Minimum steps required to visit all corners of an N * M grid

• Last Updated : 14 Sep, 2021

Given two integers N and M representing the dimensions of a 2D grid, and two integers R and C, representing the position of a block in that grid, the task is to find the minimum number of steps required to visit all the corners of the grid, starting from (R, C). In each step, it is allowed to move the side-adjacent block in the grid.

Examples:

Input: N = 2, M = 2, R = 1, C = 2
Output: 3

Explanation:
(1, 2) -> (1, 1) -> (2, 1) -> (2, 2)
Therefore, the required output is 3.

Input: N = 2, M = 3, R = 2, C = 2
Output:
Explanation:
(2, 2) -> (2, 3) -> (1, 3) -> (1, 2) -> (1, 1) -> (2, 1)
Therefore, the required output is 5.

Approach: The problem can be solved based on the following observations.

Minimum count of steps required to visit the block (i2, j2) starting from (i1, j1) is equal to abs(i2 – i1) + abs(j2 – j1)

Follow the steps given below to solve the problem:

• First visit the corner which takes minimum count of steps using the above observations.
• Visit the other corners of the grid by traversing the boundary of the grid either in clockwise or anticlockwise, depending on which will take the minimum count of steps to visit the corners.
• Finally, print the minimum count of steps obtained.

Below is the implementation of the above approach:

C++

 // C++ program to implement// the above approach#include using namespace std; // Function to find the minimum count of steps// required to visit all the corners of the gridint min_steps_required(int n, int m, int r, int c){     // Stores corner of the grid    int i, j;     // Stores minimum count of steps required    // to visit the first corner of the grid    int corner_steps_req = INT_MAX;     // Checking for leftmost upper corner    i = 1;    j = 1;    corner_steps_req = min(corner_steps_req,                           abs(r - i) + abs(j - c));     // Checking for leftmost down corner    i = n;    j = 1;    corner_steps_req = min(corner_steps_req,                           abs(r - i) + abs(j - c));     // Checking for rightmost upper corner    i = 1;    j = m;    corner_steps_req = min(corner_steps_req,                           abs(r - i) + abs(j - c));     // Checking for rightmost down corner    i = n;    j = m;    corner_steps_req = min(corner_steps_req,                           abs(r - i) + abs(j - c));     // Stores minimum count of steps required    // to visit remaining three corners of the grid    int minimum_steps = min(2 * (n - 1) + m - 1,                            2 * (m - 1) + n - 1);     return minimum_steps + corner_steps_req;} // Driver Codeint main(){     int n = 3;    int m = 2;    int r = 1;    int c = 1;     cout << min_steps_required(n, m, r, c);     return 0;}

Java

 // Java Program to implement the// above approachimport java.util.*;class GFG{  // Function to find the minimum count of steps// required to visit all the corners of the gridstatic int min_steps_required(int n, int m, int r, int c){     // Stores corner of the grid    int i, j;     // Stores minimum count of steps required    // to visit the first corner of the grid    int corner_steps_req = Integer.MAX_VALUE;     // Checking for leftmost upper corner    i = 1;    j = 1;    corner_steps_req = Math.min(corner_steps_req,                           Math.abs(r - i) + Math.abs(j - c));     // Checking for leftmost down corner    i = n;    j = 1;    corner_steps_req = Math.min(corner_steps_req,                           Math.abs(r - i) + Math.abs(j - c));     // Checking for rightmost upper corner    i = 1;    j = m;    corner_steps_req = Math.min(corner_steps_req,                           Math.abs(r - i) + Math.abs(j - c));     // Checking for rightmost down corner    i = n;    j = m;    corner_steps_req = Math.min(corner_steps_req,                           Math.abs(r - i) + Math.abs(j - c));     // Stores minimum count of steps required    // to visit remaining three corners of the grid    int minimum_steps = Math.min(2 * (n - 1) + m - 1,                            2 * (m - 1) + n - 1);     return minimum_steps + corner_steps_req;}  // Driver Codepublic static void main(String[] args){    int n = 3;    int m = 2;    int r = 1;    int c = 1;     System.out.print(min_steps_required(n, m, r, c));}} // This code is contributed by code_hunt.

Python3

 # Python3 program to implement# the above approachimport sysINT_MAX = sys.maxsize; # Function to find the minimum count of steps# required to visit all the corners of the griddef min_steps_required(n, m, r, c) :     # Stores corner of the grid    i = 0; j = 0;     # Stores minimum count of steps required    # to visit the first corner of the grid    corner_steps_req = INT_MAX;     # Checking for leftmost upper corner    i = 1;    j = 1;    corner_steps_req = min(corner_steps_req,                           abs(r - i) + abs(j - c));     # Checking for leftmost down corner    i = n;    j = 1;    corner_steps_req = min(corner_steps_req,                           abs(r - i) + abs(j - c));     # Checking for rightmost upper corner    i = 1;    j = m;    corner_steps_req = min(corner_steps_req,                           abs(r - i) + abs(j - c));     # Checking for rightmost down corner    i = n;    j = m;    corner_steps_req = min(corner_steps_req,                           abs(r - i) + abs(j - c));     # Stores minimum count of steps required    # to visit remaining three corners of the grid    minimum_steps = min(2 * (n - 1) + m - 1,                            2 * (m - 1) + n - 1);     return minimum_steps + corner_steps_req; # Driver Codeif __name__ == "__main__" :    n = 3;    m = 2;    r = 1;    c = 1;    print(min_steps_required(n, m, r, c));     # This code is contributed by AnkThon

C#

 // C# program to implement the// above approachusing System; class GFG{  // Function to find the minimum count// of steps required to visit all the// corners of the gridstatic int min_steps_required(int n, int m,                              int r, int c){         // Stores corner of the grid    int i, j;     // Stores minimum count of steps required    // to visit the first corner of the grid    int corner_steps_req = int.MaxValue;     // Checking for leftmost upper corner    i = 1;    j = 1;    corner_steps_req = Math.Min(corner_steps_req,                                Math.Abs(r - i) +                                Math.Abs(j - c));     // Checking for leftmost down corner    i = n;    j = 1;    corner_steps_req = Math.Min(corner_steps_req,                                Math.Abs(r - i) +                                Math.Abs(j - c));     // Checking for rightmost upper corner    i = 1;    j = m;    corner_steps_req = Math.Min(corner_steps_req,                                Math.Abs(r - i) +                                Math.Abs(j - c));     // Checking for rightmost down corner    i = n;    j = m;    corner_steps_req = Math.Min(corner_steps_req,                                Math.Abs(r - i) +                                Math.Abs(j - c));     // Stores minimum count of steps required    // to visit remaining three corners of the grid    int minimum_steps = Math.Min(2 * (n - 1) + m - 1,                                 2 * (m - 1) + n - 1);     return minimum_steps + corner_steps_req;}  // Driver Codepublic static void Main(String[] args){    int n = 3;    int m = 2;    int r = 1;    int c = 1;     Console.Write(min_steps_required(n, m, r, c));}} // This code is contributed by shikhasingrajput

Javascript


Output:
4

Time complexity: O(1)
Auxiliary space: O(1)

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