Number of ways to form a heap with n distinct integers

• Difficulty Level : Expert
• Last Updated : 09 Jul, 2021

Given n, how many distinct Max Heap can be made from n distinct integers?
Examples:

Input : n = 3
Output : Assume the integers are 1, 2, 3.
Then the 2 possible max heaps are:
3
/ \
1   2

3
/ \
2   1

Input : n = 4
Output : Assume the integers are 1, 2, 3, 4.
Then the 3 possible max heaps are:
4
/ \
3   2
/
1

4
/ \
2   3
/
1

4
/ \
3   1
/
2

Since there is only one element as the root, it must be the largest number. Now we have n-1 remaining elements. The main observation here is that because of the max heap properties, the structure of the heap nodes will remain the same in all instances, but only the values in the nodes will change.
Assume there are l elements in the left sub-tree and r elements in the right sub-tree. Now for the root, l + r = n-1. From this we can see that we can choose any l of the remaining n-1 elements for the left sub-tree as they are all smaller than the root.
We know the there are ways to do this. Next for each instance of these, we can have many heaps with l elements and for each of those we can have many heaps with r elements. Thus we can consider them as subproblems and recur for the final answer as:
T(n) = * T(L) * T(R).
Now we have to find the values for l and r for a given n. We know that the height of the heap h = . Also the maximum number of elements that can be present in the h th level of any heap, m = , where the root is at the 0th level. Moreover the number of elements actually present in the last level of the heap p = n – ( – 1). (since number of nodes present till the penultimate level). Thus, there can be two cases: when the last level is more than or equal to half-filled:
l = – 1, if p >= m / 2
(or) the last level is less than half-filled:
l = – 1 – ((m / 2) – p), if p < m / 2
(we get – 1 here because left subtree has nodes.
From this we can also say that r = n – l – 1.
We can use the dynamic programming approach discussed in this post here to find the values of . Similarly if we look at the recursion tree for the optimal substructure recurrence formed above, we can see that it also has overlapping subproblems property, hence can be solved using dynamic programming:

T(7)
/    \
T(3)   T(3)
/  \     /  \
T(1)  T(1) T(1) T(1)

Following is the implementation of the above approach:

C++

 // CPP program to count max heaps with n distinct keys#include using namespace std; #define MAXN 105 // maximum value of n here // dp[i] = number of max heaps for i distinct integersint dp[MAXN]; // nck[i][j] = number of ways to choose j elements//             form i elements, no order */int nck[MAXN][MAXN]; // log2[i] = floor of logarithm of base 2 of iint log2[MAXN]; // to calculate nCkint choose(int n, int k){    if (k > n)        return 0;    if (n <= 1)        return 1;    if (k == 0)        return 1;     if (nck[n][k] != -1)        return nck[n][k];     int answer = choose(n - 1, k - 1) + choose(n - 1, k);    nck[n][k] = answer;    return answer;} // calculate l for give value of nint getLeft(int n){    if (n == 1)        return 0;     int h = log2[n];     // max number of elements that can be present in the    // hth level of any heap    int numh = (1 << h); //(2 ^ h)     // number of elements that are actually present in    // last level(hth level)    // (2^h - 1)    int last = n - ((1 << h) - 1);     // if more than half-filled    if (last >= (numh / 2))        return (1 << h) - 1; // (2^h) - 1    else        return (1 << h) - 1 - ((numh / 2) - last);} // find maximum number of heaps for nint numberOfHeaps(int n){    if (n <= 1)        return 1;     if (dp[n] != -1)        return dp[n];     int left = getLeft(n);    int ans = (choose(n - 1, left) * numberOfHeaps(left)) *                             (numberOfHeaps(n - 1 - left));    dp[n] = ans;    return ans;} // function to initialize arraysint solve(int n){    for (int i = 0; i <= n; i++)        dp[i] = -1;     for (int i = 0; i <= n; i++)        for (int j = 0; j <= n; j++)            nck[i][j] = -1;     int currLog2 = -1;    int currPower2 = 1;     // for each power of two find logarithm    for (int i = 1; i <= n; i++) {        if (currPower2 == i) {            currLog2++;            currPower2 *= 2;        }        log2[i] = currLog2;    }     return numberOfHeaps(n);} // driver functionint main(){    int n = 10;    cout << solve(n) << endl;    return 0;}

Java

 // Java program to count max heaps with n distinct keysclass GFG{     static int MAXN = 105; // maximum value of n here     // dp[i] = number of max heaps for i distinct integers    static int[] dp = new int[MAXN];     // nck[i][j] = number of ways to choose j elements    //         form i elements, no order */    static int[][] nck = new int[MAXN][MAXN];     // log2[i] = floor of logarithm of base 2 of i    static int[] log2 = new int[MAXN];     // to calculate nCk    public static int choose(int n, int k)    {        if (k > n)        {            return 0;        }        if (n <= 1)        {            return 1;        }        if (k == 0)        {            return 1;        }         if (nck[n][k] != -1)        {            return nck[n][k];        }         int answer = choose(n - 1, k - 1) + choose(n - 1, k);        nck[n][k] = answer;        return answer;    }     // calculate l for give value of n    public static int getLeft(int n)    {        if (n == 1)        {            return 0;        }         int h = log2[n];         // max number of elements that can be present in the        // hth level of any heap        int numh = (1 << h); //(2 ^ h)         // number of elements that are actually present in        // last level(hth level)        // (2^h - 1)        int last = n - ((1 << h) - 1);         // if more than half-filled        if (last >= (numh / 2))        {            return (1 << h) - 1; // (2^h) - 1        }        else        {            return (1 << h) - 1 - ((numh / 2) - last);        }    }     // find maximum number of heaps for n    public static int numberOfHeaps(int n)    {        if (n <= 1)        {            return 1;        }         if (dp[n] != -1)        {            return dp[n];        }         int left = getLeft(n);        int ans = (choose(n - 1, left) * numberOfHeaps(left))                * (numberOfHeaps(n - 1 - left));        dp[n] = ans;        return ans;    }     // function to initialize arrays    public static int solve(int n)    {        for (int i = 0; i <= n; i++)        {            dp[i] = -1;        }         for (int i = 0; i <= n; i++)        {            for (int j = 0; j <= n; j++)            {                nck[i][j] = -1;            }        }         int currLog2 = -1;        int currPower2 = 1;         // for each power of two find logarithm        for (int i = 1; i <= n; i++)        {            if (currPower2 == i)            {                currLog2++;                currPower2 *= 2;            }            log2[i] = currLog2;        }         return numberOfHeaps(n);    }     // Driver code    public static void main(String[] args)    {        int n = 10;        System.out.print(solve(n));    }} // This code has been contributed by 29AjayKumar

