Minimize count of unique paths from top left to bottom right of a Matrix by placing K 1s

Given two integers N and M where M and N denote a matrix of dimensions N * M consisting of 0‘s only. The task is to minimize the count of unique paths from the top left (0, 0) to bottom right (N – 1, M – 1) of the matrix across cells consisting of 0’s only by placing exactly K 1s in the matrix.

Note: Neither the bottom right nor the top-left cell can be modified to a 0. 

Examples: 

Input: N = 3, M = 3, K = 1
Output: 2
Explanation: 
Placing K(= 1) 1s in the matrix to generate the matrix [[0, 0, 0], [0, 1, 0], [0, 0, 0]] leaves only two possible paths from top-left to bottom-right cells. 
The paths are[(0, 0) → (0, 1) → (0, 2) → (1, 2) → (2, 2)] and [(0, 0) → (1, 0) → (2, 0) → (2, 1) → (2, 2)]

Input: N = 3, M = 3, K = 3
Output: 0
Explanation:
Placing K(= 3) 1s to generate a matrix [[0, 1, 1], [1, 0, 0], [0, 0, 0]] leaves no possible path from top-left to bottom-right.



Approach: The problem can be solved by considering the following possible cases. 

  1. If K ≥ 2: The count of possible paths can be reduced to 0 by placing two 1s in (0, 1) and (1, 0) cells of the matrix.
  2. If K = 0: The count remains C(N+M-2, N-1).
  3. If K = 1: Place a 1 at the Centre of the matrix, ((N-1)/2, (M-1)/2) to minimize the path count. Therefore, the count of possible paths for this case is as follows:

Result = Total number of ways to reach the bottom right from the top left – ( Number of paths to midpoint from the top left * Number of ways to reach the endpoint from the midpoint) 
where

  • Total number of ways to reach the bottom right from the top left = C(N + M – 2, N – 1)
  • Number of paths to midpoint from the top left = C((N – 1) / 2 + (M – 1) / 2, (N – 1) / 2)
  • Number of ways to reach the endpoint from the midpoint=C(((N – 1) – (N – 1 ) / 2) + ((M – 1) – (M – 1) / 2), ((N – 1) – (N – 1) / 2))

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ Program to implememnt
// the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the value of
// Binomial Coefficient C(n, k)
int ncr(int n, int k)
{
    int res = 1;
  
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
  
    // Calculate the value of
    // [n * (n-1) *---* (n-k+1)] /
    // [k * (k-1) *----* 1]
    for (int i = 0; i < k; ++i) {
        res *= (n - i);
        res /= (i + 1);
    }
  
    return res;
}
  
// Function to find the minimum
// count of paths from top
// left to bottom right by
// placing K 1s in the matrix
int countPath(int N, int M, int K)
{
    int answer;
    if (K >= 2)
        answer = 0;
    else if (K == 0)
        answer = ncr(N + M - 2, N - 1);
    else {
  
        // Count of ways without 1s
        answer = ncr(N + M - 2, N - 1);
  
        // Count of paths from starting
        // point to mid point
        int X = (N - 1) / 2 + (M - 1) / 2;
        int Y = (N - 1) / 2;
        int midCount = ncr(X, Y);
  
        // Count of paths from mid
        // point to end point
        X = ((N - 1) - (N - 1) / 2)
            + ((M - 1) - (M - 1) / 2);
  
        Y = ((N - 1) - (N - 1) / 2);
        midCount *= ncr(X, Y);
        answer -= midCount;
    }
    return answer;
}
  
// Driver Code
int main()
{
    int N = 3;
    int M = 3;
    int K = 1;
  
    cout << countPath(N, M, K);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java Program to implememnt
// the above approach
import java.util.*;
class GFG{
  
// Function to return the value of
// Binomial Coefficient C(n, k)
static int ncr(int n, int k)
{
    int res = 1;
  
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
  
    // Calculate the value of
    // [n * (n-1) *---* (n-k+1)] /
    // [k * (k-1) *----* 1]
    for (int i = 0; i < k; ++i) 
    {
        res *= (n - i);
        res /= (i + 1);
    }
    return res;
}
  
