# Submatrix of given size with maximum 1’s

Given a binary matrix mat[][] and an integer K, the task is to find the submatrix of size K*K such that it contains maximum number of 1’s in the matrix.

Examples:

Input: mat[][] = {{1, 0, 1}, {1, 1, 0}, {1, 0, 0}}, K = 2
Output: 3
Explanation:
In the given matrix, there are 4 sub-matrix of order 2*2,
|1 0| |0 1| |1 1| |1 0|
|1 1|, |1 0|, |1 0|, |0 0|
Out of these sub-matrix, two matrix contains 3, 1’s.

Input: mat[][] = {{1, 0}, {0, 1}}, K = 1
Output: 1
Explanation:
In the given matrix, there are 4 sub-matrix of order 1*1,
|1|, |0|, |1|, |0|
Out of these sub-matrix, two matrix contains 1, 1’s.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to use the sliding window technique to solve this problem, In this technique, we generally compute the value of one window and then slide the window one-by-one to compute the solution for every window of size K.

To compute the maximum 1’s submatrix, count the number of 1’s in the row for every possible window of size K using the sliding window technique and store the counts of the 1’s in the form of a matrix.
For Example:

```Let the matrix be {{1,0,1}, {1, 1, 0}} and K = 2

For Row 1 -
Subarray 1: (1, 0), Count of 1 = 1
Subarray 2: (0, 1), Count of 1 = 1

For Row 2 -
Subarray 1: (1, 1), Count of 1 = 2
Subarray 2: (1, 0), Count of 1 = 1

Then the final matrix for count of 1's will be -
[ 1, 1 ]
[ 2, 1 ]
```

Similarly, apply the sliding window technique on every column on this matrix, to compute the count of 1’s in every possible sub-matrix and take the maximum out of those counts.

Below is the implementation of the above approach:

 `// C++ implementation to find the  ` `// maximum count of 1's in  ` `// submatrix of order K  ` `#include   ` `using` `namespace` `std; ` ` `  `// Function to find the maximum  ` `// count of 1's in the  ` `// submatrix of order K  ` `int` `maxCount(vector> &mat, ``int` `k) { ` ` `  `    ``int` `n = mat.size(); ` `    ``int` `m = mat[0].size();  ` `    ``vector> a; ` ` `  `    ``// Loop to find the count of 1's  ` `    ``// in every possible windows  ` `    ``// of rows of matrix  ` `    ``for` `(``int` `e = 0; e < n; ++e){  ` `        ``vector<``int``> s = mat[e]; ` `        ``vector<``int``> q; ` `        ``int`    `c = 0; ` `         `  `        ``// Loop to find the count of  ` `        ``// 1's in the first window  ` `        ``int` `i; ` `        ``for` `(i = 0; i < k; ++i) ` `            ``if``(s[i] == 1) ` `                ``c += 1; ` ` `  `        ``q.push_back(c); ` `        ``int` `p = s[0]; ` `         `  `        ``// Loop to find the count of  ` `        ``// 1's in the remaining windows  ` `        ``for` `(``int` `j = i + 1; j < m; ++j) {  ` `            ``if``(s[j] == 1) ` `                ``c+= 1; ` `            ``if``(p == 1) ` `                ``c-= 1; ` `            ``q.push_back(c); ` `            ``p = s[j-k + 1]; ` `        ``} ` `        ``a.push_back(q); ` `    ``} ` ` `  `    ``vector> b; ` `    ``int` `max = 0; ` `     `  `    ``// Loop to find the count of 1's  ` `    ``// in every possible submatrix  ` `    ``for` `(``int` `i = 0; i < a[0].size(); ++i) {  ` `        ``int` `c = 0; ` `        ``int` `p = a[0][i]; ` `         `  `        ``// Loop to find the count of  ` `        ``// 1's in the first window  ` `        ``int` `j; ` `        ``for` `(j = 0; j < k; ++j) { ` `            ``c+= a[j][i]; ` `        ``} ` `        ``vector<``int``> q;  ` `        ``if` `(c>max)  ` `            ``max = c; ` `        ``q.push_back(c); ` `         `  `        ``// Loop to find the count of  ` `        ``// 1's in the remaining windows  ` `        ``for` `(``int` `l = j + 1; j < n; ++j) {  ` `            ``c+= a[l][i]; ` `            ``c-= p; ` `            ``p = a[l-k + 1][i]; ` `            ``q.push_back(c); ` `            ``if` `(c > max) ` `                ``max = c; ` `        ``} ` ` `  `        ``b.push_back(q); ` `    ``} ` ` `  `    ``return` `max; ` `} ` ` `  `// Driver code  ` `int` `main()  ` `{  ` `    ``vector> mat = {{1, 0, 1}, {1, 1, 0}, {0, 1, 0}}; ` `    ``int` `k = 3; ` `     `  `    ``// Function call  ` `    ``cout<< maxCount(mat, k); ` ` `  `    ``return` `0;  ` `} `

