Smallest subarray with GCD as 1 | Segment Tree

Given an array arr[], the task is to find the smallest sub-arrays with GCD equal to 1. If there is no such sub-array then print -1.

Examples:

Input: arr[] = {2, 6, 3}
Output: 3
{2, 6, 3} is the only sub-array with GCD = 1.

Input: arr[] = {2, 2, 2}
Output: -1

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 length 
int findLen(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 = INT_MAX; 
  
    // 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++; 
  
        if (j == n) 
            break; 
  
        // Updating the final answer 
        ans = min((j - i + 1), ans); 
  
        // Incrementing i 
        i++; 
  
        // Updating j 
        j = max(j, i); 
    } 
  
    // Returning the final answer 
    if (ans == INT_MAX) 
        return -1; 
    else
        return ans; 
} 
  
// Driver code 
int main() 
{ 
    int arr[] = { 2, 2, 2 }; 
    int n = sizeof(arr) / sizeof(int); 
  
    cout << findLen(arr, n); 
  
    return 0; 
} 

Java

// Java implementation of the 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 length 
static int findLen(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 = Integer.MAX_VALUE; 
  
    // 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++; 
  
        if (j == n) 
            break; 
  
        // Updating the final answer 
        ans = Math.min((j - i + 1), ans); 
  
        // Incrementing i 
        i++; 
  
        // Updating j 
        j = Math.max(j, i); 
    } 
  
    // Returning the final answer 
    if (ans == Integer.MAX_VALUE) 
        return -1; 
    else
        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[] = { 2, 2, 2 }; 
    int n = arr.length; 
  
    System.out.println(findLen(arr, n)); 
} 
} 
  
// This code is contributed by 29AjayKumar 

Python3

# Python3 implementation of the approach 
from math import gcd as __gcd 
  
maxLen = 30
  
# Array to store segment-tree 
seg = [0 for i in range(3 * maxLen)] 
  
# Function to build segment-tree to 
# answer range GCD queries 
def build(l, r, inn, arr): 
      
    # Base-case 
    if (l == r): 
        seg[inn] = arr[l] 
        return seg[inn] 
  
    # Mid element of the range 
    mid = (l + r) // 2
  
    # Merging the result of  
    # left and right sub-tree 
    seg[inn] = __gcd(build(l, mid, 
                           2 * inn + 1, arr),  
                     build(mid + 1, r,  
                           2 * inn + 2, arr)) 
  
    return seg[inn] 
  
# Function to perform range GCD queries 
def query(l, r, l1, r1, inn): 
      
    # Base-cases 
    if (l1 <= l and r <= r1): 
        return seg[inn] 
    if (l > r1 or r < l1): 
        return 0
  
    # Mid-element 
    mid = (l + r) // 2
  
    # Calling left and right child 
    x=__gcd(query(l, mid, l1, r1, 
                  2 * inn + 1),  
            query(mid + 1, r, l1, r1, 
                  2 * inn + 2)) 
    return x 
  
# Function to find the required length 
def findLen(arr, n): 
      
    # Buildinng the segment tree 
    build(0, n - 1, 0, arr) 
  
    # Two pointer variables 
    i = 0
    j = 0
  
    # To store the finnal answer 
    ans = 10**9
  
    # Loopinng 
    while (i < n): 
  
        # Incrementinng 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
  
        if (j == n): 
            break; 
  
        # Updatinng the finnal answer 
        ans = minn((j - i + 1), ans) 
  
        # Incrementinng i 
        i += 1
  
        # Updatinng j 
        j = max(j, i) 
  
    # Returninng the finnal answer 
    if (ans == 10**9): 
        return -1
    else: 
        return ans 
  
# Driver code 
arr = [2, 2, 2] 
n = len(arr) 
  
print(findLen(arr, n)) 
  
# This code is contributed by Mohit Kumar 

C#

// C# implementation of the 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 ind, int[] arr) 
    { 
        // Base-case 
        if (l == r) 
            return seg[ind] = arr[l]; 
      
        // Mid element of the range 
        int mid = (l + r) / 2; 
      
        // Merging the result of left and right sub-tree 
        return seg[ind] = __gcd(build(l, mid, 2 * ind + 1, arr), 
                                build(mid + 1, r, 2 * ind + 2, arr)); 
    } 
      
    // Function to perform range GCD queries 
    static int query(int l, int r,  
                     int l1, int r1, int ind) 
    { 
        // Base-cases 
        if (l1 <= l && r <= r1) 
            return seg[ind]; 
        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 * ind + 1), 
                     query(mid + 1, r, l1, r1, 2 * ind + 2)); 
    } 
      
    // Function to find the required length 
    static int findLen(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 = int.MaxValue; 
      
        // 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++; 
      
            if (j == n) 
                break; 
      
            // Updating the final answer 
            ans = Math.Min((j - i + 1), ans); 
      
            // Incrementing i 
            i++; 
      
            // Updating j 
            j = Math.Max(j, i); 
        } 
      
        // Returning the final answer 
        if (ans == int.MaxValue) 
            return -1; 
        else
            return ans; 
    } 
      
    static int __gcd(int a, int b)  
    {  
        return b == 0 ? a : __gcd(b, a % b);      
    }  
      
    // Driver code 
    public static void Main() 
    { 
        int []arr = { 2, 2, 2 }; 
        int n = arr.Length; 
      
        Console.WriteLine(findLen(arr, n)); 
    } 
} 
  
// This code is contributed by kanugargng 

Output:

-1

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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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