Given a matrix mat[][], the task is to check whether the matrix is valid or not. A valid matrix is the matrix that satisfies the given conditions:
- For every row, it must contain a single distinct character.
- No two consecutive rows have a character in common.
Examples:
Input: mat[][] = {
{0, 0, 0},
{1, 1, 1},
{0, 0, 2}}
Output: No
The last row doesn’t consist of a single distinct character.Input: mat[][] = {
{8, 8, 8, 8, 8},
{4, 4, 4, 4, 4},
{6, 6, 6, 6, 6},
{5, 5, 5, 5, 5},
{8, 8, 8, 8, 8}}
Output: Yes
Approach: First of all check the first row if it contains same characters or not. If it contains same characters then iterate the second row and compare the characters of the current row with the first element of the previous row, if the two elements are equal or the characters of the current row are different, return false. Repeat the above process for all the consecutive row.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Storing the size of the matrixconst int M = 5, N = 5;// Function that returns true if// the matrix is validbool isPerfect(char board[M][N + 1]){ char m; for (int i = 0; i < M; i++) { // Get the current row string s = board[i]; // Comparing first element // of the row with the element // of previous row if (i > 0 && s[0] == m) return false; // Checking if all the characters of the // current row are same or not for (int j = 0; j < N; j++) { // Storing the first character if (j == 0) m = s[0]; // Comparing all the elements // with the first element else { if (m != s[j]) return false; } } } return true;}// Driver codeint main(){ char board[M][N + 1] = { "88888", "44444", "66666", "55555", "88888" }; if (isPerfect(board)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation of the approach class GFG { // Storing the size of the matrix static final int M = 5, N = 5; // Function that returns true if // the matrix is valid static boolean isPerfect(String board[]) { char m = 0; for (int i = 0; i < M; i++) { // Get the current row String s = board[i]; // Comparing first element // of the row with the element // of previous row if (i > 0 && s.charAt(0) == m) return false; // Checking if all the characters of the // current row are same or not for (int j = 0; j < N; j++) { // Storing the first character if (j == 0) m = s.charAt(0); // Comparing all the elements // with the first element else { if (m != s.charAt(j)) return false; } } } return true; } // Driver code public static void main (String[] args) { String board[] = { "88888", "44444", "66666", "55555", "88888" }; if (isPerfect(board)) System.out.println("Yes"); else System.out.println("No"); } } // This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approach# Storing the size of the matrixM = 5N = 5# Function that returns true if# the matrix is validdef isPerfect(board): m = '' for i in range(M): # Get the current row s = board[i] # Comparing first element # of the row with the element # of previous row if (i > 0 and s[0] == m): return False # Checking if all the characters of the # current row are same or not for j in range(N): # Storing the first character if (j == 0): m = s[0] # Comparing all the elements # with the first element else: if (m != s[j]): return False return True# Driver codeboard = ["88888", "44444", "66666", "55555", "88888"]if (isPerfect(board)): print("Yes")else: print("No")# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approachusing System; class GFG { // Storing the size of the matrix static readonly int M = 5, N = 5; // Function that returns true if // the matrix is valid static bool isPerfect(String []board) { char m = '0'; for (int i = 0; i < M; i++) { // Get the current row String s = board[i]; // Comparing first element // of the row with the element // of previous row if (i > 0 && s[0] == m) return false; // Checking if all the characters of the // current row are same or not for (int j = 0; j < N; j++) { // Storing the first character if (j == 0) m = s[0]; // Comparing all the elements // with the first element else { if (m != s[j]) return false; } } } return true; } // Driver code public static void Main (String[] args) { String []board = { "88888", "44444", "66666", "55555", "88888" }; if (isPerfect(board)) Console.WriteLine("Yes"); else Console.WriteLine("No"); } }// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation of the approach// Storing the size of the matrixvar M = 5, N = 5;// Function that returns true if// the matrix is validfunction isPerfect(board){ var m; for(var i = 0; i < M; i++) { // Get the current row var s = board[i]; // Comparing first element // of the row with the element // of previous row if (i > 0 && s[0] == m) return false; // Checking if all the characters of the // current row are same or not for(var j = 0; j < N; j++) { // Storing the first character if (j == 0) m = s[0]; // Comparing all the elements // with the first element else { if (m != s[j]) return false; } } } return true;}// Driver codevar board = [ "88888", "44444", "66666", "55555", "88888" ]; if (isPerfect(board)) document.write("Yes");else document.write("No"); // This code is contributed by itsok</script> |
Output:
Yes
Time complexity: O(M*N)
Auxiliary space: O(N) because using extra space for string s
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