Maximum height when coins are arranged in a triangle - GeeksforGeeks
We have N coins which need to arrange in form of a triangle, i.e. first row will have 1 coin, second row will have 2 coins and so on, we need to tell maximum height which we can achieve by using these N coins.
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We have N coins which need to arrange in form of a triangle, i.e. first row will have 1 coin, second row will have 2 coins and so on, we need to tell maximum height which we can achieve by using these N coins.
This problem can be solved by finding a relation between height of the triangle and number of coins. Let maximum height is H, then total sum of coin should be less than N,
Sum of coins for height H <= N
H*(H + 1)/2 <= N
H*H + H – 2*N <= 0
Now by Quadratic formula
(ignoring negative root)
Maximum H can be (-1 + √(1 + 8N)) / 2
Now we just need to find the square root of (1 + 8N) for
which we can use Babylonian method of finding square root
/* Returns the square root of n. Note that the function */float squareRoot(float n){ /* We are using n itself as initial approximation This can definitely be improved */ float x = n; float y = 1; float e = 0.000001; /* e decides the accuracy level*/ while (x - y > e) { x = (x + y) / 2; y = n/x; } return x;}// Method to find maximum height of arrangement of coinsint findMaximumHeight(int N){ // calculating portion inside the square root int n = 1 + 8*N; int maxH = (-1 + squareRoot(n)) / 2; return maxH;}