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:
- 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.
- Traverse the matrix column wise.
- Insert the count of zeroes for each column in the vector of pair.
- Sort the vector of pair according to the count of zeroes.
- 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)
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