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Queries on number of Binary sub-matrices of Given size in C++
In this problem, we are given a binary matrix bin[][] of size nXm. Our task is to solve all q queries. For query(x, y), we need to find the number of submatrix of size x*x such that all the elements of array y (binary number).
Problem description
Here, we need to count the total number of sub-matrix of a given size that consists of only one of the two bits i.e. sub-matrix will all elements 0/1.
Let’s take an example to understand the problem,
Input
n = 3 , m = 4
bin[][] = {{ 1, 1, 0, 1}
{ 1, 1, 1, 0}
{ 0, 1, 1, 1}}
q = 1
q1 = (2, 1)Output
2
Explanation
All sub-matrix of size 2 with all element 1 are −
{{ 1, 1, 0, 1}
{ 1, 1, 1, 0}
{ 0, 1, 1, 1}}
{{ 1, 1, 0, 1}
{ 1, 1, 1, 0}
{ 0, 1, 1, 1}}The solution to this problem is using the Dynamic programming approach. To solve, we will maintain a 2D matric DP[][] to store the largest submatrix of the same bit. i.e. DP[i][j] will store the value of sub-matrix whose end index is (i, j), and all the elements are the same.
For your understanding, if DP[4][5] = 2, elements at bin[3][4], bin[3][5], bin[4][4] and bin[4][5] are same.
So, for finding DP[i][j], we have two cases −
Case 1 − if i = 0 orj = 0 : DP[i][j] = 1, as only one sub-matrix can be possible.
Case 2 − otherwise, bin[i-(k-1)][j] = bin[i][j - (k-1)] …: in this casae DP[i][j] = min(DP[i][j-1] , DP[i -1][j], DP[i-1][j-1] ) + 1. This will contribute to the sub-matrix like, we require. Let’s generalise, the case with k = 2, i.e. if we consider a sub-matrix of size 2X2. Then we need to check if bin[i][j] = bin[i] [j - 1] = bin[i- 1][j] = bin[i -1 ][j -1 ], if it is possible then, we will find the DP[i][j] for k =2 .
If the case 2 conditions are not met, we will set DP[i][j] = 1, which can be treated as a default value.
This value of DP[i][j] can be for a set bit or unset bit. We will check the value of bin[i][j] to see which of the set or unset values the k value belongs. To find the frequencies, we will create two arrays, zeroFrequrency to store the frequency of sub-matrix that is generated for 0. And oneFrequrency to store the frequency of sub-matrix that is generated for 1.
Program to illustrate the working of our solution,
Eample
#include <iostream>
using namespace std;
#define N 3
#define M 4
int min(int a, int b, int c) {
if (a <= b && a <= c)
return a;
else if (b <= a && b <= c)
return b;
else
return c;
}
int solveQuery(int n, int m, int bin[N][M], int x, int y){
int DP[n][m], max = 1;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (i == 0 || j == 0)
DP[i][j] = 1;
else if ((bin[i][j] == bin[i - 1][j]) && (bin[i][j] == bin[i][j - 1]) && (bin[i][j] == bin[i - 1][j - 1])) {
DP[i][j] = min(DP[i - 1][j], DP[i - 1][j - 1], DP[i][j - 1]) + 1;
if (max < DP[i][j])
max = DP[i][j];
}
else
DP[i][j] = 1;
}
}
int zeroFrequency[n+m] = { 0 }, oneFrequency[n+m] = { 0 };
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (bin[i][j] == 0)
zeroFrequency[DP[i][j]]++;
else
oneFrequency[DP[i][j]]++;
}
}
for (int i = max - 1; i >= 0; i--) {
zeroFrequency[i] += zeroFrequency[i + 1];
oneFrequency[i] += oneFrequency[i + 1];
}
if (y == 0)
return zeroFrequency[x];
else
return oneFrequency[x];
}
int main(){
int n = 3, m = 4;
int mat[N][M] =
{{ 1, 1, 0, 1},
{ 1, 1, 1, 0},
{ 0, 1, 1, 1}};
int Q = 2;
int query[Q][2] = {{ 2, 1}, { 1, 0}};
for(int i = 0; i < Q; i++){
cout<<"For Query "<<(i+1)<<": The number of Binary sub-matrices of Given size is " <<solveQuery(n, m, mat, query[i][0], query[i][1])<<"\n";
}
return 0;
}Output
For Query 1: The number of Binary sub-matrices of Given size is 2 For Query 2: The number of Binary sub-matrices of Given size is 3
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