Maximize count of corresponding same elements in given permutations using cyclic rotations

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Given two permutations P1 and P2 of numbers from 1 to N, the task is to find the maximum count of corresponding same elements in the given permutations by performing a cyclic left or right shift on P1.
Examples:

Input: P1 = [5 4 3 2 1], P2 = [1 2 3 4 5]
Output: 1
Explanation:
We have a matching pair at index 2 for element 3.
Input: P1 = [1 3 5 2 4 6], P2 = [1 5 2 4 3 6]
Output: 3
Explanation:
Cyclic shift of second permutation towards right would give 6 1 5 2 4 3, and we get a match of 5, 2, 4. Hence, the answer is 3 matching pairs.

Naive Approach: The naive approach is to check for every possible shift in both the left and right direction count the number of matching pairs by looping through all the permutations formed.
Time Complexity: O(N²)
Auxiliary Space: O(1)
Efficient Approach: The above naive approach can be optimized. The idea is for every element to store the smaller distance between positions of this element from the left and right sides in separate arrays. Hence, the solution to the problem will be calculated as the maximum frequency of an element from the two separated arrays. Below are the steps:

Store the position of all the elements of the permutation P2 in an array(say store[]).
For each element in the permutation P1, do the following:
- Find the difference(say diff) between the position of the current element in P2 with the position in P1.
- If diff is less than 0 then update diff to (N – diff).
- Store the frequency of current difference diff in a map.
After the above steps, the maximum frequency stored in the map is the maximum number of equal elements after rotation on P1.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to maximize the matching
// pairs between two permutation
// using left and right rotation
int maximumMatchingPairs(int perm1[],
                         int perm2[],
                         int n)
{
    // Left array store distance of element
    // from left side and right array store
    // distance of element from right side
    int left[n], right[n];
 
    // Map to store index of elements
    map<int, int> mp1, mp2;
    for (int i = 0; i < n; i++) {
        mp1[perm1[i]] = i;
    }
    for (int j = 0; j < n; j++) {
        mp2[perm2[j]] = j;
    }
 
    for (int i = 0; i < n; i++) {
 
        // idx1 is index of element
        // in first permutation
 
        // idx2 is index of element
        // in second permutation
        int idx2 = mp2[perm1[i]];
        int idx1 = i;
 
        if (idx1 == idx2) {
 
            // If element if present on same
            // index on both permutations then
            // distance is zero
            left[i] = 0;
            right[i] = 0;
        }
        else if (idx1 < idx2) {
 
            // Calculate distance from left
            // and right side
            left[i] = (n - (idx2 - idx1));
            right[i] = (idx2 - idx1);
        }
        else {
 
            // Calculate distance from left
            // and right side
            left[i] = (idx1 - idx2);
            right[i] = (n - (idx1 - idx2));
        }
    }
 
    // Maps to store frequencies of elements
    // present in left and right arrays
    map<int, int> freq1, freq2;
    for (int i = 0; i < n; i++) {
        freq1[left[i]]++;
        freq2[right[i]]++;
    }
 
    int ans = 0;
 
    for (int i = 0; i < n; i++) {
 
        // Find maximum frequency
        ans = max(ans, max(freq1[left[i]],
                           freq2[right[i]]));
    }
 
    // Return the result
    return ans;
}
 
// Driver Code
int main()
{
    // Given permutations P1 and P2
    int P1[] = { 5, 4, 3, 2, 1 };
    int P2[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(P1) / sizeof(P1[0]);
 
    // Function Call
    cout << maximumMatchingPairs(P1, P2, n);
    return 0;
}

Java

// Java program for 
// the above approach
import java.util.*;
class GFG{
 
// Function to maximize the matching
// pairs between two permutation
// using left and right rotation
static int maximumMatchingPairs(int perm1[],
                                int perm2[],
                                int n)
{
  // Left array store distance of element
  // from left side and right array store
  // distance of element from right side
  int []left = new int[n];
  int []right = new int[n];
 
  // Map to store index of elements
  HashMap<Integer,
          Integer> mp1 =  new HashMap<>();
  HashMap<Integer,
          Integer> mp2 =  new HashMap<>();
   
  for (int i = 0; i < n; i++) 
  {
    mp1.put(perm1[i], i);
  }
  for (int j = 0; j < n; j++) 
  {
    mp2.put(perm2[j], j);
  }
 
  for (int i = 0; i < n; i++) 
  {
    // idx1 is index of element
    // in first permutation
    // idx2 is index of element
    // in second permutation
    int idx2 = mp2.get(perm1[i]);
    int idx1 = i;
 
    if (idx1 == idx2) 
    {
      // If element if present on same
      // index on both permutations then
      // distance is zero
      left[i] = 0;
      right[i] = 0;
    }
    else if (idx1 < idx2) 
    {
      // Calculate distance from left
      // and right side
      left[i] = (n - (idx2 - idx1));
      right[i] = (idx2 - idx1);
    }
    else
    {
      // Calculate distance from left
      // and right side
      left[i] = (idx1 - idx2);
      right[i] = (n - (idx1 - idx2));
    }
  }
 
