Number of subarrays with GCD = 1 | Segment tree

Given an array arr[], the task is to find the count of sub-arrays with GCD equal to 1.

Examples:

Input: arr[] = {1, 1, 1}
Output: 6
Every single subarray of the given array has GCD
of 1 and there are a total of 6 subarrays.



Input: arr[] = {2, 2, 2}
Output: 0

Approach: This problem can be solved in O(NlogN) using segment-tree data structure. The segment that will be built can be used to answer range-gcd queries.

Let’s understand the algorithm now. Use the two-pointer technique to solve this problem. Let’s make a few observations before discussing the algorithm.

  • Let’s say G is the GCD of the subarray arr[l…r] and G1 is the GCD of the subarray arr[l+1…r]. G smaller than or equal to G1 always.
  • Let’s say for the given L1, R1 is the first index such that GCD of the range [L, R] is 1 then for any L2 greater than or equal to L1, R2 will also be greater than or equal to R1.

After the above observation, two-pointer technique makes perfect sense i.e. if the length
of the smallest R is known for an index L then for an index L + 1, the search needs to be started from R on-wards.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define maxLen 30
  
// Array to store segment-tree
int seg[3 * maxLen];
  
// Function to build segment-tree to
// answer range GCD queries
int build(int l, int r, int in, int* arr)
{
    // Base-case
    if (l == r)
        return seg[in] = arr[l];
  
    // Mid element of the range
    int mid = (l + r) / 2;
  
    // Merging the result of left and right sub-tree
    return seg[in] = __gcd(build(l, mid, 2 * in + 1, arr),
                           build(mid + 1, r, 2 * in + 2, arr));
}
  
// Function to perform range GCD queries
int query(int l, int r, int l1, int r1, int in)
{
    // Base-cases
    if (l1 <= l and r <= r1)
        return seg[in];
    if (l > r1 or r < l1)
        return 0;
  
    // Mid-element
    int mid = (l + r) / 2;
  
    // Calling left and right child
    return __gcd(query(l, mid, l1, r1, 2 * in + 1),
                 query(mid + 1, r, l1, r1, 2 * in + 2));
}
  
// Function to find the required count
int findCnt(int* arr, int n)
{
    // Building the segment tree
    build(0, n - 1, 0, arr);
  
    // Two pointer variables
    int i = 0, j = 0;
  
    // To store the final answer
    int ans = 0;
  
    // Looping
    while (i < n) {
  
        // Incrementing j till we don't get
        // a gcd value of 1
        while (j < n and query(0, n - 1, i, j, 0) != 1)
            j++;
  
        // Updating the final answer
        ans += (n - j);
  
        // Increment i
        i++;
  
        // Update j
        j = max(j, i);
    }
  
    // Returning the final answer
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 1, 1, 1 };
    int n = sizeof(arr) / sizeof(int);
  
    cout << findCnt(arr, n);
  
    return 0;
}

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Java

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// Java implementation of the above approach
class GFG
{
static int maxLen = 30;
  
// Array to store segment-tree
static int []seg = new int[3 * maxLen];
  
// Function to build segment-tree to
// answer range GCD queries
static int build(int l, int r, 
                 int in, int[] arr)
{
    // Base-case
    if (l == r)
        return seg[in] = arr[l];
  
    // Mid element of the range
    int mid = (l + r) / 2;
  
    // Merging the result of left and right sub-tree
    return seg[in] = __gcd(build(l, mid, 2 * in + 1, arr),
                           build(mid + 1, r, 2 * in + 2, arr));
}
  
// Function to perform range GCD queries
static int query(int l, int r, int l1, 
                        int r1, int in)
{
    // Base-cases
    if (l1 <= l && r <= r1)
        return seg[in];
    if (l > r1 || r < l1)
        return 0;
  
    // Mid-element
    int mid = (l + r) / 2;
  
    // Calling left and right child
    return __gcd(query(l, mid, l1, r1, 2 * in + 1),
                 query(mid + 1, r, l1, r1, 2 * in + 2));
}
  
// Function to find the required count
static int findCnt(int[] arr, int n)
{
    // Building the segment tree
    build(0, n - 1, 0, arr);
  
    // Two pointer variables
    int i = 0, j = 0;
  
    // To store the final answer
    int ans = 0;
  
    // Looping
    while (i < n)
    {
  
        // Incrementing j till we don't get
        // a gcd value of 1
        while (j < n && query(0, n - 1
                                 i, j, 0) != 1)
            j++;
  
        // Updating the final answer
        ans += (n - j);
  
        // Increment i
        i++;
  
