# Check if two people starting from different points ever meet

• Difficulty Level : Basic
• Last Updated : 02 Dec, 2022

There are two people that start from two different positions, let’s say x1 and x2. Both can jump v1 and v2 meters ahead respectively. We have to find if both will ever meet given that the number of jumps taken by both has to be same.
Print ‘Yes’ if they will meet,
print ‘No’ they will not.

Examples :

Input  : x1 = 5, v1 = 8, x2 = 4, v2 = 7
Output : No
Explanation: The first person is starting ahead of the second one.
and his speed is also greater than the second one, so they will never meet.

Input  : x1 = 6, v1 = 6, x2 = 4, v2 = 8
Output : Yes

Naive Approach: In this, we calculate the position of each person after each jump and checks if they have landed in the same spot or not. This one having complexity O(n).

## C++

 // C++ program to find if two people// starting from different positions// ever meet or not.#include using namespace std; bool everMeet(int x1, int x2, int v1, int v2){    // If speed of a person at a position before    // other person is smaller, then return false.    if (x1 < x2 && v1 <= v2)       return false;    if (x1 > x2 && v1 >=v2)       return false;        // Making sure that x1 is greater    if (x1 < x2)    {        swap(x1, x2);        swap(v1, v2);    }            // Until one person crosses other     while (x1 >= x2) {        if (x1 == x2)            return true;                 // first person taking one        // jump in each iteration        x1 = x1 + v1;                 // second person taking        // one jump in each iteration        x2 = x2 + v2;    }     return false;  } // Driver codeint main(){    int x1 = 5, v1 = 8, x2 = 4, v2 = 7;    if (everMeet(x1, x2, v1, v2))        printf("Yes");       else        printf("No");    return 0;}

## Java

 // Java program to find// if two people starting// from different positions// ever meet or not.import java.io.*; class GFG{    static void swap(int a, int b)    {        int t = a;        a = b;        b = t;    }    static boolean everMeet(int x1, int x2,                            int v1, int v2)    {        // If speed of a person        // at a position before        // other person is smaller,        // then return false.        if (x1 < x2 && v1 <= v2)        return false;        if (x1 > x2 && v1 >= v2)        return false;                 // Making sure that        // x1 is greater        if (x1 < x2)        {            swap(x1, x2);            swap(v1, v2);        }             // Until one person        // crosses other        while (x1 >= x2)        {            if (x1 == x2)                return true;                         // first person taking one            // jump in each iteration            x1 = x1 + v1;                         // second person taking            // one jump in each iteration            x2 = x2 + v2;        }             return false;    }         // Driver code    public static void main (String[] args)    {        int x1 = 5, v1 = 8,            x2 = 4, v2 = 7;        if (everMeet(x1, x2, v1, v2))            System.out.println("Yes");        else            System.out.println("No");    }} // This code is contributed// by akt_mit

## Python3

 # Python3 program to find if two# people starting from different# positions ever meet or not. def everMeet(x1, x2, v1, v2):     # If speed of a person at    # a position before other    # person is smaller, then    # return false.    if (x1 < x2 and v1 <= v2):        return False;    if (x1 > x2 and v1 >= v2):        return False;     # Making sure that    # x1 is greater    if (x1 < x2):         x1, x2 = x2,x1;        v1, v2 = v2,v1;          # Until one person    # crosses other    while (x1 >= x2):             if (x1 == x2):            return True;                 # first person taking one        # jump in each iteration        x1 = x1 + v1;                 # second person taking        # one jump in each iteration        x2 = x2 + v2;          return False;  # Driver codex1 = 5;v1 = 8;x2 = 4;v2 = 7;if (everMeet(x1, x2,v1, v2)):    print("Yes");else:    print("No"); # This code is contributed by mits

## C#

 // C# program to find if two// people starting from different// positions ever meet or not.using System; class GFG{    static void swap(ref int a,                     ref int b)    {        int t = a;        a = b;        b = t;    }    static bool everMeet(int x1, int x2,                         int v1, int v2)    {        // If speed of a person at a        // position before other person        // is smaller, then return false.        if (x1 < x2 && v1 <= v2)        return false;        if (x1 > x2 && v1 >= v2)        return false;                 // Making sure that x1 is greater        if (x1 < x2)        {            swap(ref x1, ref x2);            swap(ref v1, ref v2);        }             // Until one person crosses other        while (x1 >= x2)        {            if (x1 == x2)                return true;                         // first person taking one            // jump in each iteration            x1 = x1 + v1;                         // second person taking            // one jump in each iteration            x2 = x2 + v2;        }             return false;    }         // Driver code    static void Main()    {        int x1 = 5, v1 = 8,            x2 = 4, v2 = 7;        if (everMeet(x1, x2, v1, v2))            Console.Write("Yes");        else            Console.Write("No");    }} // This code is contributed by// Manish Shaw(manishshaw1)