// Function to find the minimum
// count of paths from top
// left to bottom right by
// placing K 1s in the matrix
static int countPath(int N, int M, int K)
{
    int answer;
    if (K >= 2)
        answer = 0;
    else if (K == 0)
        answer = ncr(N + M - 2, N - 1);
    else 
    {
        // Count of ways without 1s
        answer = ncr(N + M - 2, N - 1);
  
        // Count of paths from starting
        // point to mid point
        int X = (N - 1) / 2 + (M - 1) / 2;
        int Y = (N - 1) / 2;
        int midCount = ncr(X, Y);
  
        // Count of paths from mid
        // point to end point
        X = ((N - 1) - (N - 1) / 2) + 
            ((M - 1) - (M - 1) / 2);
        Y = ((N - 1) - (N - 1) / 2);
        midCount *= ncr(X, Y);
        answer -= midCount;
    }
    return answer;
}
  
// Driver Code
public static void main(String[] args)
{
    int N = 3;
    int M = 3;
    int K = 1;
    System.out.print(countPath(N, M, K));
}
}
  
// This code is contributed by shikhasingrajput

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

#Python3 Program to implememnt
#the above approach
#Function to return the value of
#Binomial Coefficient C(n, k)
def ncr(n, k):
    res = 1
  
    #Since C(n, k) = C(n, n-k)
    if (k > n - k):
        k = n - k
  
    #Calculate the value of
    #[n * (n-1) *---* (n-k+1)] /
    #[k * (k-1) *----* 1]
    for i in range(k):
        res *= (n - i)
        res //= (i + 1)
  
    return res
  
#Function to find the minimum
#count of paths from top
#left to bottom right by
#placing K 1s in the matrix
def countPath(N, M, K):
    answer = 0
    if (K >= 2):
        answer = 0
    elif (K == 0):
        answer = ncr(N + M - 2, N - 1)
    else:
  
        #Count of ways without 1s
        answer = ncr(N + M - 2, N - 1)
  
        #Count of paths from starting
        #poto mid point
        X = (N - 1) // 2 + (M - 1) // 2
        Y = (N - 1) // 2
        midCount = ncr(X, Y)
  
        #Count of paths from mid
        #poto end point
        X = ((N - 1) - (N - 1) // 2)+ 
            ((M - 1) - (M - 1) // 2)
  
        Y = ((N - 1) - (N - 1) // 2)
        midCount *= ncr(X, Y)
        answer -= midCount
  
    return answer
  
#Driver Code
if __name__ == '__main__':
    N = 3
    M = 3
    K = 1
    print(countPath(N, M, K))
  
# This code is contributed by Mohit Kumar 29

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to implememnt
// the above approach
using System;
  
class GFG{
  
// Function to return the value of
// Binomial Coefficient C(n, k)
static int ncr(int n, int k)
{
    int res = 1;
  
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
  
    // Calculate the value of
    // [n * (n-1) *---* (n-k+1)] /
    // [k * (k-1) *----* 1]
    for(int i = 0; i < k; ++i) 
    {
        res *= (n - i);
        res /= (i + 1);
    }
    return res;
}
  
// Function to find the minimum
// count of paths from top
// left to bottom right by
// placing K 1s in the matrix
static int countPath(int N, int M, int K)
{
    int answer;
      
    if (K >= 2)
        answer = 0;
    else if (K == 0)
        answer = ncr(N + M - 2, N - 1);
    else
    {
          
        // Count of ways without 1s
        answer = ncr(N + M - 2, N - 1);
  
        // Count of paths from starting
        // point to mid point
        int X = (N - 1) / 2 + (M - 1) / 2;
        int Y = (N - 1) / 2;
        int midCount = ncr(X, Y);
  
        // Count of paths from mid
        // point to end point
        X = ((N - 1) - (N - 1) / 2) + 
            ((M - 1) - (M - 1) / 2);
        Y = ((N - 1) - (N - 1) / 2);
          
        midCount *= ncr(X, Y);
        answer -= midCount;
    }
    return answer;
}
  
// Driver Code
public static void Main(String[] args)
{
    int N = 3;
    int M = 3;
    int K = 1;
      
    Console.Write(countPath(N, M, K));
}
}
  
// This code is contributed by Amit Katiyar

chevron_right


Output: 

2

Time Complexity: O(N+M)
Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Recommended Posts:


Active and well versed member of Competitive Programming

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.