Given a matrix m[][], the task is to check if the given matrix is Bitonic or not. If the given matrix is Bitonic, then print YES. Otherwise, print NO.
If all the rows and the columns of the given matrix have elements in one of the following orders:
- Strictly increasing
- Strictly decreasing
- Strictly increasing followed by strictly decreasing
Then the given matrix is said to be a Bitonic Matrix
Example:
Input: m[][] = {{1, 2, 3}, {4, 5, 6}, {2, 3, 4}}
Output: YES
Explanation:
All the columns of the given matrix {1, 4, 2}, {2, 5, 3}, {3, 6, 4} forms an increasing followed by decreasing sequence
All the rows of the given matrix have an increasing sequence.
Therefore, the matrix is Bitonic.
Input: m[][] = {{1, 2, 3}, {4, 5, 6}, {2, 5, 4}}
Output: NO
Explanation:
Since the column {2, 5, 5} does not satisfy any of the three conditions, the given matrix is not Bitonic.
Approach:
Follow the steps below to solve the problem:
- Check the elements of each row of the matrix one by one, if it forms a Bitonic sequence or not. If any row is found to be not Bitonic, print NO.
- Similarly, check the elements of each column one by one,, if it forms a Bitonic sequence or not. If any row is found to be not Bitonic, print NO.
- If all the rows and columns are found to be Bitonic, then print YES
Below is the implementation of the above approach:
C++
// C++ Program to check if a// matrix is Bitonic or not#include <bits/stdc++.h>using namespace std;const int N = 3, M = 3;// Function to check if an// array is Bitonic or notbool checkBitonic(int arr[], int n){ int i, j, f = 0; // Check for increasing sequence for (i = 1; i < n; i++) { if (arr[i] > arr[i - 1]) continue; if (arr[i] == arr[i - 1]) return false; else { f = 1; break; } } if (i == n) return true; // Check for decreasing sequence for (j = i + 1; j < n; j++) { if (arr[j] < arr[j - 1]) continue; if (arr[i] == arr[i - 1]) return false; else { if (f == 1) return false; } } return true;}// Function to check whether given// matrix is bitonic or notvoid check(int arr[N][M]){ int f = 0; // Check row-wise for (int i = 0; i < N; i++) { if (!checkBitonic(arr[i], M)) { cout << "NO" << endl; return; } } // Check column wise for (int i = 0; i < N; i++) { // Generate an array // consisting of elements // of the current column int temp[N]; for (int j = 0; j < N; j++) { temp[j] = arr[j][i]; } if (!checkBitonic(temp, N)) { cout << "NO" << endl; return; } } cout << "YES";}// Driver Codeint main(){ int m[N][M] = { { 1, 2, 3 }, { 3, 4, 5 }, { 2, 6, 4 } }; check(m); return 0;} |
Java
// Java program to check if a// matrix is Bitonic or notclass GFG{final static int N = 3, M = 3;// Function to check if an// array is Bitonic or notstatic boolean checkBitonic(int arr[], int n){ int i, j, f = 0; // Check for increasing sequence for(i = 1; i < n; i++) { if (arr[i] > arr[i - 1]) continue; if (arr[i] == arr[i - 1]) return false; else { f = 1; break; } } if (i == n) return true; // Check for decreasing sequence for(j = i + 1; j < n; j++) { if (arr[j] < arr[j - 1]) continue; if (arr[i] == arr[i - 1]) return false; else { if (f == 1) return false; } } return true;}// Function to check whether given// matrix is bitonic or notstatic void check(int arr[][]){ int f = 0; // Check row-wise for(int i = 0; i < N; i++) { if (!checkBitonic(arr[i], M)) { System.out.println("NO"); return; } } // Check column wise for(int i = 0; i < N; i++) { // Generate an array // consisting of elements // of the current column int temp[] = new int[N]; for(int j = 0; j < N; j++) { temp[j] = arr[j][i]; } if (!checkBitonic(temp, N)) { System.out.println("NO"); return; } } System.out.println("YES");} // Driver Codepublic static void main(String[] args) { int m[][] = { { 1, 2, 3 }, { 3, 4, 5 }, { 2, 6, 4 } }; check(m);}}// This code is contributed by rutvik_56 |
