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Find the person who will finish last
  • Last Updated : 03 May, 2021

Given a binary matrix mat[][] of dimensions M x N and two-person P1, P2, the task is to find the person who finishes last in choosing a 0 from the matrix which changes to 1 only if the row or column of the cell consisting of 0 has one or more than one 1 in the matrix.
Note: P1 starts picking the 0s first and both the persons want to finish last. The given matrix will always have at least one 0 which could be chosen.

Examples:

Input: mat[][] = {{1, 0, 0}, {0, 0, 0}, {0, 0, 1}}
Output: P1
Explanation
P1 chooses mat[1][1], then the matrix becomes {{1, 0, 0}, {0, 1, 0}, {0, 0,1}}.
P2 has no 0 left to choose from. So, P1 finishes last.

Input: mat[][] = {{0, 0}, {0, 0}}
Output: P2
Explanation
No matter P1 chooses which 0 P2 will always have a 0 to choose and
after P2 picks a 0 there will not be any other 0 to choose from.

Approach: The idea is based on the observation that a 0 can’t be taken if either of its row or column has 1. Follow the steps below to solve this problem:



  • Initialize two sets, rows & cols to count the number of rows and columns which does not contain any 1.
  • Traverse the matrix and add rows and columns having 1 in it in the set.
  • Take the minimum number of rows or columns as if either of them becomes zero so that no more 0s can be taken.
  • After finding the minimum number of rows and columns available, if the number of choices made is odd then P1 finishes last otherwise, P2 finishes last.

Below is the implementation of the above approach:

Python3




# Python3 program for the above approach
  
# Function to find the person
# who will finish last
def findLast(mat):
  
    m = len(mat)
    n = len(mat[0])
  
    # To keep track of rows
    # and columns having 1
    rows = set()
    cols = set()
  
    for i in range(m):
        for j in range(n):
            if mat[i][j]:
                rows.add(i)
                cols.add(j)
  
    # Available rows and columns
    avRows = m-len(list(rows))
    avCols = n-len(list(cols))
  
    # Minimum number of choices we have
    choices = min(avRows, avCols)
  
    # If number of choices are odd
    if choices & 1:
      
        # P1 will finish last
        print('P1')
  
    # Otherwise, P2 will finish last
    else:
        print('P2')
  
  
# Given matrix
mat = [[1, 0, 0], [0, 0, 0], [0, 0, 1]]
  
findLast(mat)
Output:
P1

Time Complexity: O(M*N)
Auxiliary Space: O(M*N)

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