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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

// 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; 
} 

Java

// 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 

Python3

# 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 

C#

// 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 

Output:

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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