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

• Last Updated : 07 Dec, 2022

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 ` `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 > 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> colZeroCount = ``new` `ArrayList<``                                           ``ArrayList>();``    ` `    ``// 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(``            ``Arrays.asList(count,i)));``    ``}``    ` `    ``// Sort the vector of pair according to the``    ``// count of zeroes``    ``Collections.sort(colZeroCount,``                     ``new` `Comparator>()``    ``{   ``        ``@Override``        ``public` `int` `compare(ArrayList o1,``                           ``ArrayList 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 > colZeroCount = ``new` `List >();` `    ``// 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

 ``

Output

`2 3 1 `

Complexity Analysis:

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

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