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Check if a grid can become row-wise and column-wise sorted after adjacent swaps

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Given a grid of size n x len filled with lowercase characters. We can swap two adjacent characters in the same row and column. Now we have to check whether it is possible to arrange in such a order that every row and every column in the grid is lexicographically sorted.

Examples: 

Input : abcde
        fghij
        olmkn
        trpqs
        xywuv
Output : Yes
Explanation :
The grid can be rearranged as
abcde
fghij
klmno
pqrst
uvwxy

The idea to do the above problem is really simple we can simply sort the characters in the same row and then just 
check column vise if the new grid is sorted column vise or not. Please not that sorting is possible with adjacent swaps (Bubble sort for example does only adjacent swaps)

  • Define a function named “check” that takes a vector of strings v and an integer len as input and returns a boolean value.
  • Inside the “check” function:
    • Get the size of the vector v and iterate over every string in it.
    • Sort every string in v in ascending order.
    • Iterate from 0 to len-2 and for every iteration
      • Iterate over all the strings in v.
        • If the character at position j in the i-th string is greater than the character at the same position in the i+1-th string, return false.
    • If the iterations are completed without returning false, return true

Below is the implementation of the above approach.

C++

// C++ program to check if we can make a
// grid of character sorted using adjacent
// swaps.
#include <bits/stdc++.h>
using namespace std;
 
// v[] is vector of strings. len is length
// of strings in every row.
bool check(vector<string> v, int len)
{
    int n = v.size();
    for (int i = 0; i < n; i++)
        sort(v[i].begin(), v[i].end());
     
    for (int i = 0; i < len-1; i++)
        for (int j = 0; j < n; j++)
            if (v[i][j] > v[i+1][j])
                return false;
    return true;
}
 
// Driver code
int main()
{
    vector<string> v = { "ebcda", "ihgfj", "klmno",
                               "pqrst", "yvwxu" };
    int len = 5; // Length of strings
    check(v, len)? cout << "Yes" : cout << "No";
    return 0;
}

                    

Java

// Java program to check if we can make a grid
// of character sorted using adjacent swaps.
import java.util.Arrays;
 
class GfG
{
 
    // v[] is vector of strings. len is
    // length of strings in every row.
    static boolean check(String[] v, int len)
    {
        int n = v.length;
        char[] tempArray;
         
        for (int i = 0; i < n; i++)
        {
            tempArray = v[i].toCharArray();
            Arrays.sort(tempArray);
            v[i] = new String(tempArray);
        }
         
        for (int i = 0; i < len-1; i++)
            for (int j = 0; j < n; j++)
                if (v[i].charAt(j) > v[i+1].charAt(j))
                    return false;
        return true;
    }
 
    // Driver code
    public static void main(String []args)
    {
         
        String[] v = { "ebcda", "ihgfj", "klmno",
                            "pqrst", "yvwxu" };
         
        int len = 5; // Length of strings
        if (check(v, len))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// This code is contributed by Rituraj Jain

                    

Python

# Python program to check if we can make a
# grid of character sorted using adjacent
# swaps.
 
# v[] is vector of strings. len is length
# of strings in every row.
def check(v, l):
    n = len(v)
    for i in v:
        i = ''.join(sorted(i))
      
    for i in range(l - 1):
        for j in range(n):
            if (v[i][j] > v[i + 1][j]):
                return False
    return True
  
# Driver code
v = [ "ebcda", "ihgfj", "klmno", "pqrst", "yvwxu" ]
l = 5 # Length of strings
if check(v, l):
    print "Yes"
else:
    print "No"
 
# This code is contributed by Sachin Bisht

                    

C#

// C# program to check if we can make a grid
// of character sorted using adjacent swaps.
using System;
 
class GfG
{
 
    // v[] is vector of strings. len is
    // length of strings in every row.
    static Boolean check(String[] v, int len)
    {
        int n = v.Length;
        char[] tempArray;
         
        for (int i = 0; i < n; i++)
        {
            tempArray = v[i].ToCharArray();
            Array.Sort(tempArray);
            v[i] = new String(tempArray);
        }
         
        for (int i = 0; i < len-1; i++)
            for (int j = 0; j < n; j++)
                if (v[i][j] > v[i+1][j])
                    return false;
        return true;
    }
 
    // Driver code
    public static void Main(String []args)
    {
         
        String[] v = { "ebcda", "ihgfj", "klmno",
                            "pqrst", "yvwxu" };
         
        int len = 5; // Length of strings
        if (check(v, len))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code has been contributed by 29AjayKumar

                    

Javascript

<script>
// Javascript program to check if we can make a grid
// of character sorted using adjacent swaps.
 
// v[] is vector of strings. len is
    // length of strings in every row.
function check(v,len)
{
    let n = v.length;
        let tempArray;
           
        for (let i = 0; i < n; i++)
        {
            tempArray = v[i].split("");
            (tempArray).sort();
            v[i] = (tempArray).join("");
        }
           
        for (let i = 0; i < len-1; i++)
            for (let j = 0; j < n; j++)
                if (v[i][j] > v[i+1][j])
                    return false;
        return true;
}
 
// Driver code
let v = ["ebcda", "ihgfj", "klmno",
                            "pqrst", "yvwxu" ];
let len = 5; // Length of strings
 
if (check(v, len))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by rag2127
</script>

                    

Output
Yes

Time Complexity: O(n*len*log(len))
Auxiliary Space: O(1)

 



Last Updated : 14 Mar, 2023
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