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Minimizing Swaps for Adjacent pair Sum Difference in Permutation

Given a permutation P of length n. The task is to find the minimum number of swaps required in the permutation such that the difference between the maximum value of (P[i] + P[i+1]) and the minimum value of (P[i] + P[i+1]), where 0 <= i <= n-2, is minimized.

Note: In other words, we want to rearrange the elements in the permutation to minimize the gap between the largest and smallest adjacent element sums.

Examples:

Input: n = 2, P = {1, 2}
Output: 0
Explanation: In the above permutations the gap between the largest and smallest adjacent element sums is already 0 and we cannot get lower value than 0.

Input: n = 4, P = {1, 4, 3, 2}
Output: 1
Explanation: We can swap 1st and 2nd element to get { 4, 1, 3, 2 } where the largest sum of two adjacent element is 5 and smallest sum is 4 and the difference between them is 1. We cannot get a difference lower than 1.

Approach: This can be solved with the following idea:

Doing swapping between P and a vector created by seeing indexes even or odd. Check if values are equal or not, perform swaps and increment in number of swaps done.

Below are the steps involved:

  • If size of array is 2, return 0.
  • Create a 2 vecrors p1 and p2.
  • Iterate over the array:
    • If index is odd:
      • p1.push_back(n + 1 – p1.back())
      • p2.push_back(n + 1 – p2.back())
    • If index is even:
      • p1.push_back(n – p1.back())
      • p2.push_back(n + 2 – p2.back())
  • Check swapping function between p and p1, store the minimum swaps in ans.
  • Reverse p1, call swap function, update the ans.
  • Similar to do for p2 and p.
  • Return ans.

Below is the implementation of the code:

C++

// C++ Implementation
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
 
// Function to swap
int swaps(vector<int>& b, vector<int> a)
{
    int n = a.size();
    vector<int> pos(n + 1);
 
    // Iterate over the array
    for (int i = 0; i < n; ++i) {
 
        pos[a[i]] = i;
    }
 
    int ans = 0;
    for (int i = 0; i < n; ++i) {
 
        // If values are different
        if (a[i] != b[i]) {
 
            int j = pos[b[i]];
            swap(a[i], a[j]);
            swap(pos[a[i]], pos[a[j]]);
            ++ans;
        }
    }
 
    // Return swaps
    return ans;
}
 
// Function to count minimum number of swaps
// required to minimize the difference
int minimumSwaps(int n, vector<int>& P)
{
 
    // If array size is 2, return 0
    if (n == 2) {
        return 0;
    }
    vector<int> p1;
    vector<int> p2;
    p1.push_back(n);
    p2.push_back(1);
 
    // Iterate over array
    for (int i = 1; i < n; ++i) {
 
        // If index is odd
        if (i % 2 == 1) {
            p1.push_back(n + 1 - p1.back());
            p2.push_back(n + 1 - p2.back());
        }
 
        // If index is even
        else {
            p1.push_back(n - p1.back());
            p2.push_back(n + 2 - p2.back());
        }
    }
 
    // Perform swaps
    int ans = swaps(p1, P);
 
    reverse(p1.begin(), p1.end());
 
    // Update the minimum swaps in ans
    ans = min(ans, swaps(p1, P));
 
    ans = min(ans, swaps(p2, P));
    reverse(p2.begin(), p2.end());
    ans = min(ans, swaps(p2, P));
 
    // Return ans
    return ans;
}
 
// Driver code
int main()
{
 
    int n = 2;
    vector<int> P = { 1, 2 };
 
    // Function call
    cout << minimumSwaps(n, P);
 
    return 0;
}

Java

// Java Implementation
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
 
public class MinimumSwaps {
     
    public static int minimumSwaps(int n, List<Integer> P) {
        // If array size is 2, return 0
        if (n == 2) {
            return 0;
        }
         
        List<Integer> p1 = new ArrayList<>();
        List<Integer> p2 = new ArrayList<>();
        p1.add(n);
        p2.add(1);
         
        // Iterate over array
        for (int i = 1; i < n; ++i) {
            // If index is odd
            if (i % 2 == 1) {
                p1.add(n + 1 - p1.get(p1.size() - 1));
                p2.add(n + 1 - p2.get(p2.size() - 1));
            }
            // If index is even
            else {
                p1.add(n - p1.get(p1.size() - 1));
                p2.add(n + 2 - p2.get(p2.size() - 1));
            }
        }
         
        // Perform swaps
        int ans = swaps(p1, P);
         
        Collections.reverse(p1);
         
        // Update the minimum swaps in ans
        ans = Math.min(ans, swaps(p1, P));
         
        ans = Math.min(ans, swaps(p2, P));
        Collections.reverse(p2);
        ans = Math.min(ans, swaps(p2, P));
         
        // Return ans
        return ans;
    }
     
    public static int swaps(List<Integer> b, List<Integer> a) {
        int n = a.size();
        List<Integer> pos = new ArrayList<>(Collections.nCopies(n + 1, 0));
         
        // Iterate over the array
        for (int i = 0; i < n; ++i) {
            pos.set(a.get(i), i);
        }
         
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            // If values are different
            if (!a.get(i).equals(b.get(i))) {
                int j = pos.get(b.get(i));
                Collections.swap(a, i, j);
                Collections.swap(pos, a.get(i), a.get(j));
                ++ans;
            }
        }
         
        // Return swaps
        return ans;
    }
     
    public static void main(String[] args) {
        int n = 2;
        List<Integer> P = new ArrayList<>();
        P.add(1);
        P.add(2);
         
