Given binary square matrix [n*n]. Find maximum integer value in a path from top left to bottom right. We compute integer value using bits of traversed path. We start at index [0,0] and end at index [n-1][n-1]. from index [i, j], we can move [i, j+1] or [i+1, j].
Examples:
Input : mat[][] = {{1, 1, 0, 1},
{0, 1, 1, 0},
{1, 0, 0, 1},
{1, 0, 1, 1}}
Output : 111
Explanation :
Path : (0,0) -> (0,1) -> (1,1) -> (1,2) ->
(2,2) -> (3,2) ->(3,3)
Decimal value : 1*(2^0) + 1*(2^1) + 1*(2^2) + 1*(2^3) +
0*(2^4) + 1*(2^5) + 1*(2^6) = 111
The above problem can be recursively defined as below:
// p indicates power of 2, initially p = i = j = 0
MaxDecimalValue(mat, i, j, p)
// If i or j is our of boundary
If i >= n || j >= n
return 0
// Compute rest of matrix find maximum decimal value
result max(MaxDecimalValue(mat, i, j+1, p+1),
MaxDecimalValue(mat, i+1, j, p+1))
If mat[i][j] == 1
return power(2, p) + result
Else
return result
Below is the implementation of above recursive algorithm.
C++
// C++ program to find maximum decimal value path in// binary matrix#include<bits/stdc++.h>using namespace std;#define N 4// Returns maximum decimal value in binary matrix.// Here p indicate power of 2long long int maxDecimalValue(int mat[][N], int i, int j, int p){ // Out of matrix boundary if (i >= N || j >= N ) return 0; int result = max(maxDecimalValue(mat, i, j+1, p+1), maxDecimalValue(mat, i+1, j, p+1)); // If current matrix value is 1 then return result + // power(2, p) else result if (mat[i][j] == 1) return pow(2, p) + result; else return result;}//Driver programint main(){ int mat[][4] = {{ 1 ,1 ,0 ,1 }, { 0 ,1 ,1 ,0 }, { 1 ,0 ,0 ,1 }, { 1 ,0 ,1 ,1 }, }; cout << maxDecimalValue(mat, 0, 0, 0) << endl; return 0;} |
Java
// Java program to find maximum decimal value path in// binary matrixclass GFG { static final int N = 4;// Returns maximum decimal value in binary matrix.// Here p indicate power of 2 static int maxDecimalValue(int mat[][], int i, int j, int p) { // Out of matrix boundary if (i >= N || j >= N) { return 0; } int result = Math.max(maxDecimalValue(mat, i, j + 1, p + 1), maxDecimalValue(mat, i + 1, j, p + 1)); // If current matrix value is 1 then return result + // power(2, p) else result if (mat[i][j] == 1) { return (int) (Math.pow(2, p) + result); } else { return result; } }// Driver program public static void main(String[] args) { int mat[][] = {{1, 1, 0, 1}, {0, 1, 1, 0}, {1, 0, 0, 1}, {1, 0, 1, 1},}; System.out.println(maxDecimalValue(mat, 0, 0, 0)); }}//this code contributed by Rajput-Ji |
Python3
# Python3 program to find maximum decimal# value path in binary matrixN =4# Returns maximum decimal value in binary# matrix. Here p indicate power of 2def maxDecimalValue(mat, i, j, p): # Out of matrix boundary if i >= N or j >= N: return 0 result = max( maxDecimalValue(mat, i, j+1, p+1), maxDecimalValue(mat, i+1, j, p+1)) # If current matrix value is 1 then # return result + power(2, p) else # result if mat[i][j] == 1: return pow(2, p) + result else: return result# Driver Programmat = [ [1, 1, 0, 1], [0, 1, 1, 0], [1, 0, 0, 1], [1, 0, 1, 1] ]print(maxDecimalValue(mat, 0, 0, 0))# This code is contributed by Shrikant13. |
C#
