Hello Everyone,
Given N numbers and Q queries, each query consists of L and R. Task is to write a program which prints the count of numbers which divides all numbers in the given range L-R.
Examples :
Input : a = {3, 4, 2, 2, 4, 6} Q = 2 L = 1 R = 4 L = 2 R = 6 Output : 0 2 Explanation : The range 1-4 has {3, 4, 2, 2} which does not have any number that divides all the numbers in this range. The range 2-6 has {4, 2, 2, 4, 6} which has 2 numbers {2, 2} which divides all numbers in the given range. Input: a = {1, 2, 3, 5} Q = 2 L = 1 R = 4 L = 2 R = 4 Output: 1 0
Efficient approach : Use Segment Trees to solve this problem. If an element divides all the numbers in a given range, then the element is the minimum number in that range and it is the gcd of all elements in the given range L-R. So the count of the number of minimums in range L-R, given that minimum is equal to the gcd of that range will be our answer to every query. The problem boils down to finding the GCD, MINIMUM and countMINIMUM for every range using Segment trees. On every node of the tree, three values are stored.
On querying for a given range, if the gcd and minimum of the given range are equal, countMINIMUM is returned as the answer. If they are unequal, 0 is returned as the answer.
Below is the implementation of efficient approach :
- CPP
// CPP program to Count elements
// which divides all numbers in
// range L-R efficient approach
#include <bits/stdc++.h>
using namespace std;
#define N 100005
// predefines the tree with nodes
// storing gcd, min and count
struct node
{
int gcd;
int min;
int cnt;
} tree[5 * N];
// function to construct the tree
void buildtree( int low, int high,
int pos, int a[])
{
// base condition
if (low == high)
{
// initially always gcd and min
// are same at leaf node
tree[pos].min = tree[pos].gcd = a[low];
tree[pos].cnt = 1;
return ;
}
int mid = (low + high) >> 1;
// left-subtree
buildtree(low, mid, 2 * pos + 1, a);
// right-subtree
buildtree(mid + 1, high, 2 * pos + 2, a);
// finds gcd of left and right subtree
tree[pos].gcd = __gcd(tree[2 * pos + 1].gcd,
tree[2 * pos + 2].gcd);
// left subtree has the minimum element
if (tree[2 * pos + 1].min < tree[2 * pos + 2].min)
{
tree[pos].min = tree[2 * pos + 1].min;
tree[pos].cnt = tree[2 * pos + 1].cnt;
}
// right subtree has the minimum element
else
if (tree[2 * pos + 1].min > tree[2 * pos + 2].min)
{
tree[pos].min = tree[2 * pos + 2].min;
tree[pos].cnt = tree[2 * pos + 2].cnt;
}
// both subtree has the same minimum element
else
{
tree[pos].min = tree[2 * pos + 1].min;
tree[pos].cnt = tree[2 * pos + 1].cnt +
tree[2 * pos + 2].cnt;
}
}
// function that answers every query
node query( int s, int e, int low, int high, int pos)
{
node dummy;
// out of range
if (e < low or s > high)
{
dummy.gcd = dummy.min = dummy.cnt = 0;
return dummy;
}
// in range
if (s >= low and e <= high)
{
node dummy;
dummy.gcd = tree[pos].gcd;
dummy.min = tree[pos].min;
if (dummy.gcd != dummy.min)
dummy.cnt = 0;
else
dummy.cnt = tree[pos].cnt;
return dummy;
}
int mid = (s + e) >> 1;
// left-subtree
node ans1 = query(s, mid, low,
high, 2 * pos + 1);
// right-subtree
node ans2 = query(mid + 1, e, low,
high, 2 * pos + 2);
node ans;
// when both left subtree and
// right subtree is in range
if (ans1.gcd and ans2.gcd)
{
// merge two trees
ans.gcd = __gcd(ans1.gcd, ans2.gcd);
ans.min = min(ans1.min, ans2.min);
// when gcd is not equal to min
if (ans.gcd != ans.min)
ans.cnt = 0;
else
{
// add count when min is
// same of both subtree
if (ans1.min == ans2.min)
ans.cnt = ans2.cnt + ans1.cnt;
// store the minimal's count
else
if (ans1.min < ans2.min)
ans.cnt = ans1.cnt;
else
ans.cnt = ans2.cnt;
}
return ans;
}
// only left subtree is in range
else if (ans1.gcd)
return ans1;
// only right subtree is in range
else if (ans2.gcd)
return ans2;
}
// function to answer query in range l-r
int answerQuery( int a[], int n, int l, int r)
{
// calls the function which returns
// a node this function returns the
// count which will be the answer
return query(0, n - 1, l - 1, r - 1, 0).cnt;
}
// Driver Code
int main()
{
int a[] = { 3, 4, 2, 2, 4, 6 };
int n = sizeof (a) / sizeof (a[0]);
buildtree(0, n - 1, 0, a);
int l = 1, r = 4;
// answers 1-st query
cout << answerQuery(a, n, l, r) << endl;
l = 2, r = 6;
// answers 2nd query
cout << answerQuery(a, n, l, r) << endl;
return 0;
}
Output:
0 2
Time Complexity: Time Complexity for tree construction is O(n logn) since tree construction takes O(n) and finding out gcd takes O(log n). The time taken for every query in worst case will be O(log n * log n) since the inbuilt function __gcd takes O(log n)