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Print index of columns sorted by count of zeroes in the Given Matrix

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Given a matrix with N rows and M columns. The task is to print the index of columns of the given matrix based on the increasing number of zeroes in each column. 
For Example, If the 1st column contains 2 zero’s, the 2nd column contains 1 zero, and 3rd column does not contain any zeroes. Then the output will be 3, 2, 1.

Note: The matrix is considered to be having 1-based indexing.

Examples: 

Input : mat[N][M] ={{0, 0, 0},
                    {0, 2, 0},
                    {0, 1, 1},
                    {1, 1, 1}};
Output : 2 3 1 
No. of zeroes in first col: 3
No. of zeroes in second col: 1
No of zeroes in third col: 2
Therefore, sorted order of count is 1 2 3
and their corresponding column numbers are, 2 3 1

Input: mat[N][M] ={{0, 0, 0},
                    {0, 0, 3},
                    {0, 1, 1}};
Output : 3 2 1 

Approach

  1. Create a Vector of pair to store count of zeroes in each column. Where the first element of the pair is count and the second element of the pair is corresponding column index.
  2. Traverse the matrix column wise.
  3. Insert the count of zeroes for each column in the vector of pair.
  4. Sort the vector of pair according to the count of zeroes.
  5. Print the second element of the pair which contains indexes of the columns sorted according to the count of zeroes.

Below is the implementation of the above approach: 

C++




// C++ Program to print the index of columns
// of the given matrix based on the
// increasing number of zeroes in each column
 
#include <bits/stdc++.h>
 
using namespace std;
 
#define N 4 // rows
#define M 3 // columns
 
// Function to print the index of columns
// of the given matrix based on the
// increasing number of zeroes in each column
void printColumnSorted(int mat[N][M])
{
    // Vector of pair to store count of zeroes
    // in each column.
    // First element of pair is count
    // Second element of pair is column index
    vector<pair<int, int> > colZeroCount;
 
    // Traverse the matrix column wise
    for (int i = 0; i < M; i++) {
        int count = 0;
 
        for (int j = 0; j < N; j++) {
            if (mat[j][i] == 0)
                count++;
        }
 
        // Insert the count of zeroes for each column
        // in the vector of pair
        colZeroCount.push_back(make_pair(count, i));
    }
 
    // Sort the vector of pair according to the
    // count of zeroes
    sort(colZeroCount.begin(), colZeroCount.end());
 
    // Print the second element of the pair which
    // contain indexes of the sorted vector of pair
    for (int i = 0; i < M; i++)
        cout << colZeroCount[i].second + 1 << " ";
}
 
// Driver Code
int main()
{
    int mat[N][M] = { { 0, 0, 0 },
                      { 0, 2, 0 },
                      { 0, 1, 1 },
                      { 1, 1, 1 } };
 
    printColumnSorted(mat);
 
    return 0;
}


Java




// Java program to print the index of columns
// of the given matrix based on the increasing
// number of zeroes in each column
import java.io.*;
import java.util.*;
 
class GFG{
     
static int N = 4;
static int M = 3;
 
// Function to print the index of columns
// of the given matrix based on the increasing
// number of zeroes in each column
static void printColumnSorted(int[][] mat)
{
     
    // Vector of pair to store count of zeroes
    // in each column.First element of pair is
    // count.Second element of pair is column index
    ArrayList<
    ArrayList<Integer>> colZeroCount = new ArrayList<
                                           ArrayList<Integer>>();
     
    // Traverse the matrix column wise
    for(int i = 0; i < M; i++)
    {
        int count = 0;
        for(int j = 0; j < N; j++)
        {
            if (mat[j][i] == 0)
            {
                count++;
            }
        }
         
        // Insert the count of zeroes for
        // each column in the vector of pair
        colZeroCount.add(new ArrayList<Integer>(
            Arrays.asList(count,i)));
    }
     
    // Sort the vector of pair according to the
    // count of zeroes
    Collections.sort(colZeroCount,
                     new Comparator<ArrayList<Integer>>()
    {   
        @Override
        public int compare(ArrayList<Integer> o1,
                           ArrayList<Integer> o2)
        {
            return o1.get(0).compareTo(o2.get(0));
        }              
    });
     
    // Print the second element of the pair which
    // contain indexes of the sorted vector of pair
    for(int i = 0; i < M; i++)
    {
        System.out.print(
            (colZeroCount.get(i).get(1) + 1) + " ");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int[][] mat = { { 0, 0, 0 }, { 0, 2, 0 },
                    { 0, 1, 1 }, { 1, 1, 1 } };
                     
    printColumnSorted(mat);
}
}
 
// This code is contributed by avanitrachhadiya2155


Python3




# Python3 program to print the index of
# columns of the given matrix based
# on the increasing number of zeroes
# in each column
 