Python3
# Python3 program to check if a# matrix is Bitonic or notN = 3M = 3# Function to check if an # array is Bitonic or notdef checkBitonic(arr, n): i, j, f = 0, 0, 0 # Check for increasing sequence for i in range(1, n): if (arr[i] > arr[i - 1]): continue if (arr[i] == arr[i - 1]): return False else: f = 1 break if (i == n): return True # Check for decreasing sequence for j in range(i + 1, n): if (arr[j] < arr[j - 1]): continue if (arr[i] == arr[i - 1]): return False else: if (f == 1): return False return True# Function to check whether given# matrix is bitonic or notdef check(arr): f = 0 # Check row wise for i in range(N): if (not checkBitonic(arr[i], M)): print("NO") return # Check column wise i = 0 for i in range(N): # Generate an array consisting # of elements of current column temp = [0] * N for j in range(N): temp[j] = arr[j][i] if (not checkBitonic(temp, N)): print("NO") return print("YES")# Driver Codeif __name__ == '__main__': m = [ [ 1, 2, 3 ], [ 3, 4, 5 ], [ 2, 6, 4 ] ] check(m)# This code is contributed by himanshu77 |
C#
// C# program to check if a// matrix is Bitonic or notusing System;class GFG{readonly static int N = 3, M = 3;// Function to check if an// array is Bitonic or notstatic bool checkBitonic(int []arr, int n){ int i, j, f = 0; // Check for increasing sequence for(i = 1; i < n; i++) { if (arr[i] > arr[i - 1]) continue; if (arr[i] == arr[i - 1]) return false; else { f = 1; break; } } if (i == n) return true; // Check for decreasing sequence for(j = i + 1; j < n; j++) { if (arr[j] < arr[j - 1]) continue; if (arr[i] == arr[i - 1]) return false; else { if (f == 1) return false; } } return true;}// Function to check whether given// matrix is bitonic or notstatic void check(int [,]arr){ int f = 0; // Check row-wise for(int i = 0; i < N; i++) { if (!checkBitonic(GetRow(arr, i), M)) { Console.WriteLine("NO"); return; } } // Check column wise for(int i = 0; i < N; i++) { // Generate an array // consisting of elements // of the current column int []temp = new int[N]; for(int j = 0; j < N; j++) { temp[j] = arr[j, i]; } if (!checkBitonic(temp, N)) { Console.WriteLine("NO"); return; } } Console.WriteLine("YES");}public static int[] GetRow(int[,] matrix, int row) { var rowLength = matrix.GetLength(1); var rowVector = new int[rowLength]; for (var i = 0; i < rowLength; i++) rowVector[i] = matrix[row, i]; return rowVector; } // Driver Codepublic static void Main(String[] args) { int [,]m = { { 1, 2, 3 }, { 3, 4, 5 }, { 2, 6, 4 } }; check(m);}}// This code is contributed by sapnasingh4991 |
Javascript
<script>// Java Script program to check if a// matrix is Bitonic or notlet N = 3, M = 3;// Function to check if an// array is Bitonic or notfunction checkBitonic(arr,n){ let i, j, f = 0; // Check for increasing sequence for(i = 1; i < n; i++) { if (arr[i] > arr[i - 1]) continue; if (arr[i] == arr[i - 1]) return false; else { f = 1; break; } } if (i == n) return true; // Check for decreasing sequence for(j = i + 1; j < n; j++) { if (arr[j] < arr[j - 1]) continue; if (arr[i] == arr[i - 1]) return false; else { if (f == 1) return false; } } return true;}// Function to check whether given// matrix is bitonic or notfunction check(arr){ let f = 0; // Check row-wise for(let i = 0; i < N; i++) { if (!checkBitonic(arr[i], M)) { document.write("NO"); return; } } // Check column wise for(let i = 0; i < N; i++) { // Generate an array // consisting of elements // of the current column let temp = [N]; for(let j = 0; j < N; j++) { temp[j] = arr[j][i]; } if (!checkBitonic(temp, N)) { document.write("NO"); return; } } document.write("YES");} // Driver Code let m = [[ 1, 2, 3 ], [3, 4, 5 ], [ 2, 6, 4 ]]; check(m);// This code is contributed by sravan kumar </script> |
Output:
YES
Time Complexity: O(N * N)
Auxiliary Space: O(N)