// C# program to find maximum decimal value path in// binary matrixusing System;class GFG { static int N = 4;// Returns maximum decimal value in binary matrix.// Here p indicate power of 2 static int maxDecimalValue(int[,] mat, int i, int j,int p) { // Out of matrix boundary if (i >= N || j >= N) { return 0; } int result = Math.Max(maxDecimalValue(mat, i, j + 1, p + 1), maxDecimalValue(mat, i + 1, j, p + 1)); // If current matrix value is 1 then return result + // power(2, p) else result if (mat[i,j] == 1) { return (int) (Math.Pow(2, p) + result); } else { return result; } }// Driver program public static void Main() { int[,] mat = {{1, 1, 0, 1}, {0, 1, 1, 0}, {1, 0, 0, 1}, {1, 0, 1, 1},}; Console.Write(maxDecimalValue(mat, 0, 0, 0)); }}// This code is contributed by Ita_c. |
PHP
<?php// PHP program to find maximum // decimal value path in binary// matrix// Returns maximum decimal value// in binary matrix. Here p // indicate power of 2function maxDecimalValue($mat, $i, $j, $p){ $N=4; // Out of matrix boundary if ($i >= $N || $j >= $N ) return 0; $result = max(maxDecimalValue($mat, $i, $j + 1, $p + 1), maxDecimalValue($mat, $i + 1, $j, $p + 1)); // If current matrix value // is 1 then return result + // power(2, p) else result if ($mat[$i][$j] == 1) return pow(2, $p) + $result; else return $result;} // Driver Code $mat = array(array(1 ,1 ,0 ,1), array(0 ,1 ,1 ,0), array(1 ,0 ,0 ,1), array(1 ,0 ,1 ,1)); echo maxDecimalValue($mat, 0, 0, 0) ; // This code is contributed by nitin mittal. ?> |
Javascript
<script>// JavaScript program to find maximum// decimal value path in binary matrixlet N = 4; // Returns maximum decimal value in// binary matrix.Here p indicate power of 2function maxDecimalValue(mat, i, j, p) { // Out of matrix boundary if (i >= N || j >= N) { return 0; } let result = Math.max(maxDecimalValue(mat, i, j + 1, p + 1), maxDecimalValue(mat, i + 1, j, p + 1)); // If current matrix value is 1 then // return result + power(2, p) else result if (mat[i][j] == 1) { return (Math.pow(2, p) + result); } else { return result; }}// Driver Codelet mat = [ [ 1, 1, 0, 1 ], [ 0, 1, 1, 0 ], [ 1, 0, 0, 1 ], [ 1, 0, 1, 1 ] ];document.write(maxDecimalValue(mat, 0, 0, 0));// This code is contributed by souravghosh0416 </script> |
Output:
111
The time complexity of above recursive solution is exponential.
Space Complexity:O(1),since no extra space required.
Here matrix [3][3]
(2 2)
/ \
(1 2) (2 1)
/ \ / \
(0 2) (1 1) (1 1) (2 1)
/ \ / \ / \ / \
. . . . . . . .
. . . . . . . . and so no
If we see recursion tree of above recursive solution, we can observe overlapping sub-problems. Since the problem has overlapping subproblems, we can solve it efficiently using Dynamic Programming. Below is Dynamic Programming based solution.
Below is the implementation of above problem using Dynamic Programming
C++
// C++ program to find Maximum decimal value Path in// Binary matrix#include<bits/stdc++.h>using namespace std;#define N 4// Returns maximum decimal value in binary matrix.// Here p indicate power of 2long long int MaximumDecimalValue(int mat[][N], int n){ int dp[n][n]; memset(dp, 0, sizeof(dp)); if (mat[0][0] == 1) dp[0][0] = 1 ; // 1*(2^0) // Compute binary stream of first row of matrix // and store result in dp[0][i] for (int i=1; i<n; i++) { // indicate 1*(2^i) + result of previous if (mat[0][i] == 1) dp[0][i] = dp[0][i-1] + pow(2, i); // indicate 0*(2^i) + result of previous else dp[0][i] = dp[0][i-1]; } // Compute binary stream of first column of matrix // and store result in dp[i][0] for (int i = 1 ; i <n ; i++ ) { // indicate 1*(2^i) + result of previous if (mat[i][0] == 1) dp[i][0] = dp[i-1][0] + pow(2, i); // indicate 0*(2^i) + result of previous