# Rows
N = 4
 
# Columns
M = 3
 
# Function to print the index of columns
# of the given matrix based on the
# increasing number of zeroes in
# each column
def printColumnSorted(mat):
     
    # Vector of pair to store count
    # of zeroes in each column.
    # First element of pair is count
    # Second element of pair is column index
    colZeroCount = []
 
    # Traverse the matrix column wise
    for i in range(M):
        count = 0
 
        for j in range(N):
            if (mat[j][i] == 0):
                count += 1
 
        # Insert the count of zeroes for
        # each column in the vector of pair
        colZeroCount.append((count, i))
 
    # Sort the vector of pair according
    # to the count of zeroes
    colZeroCount = sorted(colZeroCount)
 
    # Print the second element of the
    # pair which contain indexes of the
    # sorted vector of pair
    for i in range(M):
        print(colZeroCount[i][1] + 1, end = " ")
 
# Driver Code
if __name__ == '__main__':
     
    mat = [ [ 0, 0, 0 ],
            [ 0, 2, 0 ],
            [ 0, 1, 1 ],
            [ 1, 1, 1 ] ]
 
    printColumnSorted(mat)
 
# This code is contributed mohit kumar 29


C#




// C# Program to print the index of columns
// of the given matrix based on the
// increasing number of zeroes in each column
using System;
using System.Collections.Generic;
 
class GFG
{
 
  static int N = 4; // rows
  static int M = 3; // columns
 
  // Function to print the index of columns
  // of the given matrix based on the
  // increasing number of zeroes in each column
  static void printColumnSorted(int[, ] mat)
  {
    // Vector of pair to store count of zeroes
    // in each column.
    // First element of pair is count
    // Second element of pair is column index
    List<Tuple<int, int> > colZeroCount = new List<Tuple<int, int> >();
 
    // Traverse the matrix column wise
    for (int i = 0; i < M; i++) {
      int count = 0;
 
      for (int j = 0; j < N; j++) {
        if (mat[j, i] == 0)
          count++;
      }
 
      // Insert the count of zeroes for each column
      // in the vector of pair
      colZeroCount.Add(Tuple.Create(count, i));
    }
 
    // Sort the vector of pair according to the
    // count of zeroes
    colZeroCount.Sort();
 
    // Print the second element of the pair which
    // contain indexes of the sorted vector of pair
    for (int i = 0; i < M; i++)
      Console.Write(colZeroCount[i].Item2 + 1 + " ");
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
    int[,] mat = { { 0, 0, 0 },
                  { 0, 2, 0 },
                  { 0, 1, 1 },
                  { 1, 1, 1 } };
 
    printColumnSorted(mat);
 
  }
}
 
// This code is contributed by phasing17.


Javascript




<script>
// Javascript program to print the index of columns
// of the given matrix based on the increasing
// number of zeroes in each column
 
let N = 4;
let M = 3;
 
// Function to print the index of columns
// of the given matrix based on the increasing
// number of zeroes in each column
function printColumnSorted(mat)
{
    // Vector of pair to store count of zeroes
    // in each column.First element of pair is
    // count.Second element of pair is column index
    let colZeroCount = [];
      
    // Traverse the matrix column wise
    for(let i = 0; i < M; i++)
    {
        let count = 0;
        for(let j = 0; j < N; j++)
        {
            if (mat[j][i] == 0)
            {
                count++;
            }
        }
          
        // Insert the count of zeroes for
        // each column in the vector of pair
        colZeroCount.push([count,i]);
    }
      
    // Sort the vector of pair according to the
    // count of zeroes
    colZeroCount.sort(function(a,b){return a[0]-b[0];});
      
    // Print the second element of the pair which
    // contain indexes of the sorted vector of pair
    for(let i = 0; i < M; i++)
    {
        document.write(
            (colZeroCount[i][1] + 1) + " ");
    }
}
 
// Driver Code
let mat = [ [ 0, 0, 0 ],
            [ 0, 2, 0 ],
            [ 0, 1, 1 ],
            [ 1, 1, 1 ] ];
             
printColumnSorted(mat);
 
 
// This code is contributed by patel2127
</script>


Output

2 3 1 

Complexity Analysis:

  • Time Complexity: O(n*m+m*log(m))
  • Auxiliary Space: O(m)


Last Updated : 07 Dec, 2022
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