Number of square matrices with all 1s
Given an N*M matrix containing only 0s and 1s, the task is to count the number of square submatrices containing all 1s.
Examples:
Input: arr[][] = {{0, 1, 1, 1},
{1, 1, 1, 1},
{0, 1, 1, 1}}
Output: 15
Explanation:
There are 10 squares of side length 1.
There are 4 squares of side length 2.
There is 1 square of side length 3.
Total number of squares = 10 + 4 + 1 = 15.Input: arr[][] = {{1, 0, 1},
{1, 1, 0},
{1, 1, 0}}
Output: 7
Approach: This problem can be solved using Dynamic Programming.
- Let the array arr[i][j] store the number of square matrices ending at (i, j)
- The recurrence relation to find the number of squares ending at (i, j) can be given by:
- If arr[i][j] is 1:
- arr[i][j] = min( min(arr[i-1][j], arr[i][j-1]), arr[i-1][j-1]) + 1
- If arr[i][j] is 1:
- Else if arr[i][j] is 0:
- arr[i][j] = 0
- Calculate the sum of the array which is equal to the number of square submatrices with all 1s.
Below is the implementation of the above approach:
CPP
// C++ program to return the number of // square submatrices with all 1s #include <bits/stdc++.h> using namespace std; #define n 3 #define m 3 // Function to return the number of // square submatrices with all 1s int countSquareMatrices(int a[][m], int N, int M) { // Initialize count variable int count = 0; for (int i = 1; i < N; i++) { for (int j = 1; j < M; j++) { // If a[i][j] is equal to 0 if (a[i][j] == 0) continue; // Calculate number of // square submatrices // ending at (i, j) a[i][j] = min(min(a[i - 1][j], a[i][j - 1]), a[i - 1][j - 1]) + 1; } } // Calculate the sum of the array for (int i = 0; i < N; i++) for (int j = 0; j < M; j++) count += a[i][j]; return count; } // Driver code int main() { int arr[][m] = { { 1, 0, 1 }, { 1, 1, 0 }, { 1, 1, 0 } }; cout << countSquareMatrices(arr, n, m); return 0; } |
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Java
// Java program to return the number of // square submatrices with all 1s class GFG { final static int n = 3; final static int m = 3; // Function to return the number of // square submatrices with all 1s static int countSquareMatrices(int a[][], int N, int M) { // Initialize count variable int count = 0; for (int i = 1; i < N; i++) { for (int j = 1; j < M; j++) { // If a[i][j] is equal to 0 if (a[i][j] == 0) continue; // Calculate number of // square submatrices // ending at (i, j) a[i][j] = Math.min(Math.min(a[i - 1][j], a[i][j - 1]), a[i - 1][j - 1]) + 1; } } // Calculate the sum of the array for (int i = 0; i < N; i++) for (int j = 0; j < M; j++) count += a[i][j]; return count; } // Driver code public static void main (String[] args) { int arr[][] = { { 1, 0, 1 }, { 1, 1, 0 }, { 1, 1, 0 } }; System.out.println(countSquareMatrices(arr, n, m)); } } // This code is contributed by AnkitRai01 |
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Python
# Python3 program to return the number of # square submatrices with all 1s n = 3m = 3 # Function to return the number of # square submatrices with all 1s def countSquareMatrices(a, N, M): # Initialize count variable count = 0 for i in range(1, N): for j in range(1, M): # If a[i][j] is equal to 0 if (a[i][j] == 0): continue # Calculate number of # square submatrices # ending at (i, j) a[i][j] = min([a[i - 1][j], a[i][j - 1], a[i - 1][j - 1]])+1 # Calculate the sum of the array for i in range(N): for j in range(M): count += a[i][j] return count # Driver code arr = [ [ 1, 0, 1], [ 1, 1, 0 ], [ 1, 1, 0 ] ] print(countSquareMatrices(arr, n, m)) # This code is contributed by mohit kumar 29 |
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C#
// C# program to return the number of // square submatrices with all 1s using System; class GFG { static int n = 3; static int m = 3; // Function to return the number of // square submatrices with all 1s static int countSquareMatrices(int [,]a, int N, int M) { // Initialize count variable int count = 0; for (int i = 1; i < N; i++) { for (int j = 1; j < M; j++) { // If a[i][j] is equal to 0 if (a[i, j] == 0) continue; // Calculate number of // square submatrices // ending at (i, j) a[i, j] = Math.Min(Math.Min(a[i - 1, j], a[i, j - 1]), a[i - 1, j - 1]) + 1; } } // Calculate the sum of the array for (int i = 0; i < N; i++) for (int j = 0; j < M; j++) count += a[i, j]; return count; } // Driver code public static void Main() { int [,]arr = { { 1, 0, 1 }, { 1, 1, 0 }, { 1, 1, 0 } }; Console.WriteLine(countSquareMatrices(arr, n, m)); } } // This code is contributed by AnkitRai01 |
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Output :
7
Time Complexity: O(N*M)
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