# Distribute M objects starting from Sth person such that every ith person gets arr[i] objects

Distribute M objects starting from Sth person such that every ith person gets arr[i] objects

Given an array arr[] consisting of N integers (1-based indexing) and two integers M and S, the task is to distribute M objects among N persons, starting from the position S, such that the ith person gets at most arr[i] objects each time.

Examples:

Input: arr[] = {2, 3, 2, 1, 4}, M = 11, S = 2Output: 1, 3, 2, 1, 4Explanation: The distribution of M (= 11) objects starting from Sth(= 2) person is as follows:

For arr[2](= 3): Give 3 objects to the 2nd person. Now, the total number of objects reduces to (11 – 3) = 8.

For arr[3] (= 2): Give 2 objects to the 3rd person. Now, the total number of objects reduces to (8 – 2) = 6.

For arr[4] (= 1): Give 1 object to the 4th person. Now, the total number of objects reduces to (6 – 1) = 5.

For arr[5] (= 4): Give 4 objects to the 5th person. Now, the total number of objects reduces to (5 – 4) = 1.

For arr[1] (= 1): Give 1 object to the 1st person. Now, the total number of objects reduced to (1 – 1) = 0.

Therefore, the distribution of objects is {1, 3, 2, 1, 4}.

Input: arr[] = {2, 3, 2, 1, 4}, M = 3, S = 4Output: 0 0 0 1 2

Approach: The given problem can be solved by traversing the array from the given starting index S and distribute the maximum objects to each array element. Follow the steps below to solve the given problem:

Initialize an auxiliary array, say distribution[] with all elements as 0 to store the distribution of M objects.

Initialize two variables, say ptr and rem as S and M respectively, to store the starting index and remaining M objects.

Iterate until rem is positive, and perform the following steps:

If the value of rem is at least the element at index ptr i.e., arr[ptr], then increment the value of distribution[ptr] by arr[ptr] and decrement the value of rem by arr[ptr].

Otherwise, increment the distribution[ptr] by rem and update rem equal to 0.

Update ptr equal to (ptr + 1) % N to iterate the given array arr[] in a cyclic manner.

After completing the above steps, print the distribution[] as the resultant distribution of objects.

Below is the implementation of the above approach:

C++

#include

using namespace std;

void distribute(int N, int K,

int M, int arr[])

{

int distribution[N] = { 0 };

int ptr = K – 1;

int rem = M;

while (rem > 0) {

if (rem >= arr[ptr]) {

distribution[ptr] += arr[ptr];

rem -= arr[ptr];

}

else {

distribution[ptr] += rem;

rem = 0;

}

ptr = (ptr + 1) % N;

}

for (int i = 0; i < N; i++) {
cout