 `# Python implementation to find the ` `# maximum count of 1's in  ` `# submatrix of order K ` ` `  `# Function to find the maximum ` `# count of 1's in the  ` `# submatrix of order K ` `def` `maxCount(mat, k): ` `    ``n, m ``=` `len``(mat), ``len``(mat[``0``]) ` `    ``a ``=``[] ` `     `  `    ``# Loop to find the count of 1's ` `    ``# in every possible windows  ` `    ``# of rows of matrix  ` `    ``for` `e ``in` `range``(n): ` `        ``s ``=` `mat[e] ` `        ``q ``=``[] ` `        ``c ``=` `0` `         `  `        ``# Loop to find the count of  ` `        ``# 1's in the first window ` `        ``for` `i ``in` `range``(k): ` `            ``if` `s[i] ``=``=` `1``: ` `                ``c ``+``=` `1` `        ``q.append(c) ` `        ``p ``=` `s[``0``] ` `         `  `        ``# Loop to find the count of  ` `        ``# 1's in the remaining windows ` `        ``for` `j ``in` `range``(i ``+` `1``, m): ` `            ``if` `s[j]``=``=``1``: ` `                ``c``+``=` `1` `            ``if` `p ``=``=``1``: ` `                ``c``-``=` `1` `            ``q.append(c) ` `            ``p ``=` `s[j``-``k ``+` `1``] ` `        ``a.append(q) ` `    ``b ``=``[] ` `    ``max` `=` `0` `     `  `    ``# Loop to find the count of 1's  ` `    ``# in every possible submatrix ` `    ``for` `i ``in` `range``(``len``(a[``0``])): ` `        ``c ``=` `0` `        ``p ``=` `a[``0``][i] ` `         `  `        ``# Loop to find the count of ` `        ``# 1's in the first window ` `        ``for` `j ``in` `range``(k): ` `            ``c``+``=` `a[j][i] ` `        ``q ``=``[] ` `        ``if` `c>``max``: ` `            ``max` `=` `c ` `        ``q.append(c) ` `         `  `        ``# Loop to find the count of ` `        ``# 1's in the remaining windows ` `        ``for` `l ``in` `range``(j ``+` `1``, n): ` `            ``c``+``=` `a[l][i] ` `            ``c``-``=` `p ` `            ``p ``=` `a[l``-``k ``+` `1``][i] ` `            ``q.append(c) ` `            ``if` `c > ``max``: ` `                ``max` `=` `c ` `        ``b.append(q) ` `    ``return` `max` `     `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``mat ``=` `[[``1``, ``0``, ``1``], [``1``, ``1``, ``0``], [``0``, ``1``, ``0``]] ` `    ``k ``=` `3` `     `  `    ``# Function call ` `    ``print``(maxCount(mat, k)) `

Output:
```5
```

Performance Analysis:

• Time Complexity: As in the above approach, there are two loops which takes O(N*M) time, Hence the Time Complexity will be O(N*M).
• Space Complexity: As in the above approach, there is extra space used, Hence the space complexity will be O(N).

Approach 2: [Dynamic Programming method] In this technique, we compute the dp[][] matrix using given mat[][] array.In dp[][] array we compute number of 1’s till the index (i,j) using previous dp[][] value and store it in dp[i][j] .

