Lexicographically smallest array formed by at most one swap for every pair of adjacent indices

Given an array A[] of length N, the task is to find lexicographically smallest array by swapping adjacent elements for each index atmost once. Thus, for any index: 

, at most one swap between A[K] and A[K+1] is allowed.
Example:

Input: A[] = { 3, 2, 1, 4} 
Output: 1 3 2 4 
Explanation: Perform the following swaps: 
Swap A[1] and A[2], now A[] = { 3, 1, 2, 4 } 
Swap A[0] and A[1], now A[] = { 1, 3, 2, 4 } 
No further swaps are possible as A[1] and A[2] have already been swapped.
Input: A[] = { 2, 1, 4, 3, 6, 5 } 
Output: 1 2 3 4 5 6 
Explanation: Perform the following swaps: 
Swap A[0] and A[1], now A[] = { 1, 2, 4, 3, 6, 5 } 
Swap A[2] and A[3], now A[] = { 1, 2, 3, 4, 6, 5 } 
Swap A[4] and A[5], now A[] = { 1, 2, 3, 4, 5, 6 } 

Approach:
To solve the problem mentioned above we can apply Greedy method. We know that we can perform at most N – 1 swaps to make the given array as smallest as possible. 

• Create a counter variable and initialize with N-1 and a hash-map to store performed swaps.
• Find the position of the minimum element from current index onward.
• Now perform swap backwards until we reach current element position.
• Also check if current swap is possible or not and decrement the counter also at each swap.
• Finally print the required array.

Below is the implementation of above approach:

C++

// C++ implementation to find the
// lexicographically smallest
// array by at most single swap
// for every pair of adjacent indices

#include <bits/stdc++.h>
using namespace std;

// Function to find the
// lexicographically smallest array
void findSmallestArray(int A[], int n)
{
    // maximum swaps possible
    int count = n - 1;

    // hash to store swaps performed
    map<pair<int, int>, int> mp;

    for (int i = 0; i < n && count > 0; ++i) {

        // let current element be
        // the minimum possible
        int mn = A[i], pos = i;

        // Find actual position of
        // the minimum element
        for (int j = i + 1; j < n; ++j) {

            // Update minimum element and
            // its position
            if (A[j] < mn) {
                mn = A[j];
                pos = j;
            }
        }

        // Perform swaps if possible
        while (pos > i && count > 0
               && !mp[{ pos - 1, pos }]) {

            // Insert current swap in hash
            mp[{ pos - 1, pos }] = 1;

            swap(A[pos], A[pos - 1]);
            --pos;
            --count;
        }
    }

    // print the required array
    for (int i = 0; i < n; ++i)
        cout << A[i] << " ";
}

// Driver code
int main()
{

    int A[] = { 2, 1, 4, 3, 6, 5 };
    int n = sizeof(A) / sizeof(A[0]);

    findSmallestArray(A, n);

    return 0;
}

Python3

# Python3 implementation to find the
# lexicographically smallest array by
# at most single swap for every pair
# of adjacent indices

# Function to find the
# lexicographically smallest array
def findSmallestArray(A, n):

    # Maximum swaps possible
    count = n - 1

    # Hash to store swaps performed
    mp = {''}

    for i in range(0, n):
        if(count <= 0):
            break;

        # Let current element be
        # the minimum possible
        mn = A[i]
        pos = i

        # Find actual position of
        # the minimum element
        for j in range(i + 1, n):

            # Update minimum element
            # and its position
            if (A[j] < mn):
                mn = A[j]
                pos = j

        # Perform swaps if possible
        while (pos > i and count > 0 and
             ((pos - 1, pos) not in mp)):

            # Insert current swap in hash
            mp.add((pos - 1, pos))

            A[pos], A[pos - 1] = A[pos - 1], A[pos]
            pos -= 1
            count -= 1

    # Print the required array
    for i in range(0, n):
        print(A[i], end = " ")

# Driver code
A = [ 2, 1, 4, 3, 6, 5 ]
n = len(A)

findSmallestArray(A, n)

# This code is contributed by Sanjit_Prasad

Output:

1 2 3 4 5 6

Time Complexity: O(N²)