        // Update j
        j = Math.max(j, i);
    }
  
    // Returning the final answer
    return ans;
}
  
static int __gcd(int a, int b) 
    return b == 0 ? a : __gcd(b, a % b);     
}
  
// Driver code
public static void main(String []args) 
{
    int arr[] = { 1, 1, 1, 1 };
    int n = arr.length;
  
    System.out.println(findCnt(arr, n));
}
}
  
// This code is contributed by PrinciRaj1992

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Python3

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# Python3 implementation of the above approach 
from math import gcd
  
maxLen = 30;
  
# Array to store segment-tree 
seg = [0] * (3 * maxLen); 
  
# Function to build segment-tree to 
# answer range GCD queries 
def build(l, r, i, arr) :
  
    # Base-case 
    if (l == r) :
        seg[i] = arr[l]; 
        return seg[i];
  
    # Mid element of the range 
    mid = (l + r) // 2
  
    # Merging the result of left and right sub-tree 
    seg[i] = gcd(build(l, mid, 2 * i + 1, arr),
                 build(mid + 1, r, 2 * i + 2, arr)); 
    return seg[i];
  
# Function to perform range GCD queries 
def query(l, r, l1, r1, i) :
  
    # Base-cases 
    if (l1 <= l and r <= r1) :
        return seg[i]; 
          
    if (l > r1 or r < l1) :
        return 0
  
    # Mid-element 
    mid = (l + r) // 2
  
    # Calling left and right child 
    return gcd(query(l, mid, l1, r1, 2 * i + 1), 
               query(mid + 1, r, l1, r1, 2 * i + 2)); 
  
# Function to find the required count 
def findCnt(arr, n) : 
  
    # Building the segment tree 
    build(0, n - 1, 0, arr); 
  
    # Two pointer variables 
    i = 0; j = 0
  
    # To store the final answer 
    ans = 0
  
    # Looping 
    while (i < n) :
  
        # Incrementing j till we don't get 
        # a gcd value of 1 
        while (j < n and 
               query(0, n - 1, i, j, 0) != 1) :
            j += 1
  
        # Updating the final answer 
        ans += (n - j); 
  
        # Increment i 
        i += 1
  
        # Update j 
        j = max(j, i); 
  
    # Returning the final answer 
    return ans; 
  
# Driver code 
if __name__ == "__main__" :
  
    arr = [ 1, 1, 1, 1 ]; 
    n = len(arr); 
  
    print(findCnt(arr, n)); 
  
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the above approach
using System;
      
class GFG
{
static int maxLen = 30;
  
// Array to store segment-tree
static int []seg = new int[3 * maxLen];
  
// Function to build segment-tree to
// answer range GCD queries
static int build(int l, int r, 
                 int iN, int[] arr)
{
    // Base-case
    if (l == r)
        return seg[iN] = arr[l];
  
    // Mid element of the range
    int mid = (l + r) / 2;
  
    // Merging the result of left and right sub-tree
    return seg[iN] = __gcd(build(l, mid, 2 * iN + 1, arr),
                           build(mid + 1, r, 2 * iN + 2, arr));
}
  
// Function to perform range GCD queries
static int query(int l, int r, int l1, 
                        int r1, int iN)
{
    // Base-cases
    if (l1 <= l && r <= r1)
        return seg[iN];
    if (l > r1 || r < l1)
        return 0;
  
    // Mid-element
    int mid = (l + r) / 2;
  
    // Calling left and right child
    return __gcd(query(l, mid, l1, r1, 2 * iN + 1),
                 query(mid + 1, r, l1, r1, 2 * iN + 2));
}
  
// Function to find the required count
static int findCnt(int[] arr, int n)
{
    // Building the segment tree
    build(0, n - 1, 0, arr);
  
    // Two pointer variables
    int i = 0, j = 0;
  
    // To store the final answer
    int ans = 0;
  
    // Looping
    while (i < n)
    {
  
        // Incrementing j till we don't get
        // a gcd value of 1
        while (j < n && query(0, n - 1, 
                               i, j, 0) != 1)
            j++;
  
        // Updating the final answer
        ans += (n - j);
  
        // Increment i
        i++;
  
        // Update j
        j = Math.Max(j, i);
    }
  
    // Returning the final answer
    return ans;
}
  
static int __gcd(int a, int b) 
    return b == 0 ? a : __gcd(b, a % b);     
}
  
// Driver code
public static void Main(String []args) 
{
    int []arr = { 1, 1, 1, 1 };
    int n = arr.Length;
  
    Console.WriteLine(findCnt(arr, n));
}
}
  
// This code is contributed by PrinciRaj1992

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Output:

10


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Improved By : AnkitRai01, princiraj1992