# Python program to search for the minimum element occurring consecutively n times in a matrix

Given a matrix containing n rows. Each row has an equal number of m elements. The task is to find elements that come consecutively n times horizontally, vertically, diagonally in the matrix. If there are multiple such elements then print the smallest element. If there is no such element
then print -1.

Examples:

```Input : n = 4
mat  2 1 3 4 5
3 2 1 3 4
5 3 2 1 4
5 4 3 2 1
5 4 4 3 2
Output :  1

Input : n = 2
mat  2 0 5 1
0 5 3 5
8 4 1 1
3 8 5 2
Output : 0
```

Implementation:

 `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 temprory  ` `            ``# 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  ` `            ``# temprory 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) `

Output :

`0`

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