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

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 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

-1


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