# 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++

 `// C++ Program to implememnt ` `// the above approach ` ` `  `#include ` `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; ` `} `

## Java

 `// 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`

## Python3

 `#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`

## C#

 `// 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`

Output:

```2
```

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

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