Algorithm :

```
1) Construct a dp[][] matrix and assign all elements to 0

initial dp[0][0] = mat[0][0]

a) compute first row and column of the dp matrix:

i) for first row:

dp[0][i] = dp[0][i-1] + mat[0][i]

ii) for first column:

dp[i][0] = dp[i-1][0] + mat[i][0]

b) now compute remaining dp matrix from (1,1) to (n,m):

dp[i][j] = mat[i][j] + dp[i-1][j] + dp[i][j-1] - dp[i-1][j-1]

2)now, we find the maximum 1's in k X k sub matrix:

a) initially we assign max = dp[k-1][k-1]

b) now first we have to check maximum for k-1 row and k-1 column:

i) for k-1 row:

if dp[k-1][j] - dp[k-1][j-k] > max:

max = dp[k-1][j] - dp[k-1][j-k]

ii) for k-1 column:

if dp[i][k-1] - dp[i-k][k-1] > max:

max = dp[i][k-1] - dp[i-k][k-1]

c) now, we check max for (k to n) row and (k to m) column:

for i from k to n-1:

for j from k to m-1:

if dp[i][j] + dp[i-k][j-k] - dp[i-k][j] - dp[i][j-k] > max:

max = dp[i][j] + dp[i-k][j-k] - dp[i-k][j] - dp[i][j-k]

now just return the max value.

```

Below is the implementation of the above approach:

 `#python3 approach  ` ` `  `def` `findMaxK(dp,k,n,m): ` `     `  `    ``# assign first kXk matrix initial value as max ` `    ``max_ ``=` `dp[k``-``1``][k``-``1``] ` `     `  `     `  `    ``for` `i ``in` `range``(k,n): ` `        ``su ``=` `dp[i``-``k][k``-``1``] ` `        ``if` `max_ < su: ` `            ``max_ ``=` `su ` `     `  `    ``for` `j ``in` `range``(k,m): ` `        ``su ``=` `dp[k``-``1``][i``-``k] ` `        ``if` `max_< su: ` `            ``max_ ``=` `su ` `             `  `    ``for` `i ``in` `range``(k,n): ` `        ``for` `j ``in` `range``(k,m): ` `            ``su ``=` `dp[i][j] ``+` `dp[i``-``k][j``-``k] ``-` `dp[i``-``k][j] ``-` `dp[i][j``-``k] ` `            ``if` `max_ < su: ` `                ``max_ ``=` `su ` `             `  `    ``return` `max_ ` `     `  `def` `buildDPdp(mat,k,n,m): ` `     `  `    ``# assign mXn dp list to 0 ` `    ``dp ``=` `[[``0` `for` `i ``in` `range``(m)] ``for` `j ``in` `range``(n)] ` `     `  `    ``# assign initial starting value ` `    ``dp[``0``][``0``] ``=` `mat[``0``][``0``] ` `     `  `    ``for` `i ``in` `range``(``1``,m): ` `        ``dp[``0``][i] ``+``=` `(dp[``0``][i``-``1``]``+``mat[``0``][i]) ` `     `  `    ``for` `i ``in` `range``(``1``,n): ` `        ``dp[i][``0``] ``+``=` `(dp[i``-``1``][``0``]``+``mat[i][``0``]) ` `     `  `     `  `    ``for` `i ``in` `range``(``1``,n): ` `        ``for` `j ``in` `range``(``1``,m): ` `            ``dp[i][j] ``=` `dp[i``-``1``][j] ``+` `dp[i][j``-``1``] ``+` `mat[i][j] ``-` `dp[i``-``1``][j``-``1``] ` ` `  `    ``return` `dp ` ` `  `def` `maxOneInK(mat,k): ` `     `  `    ``# n is colums ` `    ``n ``=` `len``(mat) ` `     `  `    ``# m is rows ` `    ``m ``=` `len``(mat[``0``]) ` `     `  `    ``#build dp list ` `    ``dp ``=` `buildDPdp(mat,k,n,m) ` `     `  `    ``# call the function and return its value ` `    ``return` `findMaxK(dp,k,n,m) ` `     `  `     `  `         `  ` `  `def` `main(): ` `    ``# mXn matrix  ` `    ``mat ``=` `[[``1``, ``0``, ``1``], [``1``, ``1``, ``0``], [``0``, ``1``, ``0``]] ` `     `  `    ``k ``=` `3` `     `  `    ``#callind function ` `    ``print``(maxOneInK(mat,k)) ` ` `  `#driver code ` `main() ` ` `  ` `  `#This code is contributed by Tokir Manva `

Output:
```5
```

Performance Analysis:

• Time Complexity: As in the above Dynamic program approach we have to calculate N X M dp matrix which takes O(N*M) time, Hence the Time Complexity will be O(N*M).
• Space Complexity: As in the above approach, there is extra space used for making dp N X M matrix, Hence the space complexity will be O(N*M).

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