Smallest triangular number larger than p
Given n number of buckets and each bucket is numbered from 1 to n and flowers in it are equal to triangular numbers. You have to choose the bucket which is left with minimum flower after picking ‘p’ flowers from it.
First bucket contains only 1 flower, second bucket contains 3, third bucket contains 6 and so on following a pattern of n(n+1)/2.
Input : p = 4 Output : bucket 3 Explanation : Buckets with flowers : 1 3 6 10 .... So, bucket 3 is left with only two flowers after selecting p flowers from it which is minimum. Input : p = 10 Output : bucket 4 Explanation : Bucket with flowers : 1 3 6 10 15 ... So, selecting 10 flowers from 4th bucket leave it with 0 flowers.
Observing the input/output of different cases, bucket number can be calculated using formula :
n = ceil( (sqrt(8*p+1)-1)/2 ) ;
How does it work?
We need smallest n such than n*(n+1)/2 >= p
So we need to find roots of equation n2 + n – 2*p >= 0.
By applying the formula discussed here, we get n = ceil( (sqrt(8*p+1)-1)/2 )
Time Complexity : O(1)