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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define maxLen 30 // Array to store segment-treeint seg[3 * maxLen]; // Function to build segment-tree to// answer range GCD queriesint 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 queriesint 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 lengthint 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 codeint main(){ int arr[] = { 2, 2, 2 }; int n = sizeof(arr) / sizeof(int); cout << findLen(arr, n); return 0;} |
Java
// Java implementation of the approachclass GFG{static int maxLen = 30; // Array to store segment-treestatic int []seg = new int[3 * maxLen]; // Function to build segment-tree to// answer range GCD queriesstatic 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 queriesstatic 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 lengthstatic 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 codepublic 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 approachfrom math import gcd as __gcd maxLen = 30 # Array to store segment-treeseg = [0 for i in range(3 * maxLen)] # Function to build segment-tree to# answer range GCD queriesdef 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 queriesdef 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 lengthdef 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 codearr = [2, 2, 2]n = len(arr) print(findLen(arr, n)) # This code is contributed by Mohit Kumar |
C#
// C# implementation of the approachusing 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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