  // Maps to store frequencies of elements
  // present in left and right arrays
  HashMap<Integer,
          Integer> freq1 = new HashMap<>();
  HashMap<Integer,
          Integer> freq2 = new HashMap<>();
   
  for (int i = 0; i < n; i++) 
  {
    if(freq1.containsKey(left[i]))
      freq1.put(left[i], 
      freq1.get(left[i]) + 1);
    else
      freq1.put(left[i], 1);
    if(freq2.containsKey(right[i]))
      freq2.put(right[i], 
      freq2.get(right[i]) + 1);
    else
      freq2.put(right[i], 1);
  }
 
  int ans = 0;
 
  for (int i = 0; i < n; i++) 
  {
    // Find maximum frequency
    ans = Math.max(ans, 
          Math.max(freq1.get(left[i]),
                   freq2.get(right[i])));
  }
 
  // Return the result
  return ans;
}
 
// Driver Code
public static void main(String[] args)
{
  // Given permutations P1 and P2
  int P1[] = {5, 4, 3, 2, 1};
  int P2[] = {1, 2, 3, 4, 5};
  int n = P1.length;
 
  // Function Call
  System.out.print(maximumMatchingPairs(P1, P2, n));
}
}
 
// This code is contributed by gauravrajput1

Python3

# Python3 program for the above approach
from collections import defaultdict
 
# Function to maximize the matching
# pairs between two permutation
# using left and right rotation
def maximumMatchingPairs(perm1, perm2, n):
 
    # Left array store distance of element
    # from left side and right array store
    # distance of element from right side
    left = [0] * n
    right = [0] * n
 
    # Map to store index of elements
    mp1 = {}
    mp2 = {}
    for i in range (n):
        mp1[perm1[i]] = i
     
    for j in range (n):
        mp2[perm2[j]] = j
     
    for i in range (n):
 
        # idx1 is index of element
        # in first permutation
 
        # idx2 is index of element
        # in second permutation
        idx2 = mp2[perm1[i]]
        idx1 = i
 
        if (idx1 == idx2):
 
            # If element if present on same
            # index on both permutations then
            # distance is zero
            left[i] = 0
            right[i] = 0
         
        elif (idx1 < idx2):
 
            # Calculate distance from left
            # and right side
            left[i] = (n - (idx2 - idx1))
            right[i] = (idx2 - idx1)
         
        else :
 
            # Calculate distance from left
            # and right side
            left[i] = (idx1 - idx2)
            right[i] = (n - (idx1 - idx2))
 
    # Maps to store frequencies of elements
    # present in left and right arrays
    freq1 = defaultdict (int)
    freq2 = defaultdict (int)
    for i in range (n):
        freq1[left[i]] += 1
        freq2[right[i]] += 1
 
    ans = 0
 
    for i in range( n):
 
        # Find maximum frequency
        ans = max(ans, max(freq1[left[i]],
                           freq2[right[i]]))
 
    # Return the result
    return ans
 
# Driver Code
if __name__ == "__main__":
     
    # Given permutations P1 and P2
    P1 = [ 5, 4, 3, 2, 1 ]
    P2 = [ 1, 2, 3, 4, 5 ]
    n = len(P1)
 
    # Function Call
    print(maximumMatchingPairs(P1, P2, n))
 
# This code is contributed by chitranayal

C#

// C# program for 
// the above approach
using System;
using System.Collections.Generic;
class GFG{
  
// Function to maximize the matching
// pairs between two permutation
// using left and right rotation
static int maximumMatchingPairs(int []perm1,
                                int []perm2,
                                int n)
{
  // Left array store distance of element
  // from left side and right array store
  // distance of element from right side
  int []left = new int[n];
  int []right = new int[n];
  
