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Count of moves to escape given Matrix from given position based on given conditions

  • Last Updated : 03 Dec, 2021

Given an N x M matrix mat[][] where initially we are standing at the cell with index (i, j), the task is to find the number of operations required to escape the given matrix where at each operation mat[x][y] can be reached from mat[i][j] such that x represents the count of 0’s in the binary representation of mat[i][j] and y represents the count of 1’s in the binary representation of mat[i][j].

Examples:

Input: mat[][] = {{5, 13, 8, 1}, {7, 9, 2, 15}, {12, 4, 8, 3}}, i = 0, j = 0
Output: 4
Explanation: Initially, the current position is mat[0][0] = 5. Therefore, the index reachable from 5 is mat[1][2] as count of 0’s in 5 is 1 and count of 1’s in 5 is 2. Hence the first jump is from mat[0][0] => mat[1][2]. Similarly, the jumps are in the following order: mat[0][0] => mat[1][2] => mat[1][1] => mat[2][2] => mat[3][1]. Since, the index (3, 1) does not exist in the given matrix, the number of required operations are 4.

Input: mat[][] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}, i = 1, j = 2
Output: -1
Explanation: It is not possible to escape the matrix from the index.

 

Approach: The given problem can be solved with the help of Brian Kernighan’s Algorithm and the inbuilt log function. Below are the steps to follow:

  • Create a variable Jump, which stores the required number of jump operations to escape the given 2D array. Initially, Jump = 0.
  • If the initial cell is already out of bounds of the given matrix, return 0.
  • Calculate the count of 0’s and 1’s in the binary representation of mat[i][j].
  • Jump to the next valid index and set mat[i][j] as -1. Also, increment the value of Jump by 1.
  • Check if the current value of mat[i][j] = -1 after the jump. It means that the current index is already visited, hence creating a never-ending loop. Therefore, return -1.
  • Repeat the above process until the current index is already visited or it is out of bounds of the given matrix.

Below is the implementation of the above approach: 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
#define Max 100
using namespace std;
 
// Function to find total count of
// set bits in any number
int brianKernighan(int N)
{
    int Count = 0;
    while (N) {
        N = N & (N - 1);
        Count += 1;
    }
    return Count;
}
 
// Function to find left most Set bit
// position in any number
int getMsbIndex(int N)
{
    return ((int)log2(N) + 1);
}
 
// Function to find the number of
// jumps to escape the matrix from
// the given index [i, j]
int countJumps(int Mat[][Max],
               int N, int M,
               int i, int j)
{
    // Initialize the variable Jump
    int C0, C1, Jump = 0;
 
    while (i < N && j < M) {
 
        // When the element is already visited
        // then a closed loop is formed and
        // it is impossible to escape then
        // return -1.
        if (Mat[i][j] == -1)
            return -1;
 
        // Calculate Count of 1 in Mat[i][j]
        C1 = brianKernighan(Mat[i][j]);
 
        // Calculate Count of 0 in Mat[i][j]
        C0 = getMsbIndex(Mat[i][j]) - C1;
 
        // Set the element Mat[i][j] visited
        Mat[i][j] = -1;
 
        // Set i and j to count if 0 and 1
        i = C0, j = C1;
 
        // Increment Jump by 1
        Jump++;
    }
 
    // Return number of Jumps to escape
    // the matrix if it is possible to
    // escape
    return Jump;
}
 
// Driver Code
int main()
{
    int N = 3, M = 4;
    int i = 0, j = 0;
    int Mat[][Max] = { { 5, 13, 8, 1 },
                       { 7, 9, 2, 15 },
                       { 12, 4, 8, 3 } };
 
    cout << countJumps(Mat, N, M, i, j);
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG {
 
// Function to find total count of
// set bits in any number
static int brianKernighan(int N)
{
    int Count = 0;
    while (N != 0) {
        N = N & (N - 1);
        Count += 1;
    }
    return Count;
}
 
// Function to find left most Set bit
// position in any number
static int getMsbIndex(int N)
{
    return ((int)(Math.log(N) / Math.log(2)) + 1);
}
 
// Function to find the number of
// jumps to escape the matrix from
// the given index [i, j]
static int countJumps(int Mat[][],
               int N, int M,
               int i, int j)
{
   
    // Initialize the variable Jump
    int C0, C1, Jump = 0;
 
    while (i < N && j < M) {
 
        // When the element is already visited
        // then a closed loop is formed and
        // it is impossible to escape then
        // return -1.
        if (Mat[i][j] == -1)
            return -1;
 
        // Calculate Count of 1 in Mat[i][j]
        C1 = brianKernighan(Mat[i][j]);
 
        // Calculate Count of 0 in Mat[i][j]
        C0 = getMsbIndex(Mat[i][j]) - C1;
 
        // Set the element Mat[i][j] visited
        Mat[i][j] = -1;
 
        // Set i and j to count if 0 and 1
        i = C0; j = C1;
 
        // Increment Jump by 1
        Jump++;
    }
 
    // Return number of Jumps to escape
    // the matrix if it is possible to
    // escape
    return Jump;
}
 
// Driver Code
public static void main (String[] args) {
         
    int N = 3, M = 4;
    int i = 0, j = 0;
    int Mat[][] = { { 5, 13, 8, 1 },
                       { 7, 9, 2, 15 },
                       { 12, 4, 8, 3 } };
 
    System.out.println(countJumps(Mat, N, M, i, j));
}
}
 
// This code is contributed by target_2.

