On a two-dimensional plane, the number of jumps required to reach point (d, 0) from the origin

In this article we will discuss a possible solution to an exciting analytical problem, namely, reaching the point (d, 0) How many jumps are required. We will use a fixed jump length and target coordinates to find the minimum number of jumps required.

Input and output scenarios

Assume that the jump length can be a or b, and the target point is (d,0). The given output is then the minimum number of jumps required to reach the goal.

Input: a = 7, b = 5, d = 9
Output: 2
Input: a = 7, b = 5, d = 5
Output: 1
Input: a = 7, b = 5, d = 24
Output: 4

Suppose you are standing at the origin (0, 0) of the 2D plane. Your target coordinates are (d, 0). The only way to reach the target coordinates is to make a fixed-length jump. Your goal is to find an efficient way to reach your goal with the fewest number of jumps.

Use If statement

We will use an if statement to find the minimum number of jumps required to reach (d, 0).

• First, we need to ensure that a is always greater than b, so that a represents a longer jump length, and b b> represents a shorter jump length. Therefore, if b > a, , then we assign the maximum value among a and b to a.

• Next, we check if d is greater than or equal to a. If this condition is met, then we can simply calculate the minimum number of jumps using (d a - 1) / a. Here, (d a - 1) means the total distance with a jump length of "a" divided by a (i.e. the length of each jump) gives the number of jumps.

• If d = 0, no jump is required.

• If d = b, then we can directly reach the point by jumping the length of b.

• If d > b and d , the minimum number of jumps is 2. This is because if we take a triangle XYZ such that X is the origin, Z is the target point, and Y is the point that satisfies XY = YZ = max(a, b). Then, the minimum jump will be 2, i.e. from X to Y and Y to Z.

Example

#include <iostream>
using namespace std;

int minJumps(int a, int b, int d) {
   // Check if b > a, then interchange the values of a and b
   if (b > a) {
      int cont = a;
      a = b;
      b = cont;
   }
    
   // When d >= a
   if (d >= a)
      return (d + a - 1) / a;

   // When the target point is 0
   if (d == 0)
      return 0;

   // When d is equal to b.
   if (d == b)
      return 1;
     
   // When distance to be covered is not equal to b.    
   return 2;  
    
}

int main() {
   int a = 3, b = 5, d = 9;
   int result = minJumps(a, b, d);
   cout << "Minimum number of jumps required to reach (d, 0) from (0, 0) is: " << result << endl;
   return 0;
}

Output

Minimum number of jumps required to reach (d, 0) from (0, 0) is: 2

Use division and modulo operators

If a or b has the value 0, then we can simply use the division and modulo operators to find the minimum number of jumps. Here, we divide the distance d by the hop length (since one of the hop lengths is 0) to get the number of hops.

Example

#include <iostream>
using namespace std;

int minJumps(int d, int jumpLength) {
   // To find number of complete jumps
   int numJumps = d / jumpLength;
   // If distance is not divisible by jump length
   if (d % jumpLength != 0) {
      numJumps++;  
   }
   return numJumps;
}
int main() {
   int d = 24, jumpLength = 4;
   int result = minJumps(d, jumpLength);
   cout << "Minimum number of jumps required to reach (d, 0) from (0, 0) is: " << result << endl;
   return 0;
}

Output

Minimum number of jumps required to reach (d, 0) from (0, 0) is: 6

Note - We can also use the ternary operator to write code in a concise way.

int minJumps(int d, int jumpLength) {
   int numJumps = (d % jumpLength == 0) ? (d / jumpLength) : (d / jumpLength) + 1;
   return numJumps;
}

in conclusion

We discussed how to find the minimum number of jumps required to reach the target point (d, 0) from the origin in the 2D plane. We use an if statement to find the number of jumps where a and b are non-zero values ​​(a and b b> are the jump lengths). If a or b is zero, then we can use division and modulo operators. To write code succinctly, we can use the ternary operator.

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