Count non-negative triplets with sum equal to N
Given an integer N. The task is to find the number of different ordered triplets(a, b, c) of non-negative integers such that a + b + c = N .
Input : N = 2
Output : 6
Triplets are : (0, 0, 2), (1, 0, 1), (0, 1, 1), (2, 0, 0), (0, 2, 0), (1, 1, 0)
Input : N = 50
Output : 1326
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First, it is easy to see that for each non-negative integer N, the equation a + b = N can be satisfied by (N+1) different ordered pairs of (a, b). Now we can assign c values from 0 to N then the ordered pairs for a+b can be found. It will form a series of N+1 natural numbers and its sum will give the count of triplets.
Below is the implementation of the above approach :
Time Complexity : O(1)