Python3




# Python Program to implement
# the above approach
import math as Math
Max = 100
 
# Function to find total count of
# set bits in any number
def brianKernighan(N):
    Count = 0
    while (N):
        N = N & (N - 1)
        Count += 1
    return Count
 
# Function to find left most Set bit
# position in any number
def getMsbIndex(N):
    return Math.floor(Math.log2(N) + 1)
 
# Function to find the number of
# jumps to escape the matrix from
# the given index [i, j]
def countJumps(Mat, N, M, i, j):
   
    # Initialize the variable Jump
    Jump = 0
 
    while (i < N and j < M):
 
        # When the element is already visited
        # then a closed loop is formed and
        # it is impossible to escape then
        # return -1.
        if (Mat[i][j] == -1):
            return -1
 
        # Calculate Count of 1 in Mat[i][j]
        C1 = brianKernighan(Mat[i][j])
 
        # Calculate Count of 0 in Mat[i][j]
        C0 = getMsbIndex(Mat[i][j]) - C1
 
        # Set the element Mat[i][j] visited
        Mat[i][j] = -1
 
        # Set i and j to count if 0 and 1
        i = C0
        j = C1
 
        # Increment Jump by 1
        Jump += 1
 
    # Return number of Jumps to escape
    # the matrix if it is possible to
    # escape
    return Jump
 
# Driver Code
N = 3
M = 4
i = 0
j = 0
Mat = [[5, 13, 8, 1],
       [7, 9, 2, 15],
       [12, 4, 8, 3]]
 
print(countJumps(Mat, N, M, i, j))
 
# This code is contributed by gfgking.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find total count of
// set bits in any number
static int brianKernighan(int N)
{
    int Count = 0;
    while (N != 0) {
        N = N & (N - 1);
        Count += 1;
    }
    return Count;
}
 
// Function to find left most Set bit
// position in any number
static int getMsbIndex(int N)
{
    return ((int)(Math.Log(N) / Math.Log(2)) + 1);
}
 
// Function to find the number of
// jumps to escape the matrix from
// the given index [i, j]
static int countJumps(int[,] Mat,
               int N, int M,
               int i, int j)
{
   
    // Initialize the variable Jump
    int C0, C1, Jump = 0;
 
    while (i < N && j < M) {
 
        // When the element is already visited
        // then a closed loop is formed and
        // it is impossible to escape then
        // return -1.
        if (Mat[i, j] == -1)
            return -1;
 
        // Calculate Count of 1 in Mat[i][j]
        C1 = brianKernighan(Mat[i, j]);
 
        // Calculate Count of 0 in Mat[i][j]
        C0 = getMsbIndex(Mat[i, j]) - C1;
 
        // Set the element Mat[i][j] visited
        Mat[i, j] = -1;
 
        // Set i and j to count if 0 and 1
        i = C0; j = C1;
 
        // Increment Jump by 1
        Jump++;
    }
 
    // Return number of Jumps to escape
    // the matrix if it is possible to
    // escape
    return Jump;
}
 
// Driver Code
public static void Main()
{
    int N = 3, M = 4;
    int i = 0, j = 0;
    int[,] Mat = {{ 5, 13, 8, 1 },
                       { 7, 9, 2, 15 },
                       { 12, 4, 8, 3 } };
 
    Console.Write(countJumps(Mat, N, M, i, j));
}
}
 
// This code is contributed by sanjoy_62.

Javascript




<script>
 
        // JavaScript Program to implement
        // the above approach   
        var Max = 100
 
        // Function to find total count of
        // set bits in any number
        function brianKernighan(N) {
            let Count = 0;
            while (N) {
                N = N & (N - 1);
                Count += 1;
            }
            return Count;
        }
 
        // Function to find left most Set bit
        // position in any number
        function getMsbIndex(N) {
            return Math.floor(Math.log2(N) + 1);
        }
 
        // Function to find the number of
        // jumps to escape the matrix from
        // the given index [i, j]
        function countJumps(Mat,
            N, M,
            i, j) {
            // Initialize the variable Jump
            let C0, C1, Jump = 0;
 
            while (i < N && j < M) {
 
                // When the element is already visited
                // then a closed loop is formed and
                // it is impossible to escape then
                // return -1.
                if (Mat[i][j] == -1)
                    return -1;
 
                // Calculate Count of 1 in Mat[i][j]
                C1 = brianKernighan(Mat[i][j]);
 
                // Calculate Count of 0 in Mat[i][j]
                C0 = getMsbIndex(Mat[i][j]) - C1;
 
                // Set the element Mat[i][j] visited
                Mat[i][j] = -1;
 
                // Set i and j to count if 0 and 1
                i = C0, j = C1;
 
                // Increment Jump by 1
                Jump++;
            }
 
            // Return number of Jumps to escape
            // the matrix if it is possible to
            // escape
            return Jump;
        }
 
        // Driver Code
        let N = 3, M = 4;
        let i = 0, j = 0;
        let Mat = [[5, 13, 8, 1],
        [7, 9, 2, 15],
        [12, 4, 8, 3]];
 
        document.write(countJumps(Mat, N, M, i, j));
 
    // This code is contributed by Potta Lokesh
    </script>

 
 

Output
4

 

Time Complexity: O(N * M * log(M * N))
Auxiliary Space: O(1)

 


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