Python3

 # Python program to count max heaps with n distinct keys MAXN = 105 # maximum value of n here # dp[i] = number of max heaps for i distinct integersdp = *MAXN # nck[i][j] = number of ways to choose j elements#             form i elements, no order */nck = [[0 for i in range(MAXN)] for j in range(MAXN)] # log2[i] = floor of logarithm of base 2 of ilog2 = *MAXN # to calculate nCkdef choose(n, k):    if (k > n):        return 0    if (n <= 1):        return 1    if (k == 0):        return 1     if (nck[n][k] != -1):        return nck[n][k]     answer = choose(n - 1, k - 1) + choose(n - 1, k)    nck[n][k] = answer    return answer  # calculate l for give value of ndef getLeft(n):    if (n == 1):        return 0     h = log2[n]     # max number of elements that can be present in the    # hth level of any heap    numh = (1 << h) #(2 ^ h)     # number of elements that are actually present in    # last level(hth level)    # (2^h - 1)    last = n - ((1 << h) - 1)     # if more than half-filled    if (last >= (numh // 2)):        return (1 << h) - 1 # (2^h) - 1    else:        return (1 << h) - 1 - ((numh // 2) - last)  # find maximum number of heaps for ndef numberOfHeaps(n):    if (n <= 1):        return 1     if (dp[n] != -1):        return dp[n]     left = getLeft(n)    ans = (choose(n - 1, left) * numberOfHeaps(left)) * (numberOfHeaps(n - 1 - left))    dp[n] = ans    return ans  # function to initialize arraysdef solve(n):    for i in range(n+1):        dp[i] = -1     for i in range(n+1):        for j in range(n+1):            nck[i][j] = -1     currLog2 = -1    currPower2 = 1     # for each power of two find logarithm    for i in range(1,n+1):        if (currPower2 == i):            currLog2 += 1            currPower2 *= 2        log2[i] = currLog2    return numberOfHeaps(n)  # Driver coden = 10print(solve(n)) # This code is contributed by ankush_953

C#

 // C# program to count max heaps with n distinct keysusing System;   class GFG{    static int MAXN = 105; // maximum value of n here           // dp[i] = number of max heaps for i distinct integers    static int[] dp = new int[MAXN];            // nck[i][j] = number of ways to choose j elements    //             form i elements, no order */    static int[,] nck = new int[MAXN,MAXN];            // log2[i] = floor of logarithm of base 2 of i    static int[] log2 = new int[MAXN];            // to calculate nCk    public static int choose(int n, int k)    {        if (k > n)            return 0;        if (n <= 1)            return 1;        if (k == 0)            return 1;               if (nck[n,k] != -1)            return nck[n,k];               int answer = choose(n - 1, k - 1) + choose(n - 1, k);        nck[n,k] = answer;        return answer;    }           // calculate l for give value of n    public static int getLeft(int n)    {        if (n == 1)            return 0;               int h = log2[n];               // max number of elements that can be present in the         // hth level of any heap        int numh = (1 << h); //(2 ^ h)               // number of elements that are actually present in        // last level(hth level)        // (2^h - 1)        int last = n - ((1 << h) - 1);               // if more than half-filled        if (last >= (numh / 2))            return (1 << h) - 1; // (2^h) - 1        else            return (1 << h) - 1 - ((numh / 2) - last);    }           // find maximum number of heaps for n    public static int numberOfHeaps(int n)    {        if (n <= 1)            return 1;               if (dp[n] != -1)            return dp[n];               int left = getLeft(n);        int ans = (choose(n - 1, left) * numberOfHeaps(left)) *                                  (numberOfHeaps(n - 1 - left));        dp[n] = ans;        return ans;    }           // function to initialize arrays    public static int solve(int n)    {        for (int i = 0; i <= n; i++)            dp[i] = -1;               for (int i = 0; i <= n; i++)            for (int j = 0; j <= n; j++)                nck[i,j] = -1;               int currLog2 = -1;        int currPower2 = 1;               // for each power of two find logarithm        for (int i = 1; i <= n; i++) {            if (currPower2 == i) {                currLog2++;                currPower2 *= 2;            }            log2[i] = currLog2;        }               return numberOfHeaps(n);    }           // driver function    static void Main()    {        int n = 10;        Console.Write(solve(n));    }    //This code is contributed by DrRoot_}

Javascript



Output:

3360

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