# Distance covered after last breakpoint with at most K cost

Given an array arr[] of size Nand integers K and Dconsisting of breakpoints in the journey, the task is to find the distance covered after the last breakpoint with at most K cost following the given conditions:

• Starting point is 0
• 1 unit distance is covered for 1 cost.
• For moving D continuous distance there is an extra cost of D ie, total cost becomes D*1 + D = 2D.

Note: If N = 0 that means there is no breakpoint and starting is from 0.

Examples:

Input: arr[] = {17, 12, 18}, K = 28, D = 8
Output: 2
Explanation: Movement from 0 to 8. Covered distance cost = 8*1 + 8 = 16
Movement from 8 to 12 as 12 is a break point. Cost = 12 – 8 = 4. Total cost = 16 + 4 = 20.
Movement from 12 to 17. Cost = 17 – 12 = 5. Total cost = 20 + 5 = 25.
Movement from 17 to 18. Cost = 18 – 17 = 1. Total Cost = 26.
Remaining cost = 2.
So extra distance that can be covered after last breakpoint(18) = 2.

Input: arr[] = {}, K = 100, D = 8
Output: 52
Explanation: Movement from 0 to 8. Distance = 8. Cost = 8*1 + 8 = 16. Total = 16
Movement from 8 to 16. Distance = 8. Cost = 8*1 + 8 = 16. Total = 32.
Movement from 16 to 24. Distance = 8. Cost = 8*1 + 8 = 16. Total = 48.
Movement from 24 to 32. Distance = 8. Cost = 8*1 + 8 = 16. Total = 64.
Movement from 32 to 40. Distance = 8. Cost = 8*1 + 8 = 16. Total = 80.
Movement from 40 to 48. Distance = 8. Cost = 8*1 + 8 = 16. Total = 96.
Movement from 48 to 52. Distance = 4. Cost = 4*1 = 4. Total = 100.
Distance covered after last breakpoint (no breakpoint here) = 52.

Approach: The given problem can be solved based on the following mathematical observation:

Distance between two consecutive breakpoints X and Y is (X – Y).
The number of times D consecutive movements are made = floor((X – Y)/D).
The cost for these consecutive movements is 2*D * floor((XY)/D).
Say M cost is paid to reach the last breakpoint.
Then the extra D consecutive distances that can be covered is floor((K – M)/2*D).
The cost for this is D* floor((K – M)/2*D). If the remaining cost is > D then only D additional distance can be covered.

Follow the steps to solve the problem:

• First check for Edge Case N = 0
• If N!=0 then sort the given breakpoints.
• Calculate the distance from the 1st point to 0th point.
• Traverse the array from i = 1 to N – 1.
• Calculate the distance between arr[i] and arr[i-1].
• Calculate the cost to cover this distance (say Count).
• If, K<= Countreturn 0
• Calculate the extra distance by using the above observation:
• Res = D*(( K – Count) / (2*D) ) (number of times continuous D distances covered)
• Rem = (K – Count) % (2*D) (Remaining cost)
• If, Rem >= D && Rem <=(2*D-1) then Rem = D
• return, Res + Rem as the final extra distance covered.

Below is the implementation of the above approach:

## C++

 ` ` `#include ` `using` `namespace` `std;` ` ` `int` `countDays(``int``* arr, ``int` `N, ``int` `K, ``int` `D)` `{` `    ` `    ``if` `(N == 0) {` `        ``int` `Res = D * (K / (2 * D));` `        ``int` `Rem = K % (2 * D);` `        ``if` `(Rem >= D && Rem <= (2 * D - 1))` `            ``Rem = D;` `        ``return` `(Res + Rem);` `    ``}` ` ` `    ` `    ``sort(arr, arr + N);` ` ` `    ` `    ` `    ``int` `Count = 2 * D * (arr / D)` `                ``+ arr % D;` `    ``int` `Curr = 0;` ` ` `    ` `    ``for` `(``int` `i = 1; i < N; i++) {` ` ` `        ` `        ` `        ``if` `((arr[i] - arr[i - 1]) >= 1) {` ` ` `            ` `            ` `            ``Curr = arr[i] - arr[i - 1];` ` ` `            ` `            ``Count += 2 * D * (Curr / D)` `                     ``+ (Curr % D);` `        ``}` `    ``}` ` ` `    ` `    ``if` `(K <= Count)` `        ``return` `0;` ` ` `    ` `    ``int` `Res = D * ((K - Count) / (2 * D));` `    ``int` `Rem = (K - Count) % (2 * D);` ` ` `    ` `    ``if` `(Rem >= D && Rem <= (2 * D - 1))` `        ``Rem = D;` ` ` `    ` `    ``return` `(Res + Rem);` `}` ` ` `int` `main()` `{` `    ``int` `arr[] = { 17, 12, 18 };` `    ``int` `D = 8;` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``int` `K = 28;` ` ` `    ` `    ``cout << countDays(arr, N, K, D);` `    ``return` `0;` `}`

Time Complexity: O(N * logN)
Auxiliary Space: O(1)