  // Map to store index of elements
  Dictionary<int,
             int> mp1=new Dictionary<int,
                                     int>();
  Dictionary<int,
             int> mp2=new Dictionary<int,
                                     int>();
    
  for (int i = 0; i < n; i++) 
  {
    mp1[perm1[i]] = i;
  }
  for (int j = 0; j < n; j++) 
  {
    mp2[perm2[j]] = j;
  }
  
  for (int i = 0; i < n; i++) 
  {
    // idx1 is index of element
    // in first permutation
    // idx2 is index of element
    // in second permutation
    int idx2 = mp2[perm1[i]];
    int idx1 = i;
  
    if (idx1 == idx2) 
    {
      // If element if present on same
      // index on both permutations then
      // distance is zero
      left[i] = 0;
      right[i] = 0;
    }
    else if (idx1 < idx2) 
    {
      // Calculate distance from left
      // and right side
      left[i] = (n - (idx2 - idx1));
      right[i] = (idx2 - idx1);
    }
    else
    {
      // Calculate distance from left
      // and right side
      left[i] = (idx1 - idx2);
      right[i] = (n - (idx1 - idx2));
    }
  }
  
  // Maps to store frequencies of elements
  // present in left and right arrays
  Dictionary<int,
             int> freq1=new Dictionary <int,
                                        int>();
  Dictionary<int,
             int> freq2=new Dictionary <int,
                                        int>();
    
  for (int i = 0; i < n; i++) 
  {
    if(freq1.ContainsKey(left[i]))
      freq1[left[i]]++;
    else
      freq1[left[i]] = 1;
       
    if(freq2.ContainsKey(right[i]))
      freq2[right[i]]++;
    else
      freq2[right[i]] = 1;
  }
  
  int ans = 0;
  
  for (int i = 0; i < n; i++) 
  {
    // Find maximum frequency
    ans = Math.Max(ans, 
          Math.Max(freq1[left[i]],
                   freq2[right[i]]));
  }
  
  // Return the result
  return ans;
}
  
// Driver Code
public static void Main(string[] args)
{
  // Given permutations P1 and P2
  int []P1 = {5, 4, 3, 2, 1};
  int []P2 = {1, 2, 3, 4, 5};
  int n = P1.Length;
  
  // Function Call
  Console.Write(maximumMatchingPairs(P1, P2, n));
}
}
 
// This code is contributed by Rutvik_56

Javascript

<script>
 
// Javascript program for the above approach
 
// Function to maximize the matching
// pairs between two permutation
// using left and right rotation
function maximumMatchingPairs(perm1, perm2, n)
{
    // Left array store distance of element
    // from left side and right array store
    // distance of element from right side
    var left = Array(n);
    var right = Array(n);
 
    // Map to store index of elements
    var mp1 = new Map(), mp2 = new Map();
    for (var i = 0; i < n; i++) {
        mp1.set(perm1[i], i);
    }
    for (var j = 0; j < n; j++) {
        mp2.set(perm2[j], j);
    }
 
    for (var i = 0; i < n; i++) {
 
        // idx1 is index of element
        // in first permutation
 
        // idx2 is index of element
        // in second permutation
        var idx2 = mp2.get(perm1[i]);
        var idx1 = i;
 
        if (idx1 == idx2) {
 
            // If element if present on same
            // index on both permutations then
            // distance is zero
            left[i] = 0;
            right[i] = 0;
        }
        else if (idx1 < idx2) {
 
            // Calculate distance from left
            // and right side
            left[i] = (n - (idx2 - idx1));
            right[i] = (idx2 - idx1);
        }
        else {
 
            // Calculate distance from left
            // and right side
            left[i] = (idx1 - idx2);
            right[i] = (n - (idx1 - idx2));
        }
    }
 
    // Maps to store frequencies of elements
    // present in left and right arrays
    var freq1 = new Map(), freq2 = new Map();
    for (var i = 0; i < n; i++) {
        if(freq1.has(left[i]))
            freq1.set(left[i], freq1.get(left[i])+1)
        else
            freq1.set(left[i], 1)
 
        if(freq2.has(right[i]))
            freq2.set(right[i], freq2.get(right[i])+1)
        else
            freq2.set(right[i], 1)
    }
 
    var ans = 0;
 
    for (var i = 0; i < n; i++) {
 
        // Find maximum frequency
        ans = Math.max(ans, Math.max(freq1.get(left[i]),
                           freq2.get(right[i])));
    }
 
    // Return the result
    return ans;
}
 
// Driver Code
// Given permutations P1 and P2
var P1 = [5, 4, 3, 2, 1];
var P2 = [1, 2, 3, 4, 5];
var n = P1.length;
// Function Call
document.write( maximumMatchingPairs(P1, P2, n));
 
</script>

Output:

Time Complexity: O(NlogN)
Auxiliary Space: O(N), since N extra space has been taken.

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