A race is going on in which several stones are placed on a road. A bucket is placed at the starting point of the race, which is 5 units away from the first stone. The other stones are 3 units apart from each other and lie straight in a line one after another. I.e., the distance between 1st and 2nd stone is 3 units, between 3rd and 4th stone, it is also 3 units and so on. The competitor starts from the bucket, picks up the nearest stone, goes back and puts that stone into the bucket, then runs again to collect the next nearest stone, runs back, and puts it in the bucket. This way the process is continued till all the stones have been put into the bucket.
Now if there are n stones lying on the ground, how much distance will be covered by the competitor in completing the whole process.
Examples:
Input : n = 3 Output : Distance = 48 Explanation = 2*5 + 2(5 + 3) + 2(5 + 3 + 3) = 10 + 16 + 22 = 48 Input : n = 5 Output : Distance = 110 Explanation = 2*5 + 2(5 + 3) + 2(5 + 3 + 3) + 2(5 + 3 + 3 + 3) + 2(5 + 3 + 3 + 3 + 3) = 10 + 16 + 22 + 28 + 34 = 110
Observing the pattern:
Distance run by competitor to pick first stone = 2 * 5
Distance run by competitor to pick second stone = 2(5 + 3)
Distance run by competitor to pick third stone = 2(5 + 3 + 3)
= 2(5 + (2 * 3))
Distance run by competitor to pick fourth stone = 2(5 + 3 + 3 + 3)
= 2(5 + (3 * 3))
Distance run by competitor to pick fifth stone = 2(5 + 3 + 3 + 3 + 3)
= 2(5 + (4 * 3))
.
.
.
Distance run by competitor to pick n-th stone = 2(5 + 3 + 3 + ……. + (n-1) times )
= 2(5 + (n-1) *3)
So total distance run by competitor = sum of all the above distances
= (2 * 5) + 2(5 + 3) + 2(5 + (2 * 3)) + 2(5 + (3 * 3)) + ………….. + 2(5 + (n-1) *3)
= 2(5 + (5 + 3) + (5 + (2 * 3)) + (5 + (3 * 3)) + ………………. + (5 + (n-1) * 3)
= 2(5 + 5 + 5 …… + n times) + (3 + (2 * 3) + (3 * 3) + ……… + (n-1) * 3)
= 2(5n + 3(1 + 2 + 3 + ……………. + n-1))
= 2(5n + 3/2[(n-1)*(n-1 + 1)] )
= 2(5n + 3/2[(n-1)*n])
= 2(5n + 3/2(n2 – n))
= 10n + 3*n2 – 3*n
= 3*n2 + 7*n
= n*((3 * n) + 7)
Below is the implementation of the approach:
C++
// C++ program to calculate // the distance for given problem #include <bits/stdc++.h> using namespace std; // function to calculate the // distance int find_distance(int n) { return n * ((3 * n) + 7); } // Driver program int main() { int n = 5; cout << "Distance = " << find_distance(n); return 0; } |
Java
// Java program to calculate the // distance for given problem class demo { // function to calculate // the distance public static int find_distance(int n) { return n * (3 * n + 7); } // Driver program public static void main(String args[]) { int n = 5; System.out.print("Distance = "); System.out.println(find_distance(n)); } } |
Python3
# Python3 code to calculate # the distance for given problem # function to calculate the # distance def find_distance(n): return n * ((3 * n) + 7) # main function n = 5ans = find_distance( n ) print (ans) # This code is contributed by Saloni Gupta |
C#
// C# program to calculate // the distance for given problem using System; class GFG { // function to calculate the // distance public static int find_distance(int n) { return n * ((3 * n) + 7); } // Driver program public static void Main() { int n = 5; Console.Write(find_distance(n)); } } // This code is contributed by // Smitha Dinesh Semwal |
PHP
<?php //PHP program to calculate // the distance for given problem // function to calculate the // distance function find_distance($n) { return $n * ((3 * $n) + 7); } // Driver program $n = 5; echo "Distance = ", find_distance($n); // This code is contributed by aj_36 ?> |
Javascript
<script> // JavaScript program to calculate the // distance for given problem // function to calculate // the distance function find_distance(n) { return n * (3 * n + 7); } // Driver code let n = 5; document.write("Distance = "); document.write(find_distance(n)); </script> |
Output:
110
Time Complexity: O(1)
Auxiliary Space: O(1)
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