        // Function call
        System.out.println(minimumSwaps(n, P));
    }
}
 
// This code is contributed by Tapesh(tapeshdua420)

Python3

def swaps(b, a):
    n = len(a)
    pos = [0] * (n + 1)
 
    # Iterate over the array
    for i in range(n):
        pos[a[i]] = i
 
    ans = 0
    for i in range(n):
        # If values are different
        if a[i] != b[i]:
            j = pos[b[i]]
            a[i], a[j] = a[j], a[i]
            pos[a[i]], pos[a[j]] = pos[a[j]], pos[a[i]]
            ans += 1
 
    # Return swaps
    return ans
 
def minimum_swaps(n, P):
    # If array size is 2, return 0
    if n == 2:
        return 0
    p1 = [n]
    p2 = [1]
 
    # Iterate over array
    for i in range(1, n):
        # If index is odd
        if i % 2 == 1:
            p1.append(n + 1 - p1[-1])
            p2.append(n + 1 - p2[-1])
        # If index is even
        else:
            p1.append(n - p1[-1])
            p2.append(n + 2 - p2[-1])
 
    # Perform swaps
    ans = swaps(p1, P)
 
    p1.reverse()
    # Update the minimum swaps in ans
    ans = min(ans, swaps(p1, P))
 
    ans = min(ans, swaps(p2, P))
    p2.reverse()
    ans = min(ans, swaps(p2, P))
 
    # Return ans
    return ans
 
# Driver code
if __name__ == "__main__":
    n = 2
    P = [1, 2]
 
    # Function call
    print(minimum_swaps(n, P))

C#

using System;
using System.Collections.Generic;
 
public class GFG {
    public static int MinimumSwaps(int n, List<int> P)
    {
        // If array size is 2, return 0
        if (n == 2) {
            return 0;
        }
 
        List<int> p1 = new List<int>();
        List<int> p2 = new List<int>();
        p1.Add(n);
        p2.Add(1);
 
        // Iterate over array
        for (int i = 1; i < n; ++i) {
            // If index is odd
            if (i % 2 == 1) {
                p1.Add(n + 1 - p1[p1.Count - 1]);
                p2.Add(n + 1 - p2[p2.Count - 1]);
            }
            // If index is even
            else {
                p1.Add(n - p1[p1.Count - 1]);
                p2.Add(n + 2 - p2[p2.Count - 1]);
            }
        }
 
        // Perform swaps
        int ans = Swaps(p1, P);
 
        p1.Reverse();
 
        // Update the minimum swaps in ans
        ans = Math.Min(ans, Swaps(p1, P));
 
        ans = Math.Min(ans, Swaps(p2, P));
        p2.Reverse();
        ans = Math.Min(ans, Swaps(p2, P));
 
        // Return ans
        return ans;
    }
 
    public static int Swaps(List<int> b, List<int> a)
    {
        int n = a.Count;
        List<int> pos = new List<int>(new int[n + 1]);
 
        // Iterate over the array
        for (int i = 0; i < n; ++i) {
            pos[a[i]] = i;
        }
 
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            // If values are different
            if (a[i] != b[i]) {
                int j = pos[b[i]];
                Swap(a, i, j);
                Swap(pos, a[i], a[j]);
                ++ans;
            }
        }
 
        // Return swaps
        return ans;
    }
 
    private static void Swap(List<int> list, int i, int j)
    {
        int temp = list[i];
        list[i] = list[j];
        list[j] = temp;
    }
 
    public static void Main(string[] args)
    {
        int n = 2;
        List<int> P = new List<int>{ 1, 2 };
 
        // Function call
        Console.WriteLine(MinimumSwaps(n, P));
    }
}

Javascript

// JS Implementation
 
// Function to calculate swaps between two arrays
function swaps(b, a) {
    const n = a.length;
    const pos = new Array(n + 1).fill(0);
 
    // Iterate over the array
    for (let i = 0; i < n; i++) {
        pos[a[i]] = i;
    }
 
    let ans = 0;
    for (let i = 0; i < n; i++) {
        // If values are different
        if (a[i] !== b[i]) {
            const j = pos[b[i]];
            [a[i], a[j]] = [a[j], a[i]];
            [pos[a[i]], pos[a[j]]] = [pos[a[j]], pos[a[i]]];
            ans += 1;
        }
    }
 
    // Return swaps
    return ans;
}
 
// Function to calculate minimum swaps for a given array size and permutation
function minimum_swaps(n, P) {
    // If array size is 2, return 0
    if (n === 2) {
        return 0;
    }
 
    const p1 = [n];
    const p2 = [1];
 
    // Iterate over array
    for (let i = 1; i < n; i++) {
        // If index is odd
        if (i % 2 === 1) {
            p1.push(n + 1 - p1[p1.length - 1]);
            p2.push(n + 1 - p2[p2.length - 1]);
        }
        // If index is even
        else {
            p1.push(n - p1[p1.length - 1]);
            p2.push(n + 2 - p2[p2.length - 1]);
        }
    }
 
    // Perform swaps
    let ans = swaps(p1, P);
 
    p1.reverse();
    // Update the minimum swaps in ans
    ans = Math.min(ans, swaps(p1, P));
 
    ans = Math.min(ans, swaps(p2, P));
    p2.reverse();
    ans = Math.min(ans, swaps(p2, P));
 
    // Return ans
    return ans;
}
 
// Driver code
const n = 2;
const P = [1, 2];
 
// Function call
console.log(minimum_swaps(n, P));
 
 
// This code is contributed by Tapesh(tapeshdua420)

Output

0

Time Complexity: O(N)
Auxiliary Space: O(N)


 


Reffered: https://www.geeksforgeeks.org

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