Given an array of N integers the place arr[i] denotes the variety of sticks of size 2i. The process is to seek out the variety of triangles attainable with given lengths having space ≥ zero.
Note: Every stick can solely be used as soon as.
Input: a = 1, 2, 2, 2, 2
All attainable triangles are:
(20, 24, 24), (21, 23, 23), (21, 22, 22).
Input: a = three, three, three
Approach: The predominant commentary is that the triangles with space ≥ zero can solely be shaped if there are three identical lengths of sticks or one totally different and two comparable lengths of sticks. Hence greedily iterate from the again and rely the variety of pairs of identical size sticks out there which is arr[i] / 2. But if there’s a stick remaining, then a pair and a stick is used to type a triangle. In the tip, the whole variety of sticks left is calculated which is 2 * pairs and the variety of triangles that may be shaped with these remaining sticks can be (2 * pairs) / three.
Below is the implementation of the above strategy:
utilizing namespace std;
int countTriangles(int a, int n)
int cnt = zero;
int pairs = zero;
for (int i = n – 1; i >= zero; i–)
cnt += (2 * pairs) / three;
Striver(underscore)79 at Codechef and codeforces D
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first_page Count of components whose absolute distinction with the sum of all the opposite components is bigger than okay
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