else dp[i][0] = dp[i-1][0]; } // Traversal rest Binary matrix and Compute maximum // decimal value for (int i=1 ; i < n ; i++ ) { for (int j=1 ; j < n ; j++ ) { // Here (i+j) indicate the current power of // 2 in path that is 2^(i+j) if (mat[i][j] == 1) dp[i][j] = max(dp[i][j-1], dp[i-1][j]) + pow(2, i+j); else dp[i][j] = max(dp[i][j-1], dp[i-1][j]); } } // Return maximum decimal value in binary matrix return dp[n-1][n-1];}// Driver programint main(){ int mat[][4] = {{ 1 ,1 ,0 ,1 }, { 0 ,1 ,1 ,0 }, { 1 ,0 ,0 ,1 }, { 1 ,0 ,1 ,1 }, }; cout << MaximumDecimalValue(mat, 4) << endl; return 0;} |
Java
// Java program to find Maximum decimal value Path in// Binary matrixpublic class GFG { final static int N = 4;// Returns maximum decimal value in binary matrix.// Here p indicate power of 2 static int MaximumDecimalValue(int mat[][], int n) { int dp[][] = new int[n][n]; if (mat[0][0] == 1) { dp[0][0] = 1; // 1*(2^0) } // Compute binary stream of first row of matrix // and store result in dp[0][i] for (int i = 1; i < n; i++) { // indicate 1*(2^i) + result of previous if (mat[0][i] == 1) { dp[0][i] = (int) (dp[0][i - 1] + Math.pow(2, i)); } // indicate 0*(2^i) + result of previous else { dp[0][i] = dp[0][i - 1]; } } // Compute binary stream of first column of matrix // and store result in dp[i][0] for (int i = 1; i < n; i++) { // indicate 1*(2^i) + result of previous if (mat[i][0] == 1) { dp[i][0] = (int) (dp[i - 1][0] + Math.pow(2, i)); } // indicate 0*(2^i) + result of previous else { dp[i][0] = dp[i - 1][0]; } } // Traversal rest Binary matrix and Compute maximum // decimal value for (int i = 1; i < n; i++) { for (int j = 1; j < n; j++) { // Here (i+j) indicate the current power of // 2 in path that is 2^(i+j) if (mat[i][j] == 1) { dp[i][j] = (int) (Math.max(dp[i][j - 1], dp[i - 1][j]) + Math.pow(2, i + j)); } else { dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]); } } } // Return maximum decimal value in binary matrix return dp[n - 1][n - 1]; }// Driver program public static void main(String[] args) { int mat[][] = {{1, 1, 0, 1}, {0, 1, 1, 0}, {1, 0, 0, 1}, {1, 0, 1, 1},}; System.out.println(MaximumDecimalValue(mat, 4)); }}/*This code is contributed by Rajput-Ji*/ |
Python3
# Python3 program to find Maximum decimal # value Path in# Binary matrixN=4# Returns maximum decimal value in binary matrix.# Here p indicate power of 2def MaximumDecimalValue(mat, n): dp=[[0 for i in range(n)] for i in range(n)] if (mat[0][0] == 1): dp[0][0] = 1 # 1*(2^0) # Compute binary stream of first row of matrix # and store result in dp[0][i] for i in range(1,n): # indicate 1*(2^i) + result of previous if (mat[0][i] == 1): dp[0][i] = dp[0][i-1] + 2**i # indicate 0*(2^i) + result of previous else: dp[0][i] = dp[0][i-1] # Compute binary stream of first column of matrix # and store result in dp[i][0] for i in range(1,n): # indicate 1*(2^i) + result of previous if (mat[i][0] == 1): dp[i][0] = dp[i-1][0] + 2**i # indicate 0*(2^i) + result of previous else: dp[i][0] = dp[i-1][0] # Traversal rest Binary matrix and Compute maximum # decimal value for i in range(1,n): for j in range(1,n): # Here (i+j) indicate the current power of # 2 in path that is 2^(i+j) if (mat[i][j] == 1): dp[i][j] = max(dp[i][j-1], dp[i-1][j])+(2**(i+j)) else: dp[i][j] = max(dp[i][j-1], dp[i-1][j]) # Return maximum decimal value in binary matrix return dp[n-1][n-1]# Driver program if __name__=='__main__': mat = [[ 1 ,1 ,0 ,1 ], [ 0 ,1 ,1 ,0 ], [ 1 ,0 ,0 ,1 ], [ 1 ,0 ,1 ,1 ]] print (MaximumDecimalValue(mat, 4))#this code is contributed by sahilshelangia |
C#
