Maximum length of rod for Q-th person

Given lengths of n rods in an array a[]. If any person picks any rod, half of the longest rod (or (max + 1) / 2 ) is assigned and remaining part (max – 1) / 2 is put back. It may be assumed that sufficient number of rods are always available, answer M queries given in an array q[] to find the largest length of rod available for qith person, provided qi is a valid person number starting from 1.

Examples :

Input : a[] = {6, 5, 9, 10, 12}
        q[] = {1, 3}
Output : 12 9
The first person gets maximum length as 12. 
We remove 12 from array and put back (12 -1) / 2 = 5. 
Second person gets maximum length as 10.  
We put back (10 - 1)/2 which is 4.
Third person gets maximum length as 9.

Input : a[] = {6, 5, 9, 10, 12}
        q[] = {3, 1, 2, 7, 4, 8, 9, 5, 10, 6}
Output : 9 12 10 5 6 4 3 6 3 5

Approach :
Use a stack and a queue. First sort all the lengths and push them onto a stack. Now, take the top element of stack, and divide by 2 and push the remaining length to queue. Now, from next customer onwards :



  1. If stack is empty, pop front queue and push back to queue. It’s half (front / 2), if non zero.
  2. If queue is empty, pop from stack and push to queue it’s half (top / 2), if non zero.
  3. If both are non empty, compare top and front, which ever is larger should be popped, divided by 2 and then pushed back.
  4. If both are empty, store is empty! Stop here!

At each step above store the length available to ith customer in separate array, say “ans”. Now, start answering the queries by giving ans[Qi] as output.

Below is the implementation of above approach :

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// CPP code to find the length of largest
// rod available for Q-th customer
#include <bits/stdc++.h>
using namespace std;
  
// function to find largest length of
// rod available for Q-th customer
vector<int> maxRodLength(int ar[],
                        int n, int m)
{
    queue<int> q;
  
    // sort the rods according to lengths
    sort(ar, ar + n);
  
    // Push sorted elements to a stack
    stack<int> s;
    for (int i = 0; i < n; i++)
        s.push(ar[i]);
  
    vector<int> ans;
  
    while (!s.empty() || !q.empty()) {
        int val;
          
        // If queue is empty -> pop from stack
        // and push to queue it’s half(top/2),
        // if non zero.
        if (q.empty()) {
            val = s.top();
            ans.push_back(val);
            s.pop();
            val /= 2;
  
            if (val)
                q.push(val);
        }
        // If stack is empty -> pop front from
        // queue and push back to queue it’s
        // half(front/2), if non zero.
        else if (s.empty()) {
            val = q.front();
            ans.push_back(val);
            q.pop();
            val /= 2;
            if (val != 0)
                q.push(val);
        }
        // If both are non empty ->
        // compare top and front, whichsoever is
        // larger should be popped, divided by 2
        // and then pushed back.
        else {
            val = s.top();
            int fr = q.front();
            if (fr > val) {
                ans.push_back(fr);
                q.pop();
                fr /= 2;
                if (fr)
                    q.push(fr);
            }
            else {
                ans.push_back(val);
                s.pop();
                val /= 2;
                if (val)
                    q.push(val);
            }
        }
    }
  
    return ans;
}
  
// Driver code
int main()
{
    // n : number of rods
    // m : number of queries
    int n = 5, m = 10;
      
    int ar[n] = { 6, 5, 9, 10, 12 };
  
    vector<int> ans = maxRodLength(ar, n, m);
  
    int query[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
    int size = sizeof(query) / sizeof(query[0]);
    for (int i = 0; i < size; i++)
        cout << ans[query[i] - 1] << " ";
  
    return 0;
}

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Output:

12 10 9 6 6 5 5 4 3 3

Time complexity : O(N log(N))



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