Programa de Python para buscar el elemento mínimo que ocurre consecutivamente n veces en una array

Python3

def check(l,row,col ,n,l2):
  
    # Iterate through the matrix.
    # Check for diagonal.
    for i in range(row - n+1):
        for j in range(col - n+1):
             
            # Store the value in a temporary
            # variable.
            num = l[i][j]
             
            # Check for the condition only
            # if we have more rows and columns
            # than n.
            if(row-i >= n and col-j >= n):
                count = 0
                 
                # check if the number is present
                # n times.
                for k in range(n):
                     
                    # if number is not present n
                    # times than break
                    if(num != l[i+k][j+k]):
                        break
                         
                    # increment the count for checking
                    # the condition follows.
                    else:
                        count += 1
                         
                # if count is same or greater as that of
                # n than the number follows the condition.
                if(count == n):
                    l2.append(num)
            else:
                break
      
    #check for row condition
    for i in range(row):
        for j in range(col - (n-1) ):
             
            num = l[i][j]
            count = 0
             
            for k in range(n):
                 
                if num != l[i][j + k]:
                    break
                else:
                    count+=1
                     
            # if count is same or greater as that
            # of n than the number follows the condition.
            if(count==n):
                l2.append(num)
  
    #check for column condition.
    for i in range(row - (n-1)):
        for j in range(col):
             
            num = l[i][j]
            count = 0
             
            for k in range(n):
                 
                if num != l[i+k][j]:
                    break
                else:
                    count += 1
                     
            # if count is same or greater as that of
            # n than the number follows the condition.
            if(count == n):
                l2.append(num)
  
    # It would require a complex code to
    # check for anti-diagonal Just rotate
    # the matrix and check for diagonal
    # condition again.
  
    #matrix rotation by 90 degrees.
    for i in range(0, int(row / 2)):
  
        for j in range(i, col - i - 1):
         
            # store current cell in
            # num variable
            num = l[i][j]
             
            # move values from right to top
            l[i][j] = l[j][col - 1 - i]
             
            # move values from bottom to right
            l[j][col - 1 - i] = l[row - 1 - i][col - 1 - j]
             
            # move values from left to bottom
            l[row - 1 - i][col - 1 - j] = l[row - 1 - j][i]
             
            # assign num to left
            l[row - 1 - j][i] = num
              
    # Iterate through the rotated matrix.
    for i in range(row - n+1):
        for j in range(col - n+1):
             
            # Store the value in a
            # temporary variable.
            num = l[i][j]
             
            # Check for the condition only if
            # we have more rows and columns than n.
            if(row-i >= n and col-j >= n):
                count = 0
                 
                # check if the number is present
                # n times.
                for k in range(n):
                     
                    # if number is not present n
                    # times than break
                    if(num != l[i+k][j+k]):
                        break
                     
                    # increment the count for checking
                    # the condition follows.
                    else:
                        count += 1
                 
                # if count is same or greater as that of
                # n than the number follows the condition.
                if(count == n):
                    l2.append(num)
            else:
                break
  
    # check if any element followed the condition.
    if(len(l2) == 0):
        print(-1)
     
    # print the minimum of all the elements
    # which follow the condition.
    else:
        print(min(l2))
 
if __name__ == "__main__":
  
    # Create Matrix
    l = [[2, 1, 3, 4, 5],
        [0, 2, 1, 3, 4],
        [5, 0, 2, 1, 4],
        [5, 4, 0, 2, 1],
        [5, 4, 4, 0, 2]]
  
    # Create a list to store all the elements
    # which follow the condition.
    l2 = []
    check(l,4,4,2,l2)