// C# program to find Maximum decimal value Path in// Binary matrixusing System;public class GFG { readonly static int N = 4;// Returns maximum decimal value in binary matrix.// Here p indicate power of 2 static int MaximumDecimalValue(int [,]mat, int n) { int [,]dp = new int[n,n]; if (mat[0,0] == 1) { dp[0,0] = 1; // 1*(2^0) } // Compute binary stream of first row of matrix // and store result in dp[0,i] for (int i = 1; i < n; i++) { // indicate 1*(2^i) + result of previous if (mat[0,i] == 1) { dp[0,i] = (int) (dp[0,i - 1] + Math.Pow(2, i)); } // indicate 0*(2^i) + result of previous else { dp[0,i] = dp[0,i - 1]; } } // Compute binary stream of first column of matrix // and store result in dp[i,0] for (int i = 1; i < n; i++) { // indicate 1*(2^i) + result of previous if (mat[i,0] == 1) { dp[i,0] = (int) (dp[i - 1,0] + Math.Pow(2, i)); } // indicate 0*(2^i) + result of previous else { dp[i,0] = dp[i - 1,0]; } } // Traversal rest Binary matrix and Compute maximum // decimal value for (int i = 1; i < n; i++) { for (int j = 1; j < n; j++) { // Here (i+j) indicate the current power of // 2 in path that is 2^(i+j) if (mat[i,j] == 1) { dp[i,j] = (int) (Math.Max(dp[i,j - 1], dp[i - 1,j]) + Math.Pow(2, i + j)); } else { dp[i,j] = Math.Max(dp[i,j - 1], dp[i - 1,j]); } } } // Return maximum decimal value in binary matrix return dp[n - 1,n - 1]; } // Driver program public static void Main() { int [,]mat = {{1, 1, 0, 1}, {0, 1, 1, 0}, {1, 0, 0, 1}, {1, 0, 1, 1},}; Console.Write(MaximumDecimalValue(mat, 4)); }}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript program to find Maximum decimal value Path in// Binary matrix let N = 4; // Returns maximum decimal value in binary matrix. // Here p indicate power of 2 function MaximumDecimalValue(mat,n) { let dp=new Array(n); for(let i = 0; i < n; i++) { dp[i] = new Array(n); for(let j = 0; j < n; j++) { dp[i][j] = 0; } } if (mat[0][0] == 1) { dp[0][0] = 1; // 1*(2^0) } // Compute binary stream of first row of matrix // and store result in dp[0][i] for (let i = 1; i < n; i++) { // indicate 1*(2^i) + result of previous if (mat[0][i] == 1) { dp[0][i] = dp[0][i - 1] + Math.pow(2, i); } // indicate 0*(2^i) + result of previous else { dp[0][i] = dp[0][i - 1]; } } // Compute binary stream of first column of matrix // and store result in dp[i][0] for (let i = 1; i < n; i++) { // indicate 1*(2^i) + result of previous if (mat[i][0] == 1) { dp[i][0] = Math.floor(dp[i - 1][0] + Math.pow(2, i)); } // indicate 0*(2^i) + result of previous else { dp[i][0] = dp[i - 1][0]; } } // Traversal rest Binary matrix and Compute maximum // decimal value for (let i = 1; i < n; i++) { for (let j = 1; j < n; j++) { // Here (i+j) indicate the current power of // 2 in path that is 2^(i+j) if (mat[i][j] == 1) { dp[i][j] = Math.floor(Math.max(dp[i][j - 1], dp[i - 1][j]) + Math.pow(2, i + j)); } else { dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]); } } } // Return maximum decimal value in binary matrix return dp[n - 1][n - 1]; } // Driver program let mat = [[ 1 ,1 ,0 ,1 ], [ 0 ,1 ,1 ,0 ], [ 1 ,0 ,0 ,1 ], [ 1 ,0 ,1 ,1 ]]; document.write(MaximumDecimalValue(mat, 4)) // This code is contributed by rag2127.</script> |
Output:
111
Time Complexity : O(n2)
Auxiliary space : O(n2)
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