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Smallest triangular number larger than p
  • Last Updated : 19 Mar, 2021

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.
Examples : 
 

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.

 

Approach : 
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 ) 
 

C++




// CPP code to find the bucket to choose
// for picking flowers out of it
#include<bits/stdc++.h>
using namespace std;
 
int findBucketNo(int p)
{
   return ceil( ( sqrt( 8*p + 1 ) -1 ) / 2 ) ;
}
 
// Driver code
int main()
{
  int p = 10 ;   
  cout << findBucketNo(p);
  return 0;
}

Java




//Java code to find the bucket to choose
// for picking flowers out of it
import java.lang.System.*;
 
class GFG {
     
    static int findBucketNo(int p)
    {
        return (int)Math.ceil((
        Math.sqrt( 8*p + 1 ) -1 ) / 2 ) ;
    }
     
    // Driver code
    public static void main(String[] args)
    {
        int p = 10 ;
         
        System.out.println(findBucketNo(p));
    }
}
 
// This code is contributed by
// Smitha Dinesh Semwal

Python 3




# Python 3 code to find the bucket
# to choose for picking flowers
# out of it
import math
 
def findBucketNo(p):
 
    return math.ceil( ( math.sqrt(
               8*p + 1 ) -1 ) / 2 )
 
# Driver code
p = 10
 
print(findBucketNo(p))
 
# This code is contributed by
# Smitha Dinesh Semwal

C#




// C# code to find the bucket to choose
// for picking flowers out of it
using System;
 
class GFG {
     
    static int findBucketNo(int p)
    {
        return (int)Math.Ceiling((
        Math.Sqrt( 8*p + 1 ) -1 ) / 2 );
    }
     
    // Driver code
    static public void Main ()
    {
        int p = 10 ;
        Console.WriteLine(findBucketNo(p));
    }
}
 
// This code is contributed by Ajit.

PHP




<?php
// PHP code to find the bucket
// to choose for picking
// flowers out of it
 
function findBucketNo($p)
{
    return ceil( ( sqrt( 8 * $p +
                  1 ) -1 ) / 2 ) ;
}
 
// Driver code
$p = 10 ;
echo(findBucketNo($p));
 
// This code is contributed by Ajit.
?>

Javascript




<script>
// javascript code to find the bucket to choose
// for picking flowers out of it
function findBucketNo( p)
{
   return Math.ceil( ( Math.sqrt( 8 * p + 1 ) -1 ) / 2 ) ;
}
 
// Driver code
let p = 10 ;   
   document.write(findBucketNo(p));
    
    // This code is contributed by gauravrajput1
 
</script>
Output : 
4

 

Time Complexity : O(1)
 

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