## PHP

 \$x2 && \$v1 >= \$v2)        return false;     // Making sure that    // x1 is greater    if (\$x1 < \$x2)    {        list(\$x1, \$x2) = array(\$x2,                               \$x1);        list(\$v1, \$v2) = array(\$v2,                               \$v1);         }     // Until one person    // crosses other    while (\$x1 >= \$x2)    {        if (\$x1 == \$x2)            return true;                 // first person taking one        // jump in each iteration        \$x1 = \$x1 + \$v1;                 // second person taking        // one jump in each iteration        \$x2 = \$x2 + \$v2;    }     return false;} // Driver code\$x1 = 5;\$v1 = 8;\$x2 = 4;\$v2 = 7;if (everMeet(\$x1, \$x2,             \$v1, \$v2))    echo "Yes";else    echo "No"; // This code is contributed by ajit?>

## Javascript



Output:

No

Auxiliary Space: O(1) because it is using constant space for variables

Efficient approach: Other way is to use the concept of Relative Speed. If the X1 is D distance away from X2, then relative speed of X1 should be in factors of D so as both land on the same spot.
Example:
X1 = 10
X2 = 25
V1 = 10
V2 = 8
Since here D = 15 and relative speed = 2 which is not a factor of D they will never meet.
The time complexity of this algorithm is O(1)

Implementation is given below:

## C++

 // C++ program to find if two people// starting from different positions// ever meet or not.#include using namespace std; bool everMeet(int x1, int x2, int v1, int v2){     // If speed of a person at a position before    // other person is smaller, then return false.    if (x1 < x2 && v1 <= v2)       return false;    if (x1 > x2 && v1 >= v2)       return false;      // Making sure that x1 is greater    if (x1 < x2)    {        swap(x1, x2);        swap(v1, v2);    }        // checking if relative speed is    // a factor of relative distance or not    return ((x1 - x2) % (v1 - v2) == 0);} // Driver codeint main(){    int x1 = 5, v1 = 8, x2 = 4, v2 = 7;    if (everMeet(x1, x2, v1, v2))        printf("Yes");       else        printf("No");    return 0;}

## Java

 // Java program to find if two people// starting from different positions// ever meet or not. public class GFG {     static boolean everMeet(int x1, int x2, int v1, int v2) {         // If speed of a person at a position before        // other person is smaller, then return false.        if (x1 < x2 && v1 <= v2) {            return false;        }        if (x1 > x2 && v1 >= v2) {            return false;        }         // Making sure that x1 is greater        if (x1 < x2) {            swap(x1, x2);            swap(v1, v2);        }         // checking if relative speed is        // a factor of relative distance or not        return ((x1 - x2) % (v1 - v2) == 0);    }     static void swap(int a, int b) {        int t = a;        a = b;        b = t;    }     public static void main(String[] args) {        int x1 = 5, v1 = 8, x2 = 4, v2 = 7;        if (everMeet(x1, x2, v1, v2)) {            System.out.printf("Yes");        } else {            System.out.printf("No");        }    }} // This code is contributed by 29AjayKumar

## Python3

 # Python3 program to find if two people# starting from different positions# ever meet or not. def everMeet(x1, x2, v1, v2):     # If speed of a person at a position before    # other person is smaller, then return false.    if (x1 < x2 and v1 <= v2):        return False;        if (x1 > x2 and v1 >= v2):            return False;                         # Making sure that x1 is greater            if (x1 < x2):                swap(x1, x2)                swap(v1, v2)                                 # checking if relative speed is                # a factor of relative distance or not                return ((x1 - x2) % (v1 - v2) == 0)                def swap(a, b):                    t = a                    a = b                    b = t# Driver codedef main():         x1, v1, x2, v2 =5, 8, 4, 7    if (everMeet(x1, x2, v1, v2)):        print("Yes")    else:        print("No") if __name__ == '__main__':    main()          # This code is contributed by 29AjayKumar

## C#

 //C# program to find if two people// starting from different positions// ever meet or not.using System;public class GFG{      static bool everMeet(int x1, int x2, int v1, int v2) {         // If speed of a person at a position before        // other person is smaller, then return false.        if (x1 < x2 && v1 <= v2) {            return false;        }        if (x1 > x2 && v1 >= v2) {            return false;        }         // Making sure that x1 is greater        if (x1 < x2) {            swap(x1, x2);            swap(v1, v2);        }         // checking if relative speed is        // a factor of relative distance or not        return ((x1 - x2) % (v1 - v2) == 0);    }     static void swap(int a, int b) {        int t = a;        a = b;        b = t;    }     public static void Main() {        int x1 = 5, v1 = 8, x2 = 4, v2 = 7;        if (everMeet(x1, x2, v1, v2)) {            Console.WriteLine("Yes");        } else {            Console.WriteLine("No");        }    }}  // This code is contributed by 29AjayKumar

## PHP

 \$x2 && \$v1 >= \$v2)        return false;     // Making sure that x1 is greater    if (\$x1 < \$x2)    {        list(\$x1, \$x2) = array(\$x2, \$x1);        list(\$v2, \$v1) = array(\$v1, \$v2);    }     // checking if relative speed is    // a factor of relative distance or not    return ((\$x1 - \$x2) % (\$v1 - \$v2) == 0);} // Driver code\$x1 = 5; \$v1 = 8;\$x2 = 4; \$v2 = 7;if (everMeet(\$x1, \$x2, \$v1, \$v2))    print("Yes");else    print("No"); // This code is contributed by mits?>

## Javascript



Output :

No

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

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