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The minimum number of elements in a heap of height h is2h+12h2h -12h-1

Question

The minimum number of elements in a heap of height h is2h+12h2h -12h-1

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Solution 1

It seems like there's a mistake in your question. The minimum number of elements in a heap of height h is actually 2^h.

Here's the step-by-step explanation:

  1. A heap is a complete binary tree. In a complete binary tree, all levels except possibly the last are completely filled and all nodes are as left as possible.

  2. The height of a complete binary tree (heap) with N nodes lies between log2(N+1) and log2N + 1.

  3. In a heap of height h, the minimum number of elements it can have is when only the last level is not fully filled. This means all levels from 0 to h-1 are fully filled.

  4. The number of nodes at level i is 2^i. So, the total number of nodes in h-1 levels is 2^0 + 2^1 + 2^2 + ... + 2^(h-1). This is a sum of a geometric progression with a=1, r=2 and n=h. The sum is 2^h - 1.

  5. Since the last level can have at least one node, the minimum number of nodes in a heap of height h is (2^h - 1) + 1 = 2^h.

So, the minimum number of elements in a heap of height h is 2^h.

This problem has been solved

Solution 2

It seems like there's a mistake in your question. The minimum number of elements in a heap of height h is given by the formula 2^h.

Here's the step-by-step explanation:

  1. A heap is a complete binary tree. In a complete binary tree, all levels are completely filled except possibly for the last level, which is filled from left to right.

  2. The height of a complete binary tree with N nodes lies between log2(N+1) and log2N + 1.

  3. In a heap of height h, the minimum number of nodes it can have is when only the last level is not fully filled. This means that all levels from 0 to h-1 are fully filled.

  4. The number of nodes at level i is 2^i. So, the total number of nodes from level 0 to h-1 is 2^0 + 2^1 + 2^2 + ... + 2^(h-1).

  5. The sum of a geometric series a + ar + ar^2 + ... + ar^(n-1) is a*(r^n - 1)/(r - 1). Here, a = 1, r = 2 and n = h. So, the sum is (2^h - 1)/(2 - 1) = 2^h - 1.

  6. Since there must be at least one node at level h (otherwise, the height would be less than h), the minimum number of nodes in a heap of height h is (2^h - 1) + 1 = 2^h.

So, the minimum number of elements in a heap of height h is 2^h.

This problem has been solved

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