Distribute N candies among K people

Here we can simply use the concept of brute force. 

• We initialize an array with 0 elements.
• Then we make a variable and initialize it with 1(let’s call it “increment”). This variable is used to keep track of the number of candies to be given to the current friend.
• We start with the first element and increase it by ‘increment’.
• Now we reduce the total number of candies by the variable ‘increment’. (N = N - increment).
• Now we increase the value of ‘increment’ by 1.
• In the left ‘N’ is less than ‘increment’ we make ‘increment’ equal to ‘N’. Because as written in the question the number of candies left is less than the candies to be given then all left candies will be given to the friend.
• We go to the next element and continue doing that.
• If we reach the end of the array we will go to the first element again for the next round of distribution.
• If the total number of candies is 0 we stop the iteration.

• We initialize an array with 0 elements.
• We know that 1 + 2 + 3……+ r = ((r * (r + 1)) / 2).
• We can sum of natural numbers till the last term of the series which will be (turns * K) and subtracting the sum of natural numbers till the last term of the previous series which is (turns - 1) * K.
• Keep doing this till the sum is less than ‘N’, once it exceeds then distribute candies in the given way till possible.
• We call a turn completed if every person gets the desired number of candies he is to get in a turn.
• We will assign the left candies in brute force way only.

Here we can simply use the concept of binary search. 

• We initialize an array with 0 elements.
• We use binary search to find the largest ‘x’ which satisfies x * (x + 1) <= (2 * N). (Here x is the total number of times Sanyam will distribute the candy to his friend). A binary search will help us to locate the number of times Sanyam is giving candy to his friend.
• Now we need to calculate the number of times it will visit an ith friend in the ‘x’ iteration.
• Use arithmetic progression to find the sum in the ‘r’ iteration
  • (rth term:- i + (r - 1) * k).
• Make N = N - (x * (x + 1)) / 2 and x = x + 1. (We are doing this last part to make sure that the left part of candies are being given to the next friend, as mentioned in the question that the left candies are required to be given to the next friend.)
• Take x mod by k. (x = x % K);
• Increase ‘xth’ term by N.