt = Array.from({length: n + 1}, (_, i) => 0); for (let i = 0; i < n; ++i) { ++cnt[arr[i]]; } let ele = 1, replacement = 0; let vis = Array.from({length: n + 1}, (_, i) => 0); for (let i = 0; i < n; ++i) { // If count of element is 1, no // need to replace if (cnt[arr[i]] == 1) { continue; } // Find the element that has not // occurred in array while (cnt[ele]>0) { ++ele; } // If replacement element is greater // than current arr[i] then visit // that element for next iteration if (ele > arr[i] && !vis[arr[i]]) { vis[arr[i]] = true; } else { // Decrement count and // assign the element // to array --cnt[arr[i]]; arr[i] = ele; // Increment the // replacement count ++replacement; // Increment element // after assigning // to the array ++ele; } } document.write(replacement + "<br/>"); for (let i = 0; i < n; ++i) { document.write(arr[i] + " "); } } // driver program let arr = [2, 3, 4, 3, 2]; let sz = arr.length; lexicoSmallestPermuatation(arr, sz); </script> |
Output
2 1 3 4 5 2
Time complexity: O(n)
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