Given a matrix arr[][] of dimensions M * N, the task is to count the number of digits of every element present in the given matrix.
Examples:
Input: arr[][] = { { 27, 173, 5 }, { 21, 6, 624 }, { 5, 321, 49 } }
Output:
2 3 1
2 1 3
1 3 2Input: arr[][] = { {11, 12, 33 }, { 64, 57, 61 }, { 74, 88, 39 } }
Output:
2 2 2
2 2 2
2 2 2
Approach: The idea to solve this problem is to traverse the given matrix and for every index of the matrix, count the number of digits of the number present at that index using the following expression:
Number of digits present in any value X is given by floor(log10(X))+1.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;const int M = 3;const int N = 3;// Function to count the number of digits// in each element of the given matrixvoid countDigit(int arr[M][N]){ // Traverse each row of arr[][] for (int i = 0; i < M; i++) { // Traverse each column of arr[][] for (int j = 0; j < N; j++) { // Store the current matrix element int X = arr[i][j]; // Count the number of digits int d = floor(log10(X) * 1.0) + 1; // Print the result cout << d << " "; } cout << endl; }}// Driver Codeint main(){ // Given matrix int arr[][3] = { { 27, 173, 5 }, { 21, 6, 624 }, { 5, 321, 49 } }; countDigit(arr); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{static int M = 3;static int N = 3;// Function to count the number of digits// in each element of the given matrixstatic void countDigit(int arr[][]){ // Traverse each row of arr[][] for (int i = 0; i < M; i++) { // Traverse each column of arr[][] for (int j = 0; j < N; j++) { // Store the current matrix element int X = arr[i][j]; // Count the number of digits int d = (int) (Math.floor(Math.log10(X) * 1.0) + 1); // Print the result System.out.print(d+ " "); } System.out.println(); }}// Driver Codepublic static void main(String[] args){ // Given matrix int arr[][] = { { 27, 173, 5 }, { 21, 6, 624 }, { 5, 321, 49 } }; countDigit(arr);}}// This code is contributed by Princi Singh |
Python3
# Python3 program for the above approachfrom math import floor, log10M = 3N = 3# Function to count the number of digits# in each element of the given matrixdef countDigit(arr): # Traverse each row of arr[][] for i in range(M): # Traverse each column of arr[][] for j in range(N): # Store the current matrix element X = arr[i][j] # Count the number of digits d = floor(log10(X) * 1.0) + 1 # Print the result print(d, end = " ") print()# Driver Codeif __name__ == '__main__': # Given matrix arr = [ [ 27, 173, 5 ], [ 21, 6, 624 ], [ 5, 321, 49 ] ] countDigit(arr)# This code is contributed by mohit kumar 29 |
C#
// C# program to implement// the above approach using System;class GFG{ static int M = 3;static int N = 3; // Function to count the number of digits// in each element of the given matrixstatic void countDigit(int[,] arr){ // Traverse each row of arr[][] for (int i = 0; i < M; i++) { // Traverse each column of arr[][] for (int j = 0; j < N; j++) { // Store the current matrix element int X = arr[i, j]; // Count the number of digits int d = (int) (Math.Floor(Math.Log10(X) * 1.0) + 1); // Print the result Console.Write(d + " "); } Console.WriteLine(); }} // Driver Codepublic static void Main(){ // Given matrix int[,] arr = { { 27, 173, 5 }, { 21, 6, 624 }, { 5, 321, 49 } }; countDigit(arr);}}// This code is contributed by susmitakundugoaldanga |
Javascript
<script>// JavaScript program to implement// the above approachlet M = 3;let N = 3; // Function to count the number of digits// in each element of the given matrixfunction countDigit(arr){ // Traverse each row of arr[][] for (let i = 0; i < M; i++) { // Traverse each column of arr[][] for (let j = 0; j < N; j++) { // Store the current matrix element let X = arr[i][j]; // Count the number of digits let d = (Math.floor(Math.log10(X) * 1.0) + 1); // Print the result document.write(d+ " "); } document.write("<br/>"); }} // Driver code // Given matrix let arr = [[ 27, 173, 5 ], [ 21, 6, 624 ], [ 5, 321, 49 ]]; countDigit(arr);</script> |
Output:
2 3 1 2 1 3 1 3 2
Time Complexity: O(M * N)
Auxiliary Space: O(1)
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