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Check if a Matrix is Bitonic or not
  • Last Updated : 07 May, 2021

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 not
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 not
void 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 Code
int 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 not
class GFG{
 
final static int N = 3, M = 3;
 
// Function to check if an
// array is Bitonic or not
static 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 not
static 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 Code
public 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 not
N = 3
M = 3
 
# Function to check if an
# array is Bitonic or not
def 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 not
def 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 Code
if __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 not
using System;
class GFG{
 
readonly static int N = 3, M = 3;
 
// Function to check if an
// array is Bitonic or not
static 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 not
static 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 Code
public 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 not
let N = 3, M = 3;
 
// Function to check if an
// array is Bitonic or not
function 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 not
function 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 * M) 
Auxiliary Space: O(N)
 

 

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