# Find the smallest subarray having atleast one duplicate

Given an array arr of N elements, the task is to find the length of the smallest subarray of the given array that contains at least one duplicate element. A subarray is formed from consecutive elements of an array. If no such array exists, print “-1”.

Examples:

Input: arr = {1, 2, 3, 1, 5, 4, 5}
Output: 3
Explanation:
Input: arr = {4, 7, 11, 3, 1, 2, 4}
Output: 7
Explanation:

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

Naive Approach:

• The trick is to find all pairs of two elements with equal value. Since these two elements have equal value, the subarray enclosing them would have at least a single duplicate and will be one of the candidates for the answer.
• A simple solution is to use two nested loops to find every pair of elements.If the two elements are equal then update the maximum length obtained so far.

Below is the implementation of the above approach:

 // C++ program to find // the smallest subarray having // atleast one duplicate #include using namespace std;    // Function to calculate // SubArray Length int subArrayLength(int arr[], int n) {        int minLen = INT_MAX;        for (int i = 1; i < n; i++) {         for (int j = 0; j < i; j++) {             // If the two elements are equal,             // then the subarray arr[i..j]             // will definitely have a duplicate             if (arr[i] == arr[j]) {                 // Update the minimum length                 // obtained so far                 minLen = min(minLen, i - j + 1);             }         }     }     if (minLen == INT_MAX) {         return -1;     }        return minLen; } // Driver Code int main() {     int n = 7;     int arr[] = { 1, 2, 3, 1, 5, 4, 5 };        int ans = subArrayLength(arr, n);     cout << ans << '\n';        return 0; }

 // Java program to find  // the smallest subarray having  // atleast one duplicate    class GFG {            final static int INT_MAX = Integer.MAX_VALUE;            // Function to calculate      // SubArray Length      static int subArrayLength(int arr[], int n)      {                 int minLen = INT_MAX;                 for (int i = 1; i < n; i++)          {              for (int j = 0; j < i; j++)              {                  // If the two elements are equal,                  // then the subarray arr[i..j]                  // will definitely have a duplicate                  if (arr[i] == arr[j])                  {                      // Update the minimum length                      // obtained so far                      minLen = Math.min(minLen, i - j + 1);                  }              }          }          if (minLen == INT_MAX)         {              return -1;          }                 return minLen;      }             // Driver Code      public static void main(String[] args)     {          int n = 7;          int arr[] = { 1, 2, 3, 1, 5, 4, 5 };                 int ans = subArrayLength(arr, n);          System.out.println(ans);                 }  }    // This code is contributed by AnkitRai01

 # Python program for above approach n = 7 arr = [1, 2, 3, 1, 5, 4, 5] minLen = n + 1    for i in range(1, n):     for j in range(0, i):         if arr[i]== arr[j]:             minLen = min(minLen, i-j + 1)    if minLen == n + 1:        print("-1") else:        print(minLen)

 // C# program to find  // the smallest subarray having  // atleast one duplicate using System;    class GFG {            static int INT_MAX = int.MaxValue;            // Function to calculate      // SubArray Length      static int subArrayLength(int []arr, int n)      {                 int minLen = INT_MAX;                 for (int i = 1; i < n; i++)          {              for (int j = 0; j < i; j++)              {                  // If the two elements are equal,                  // then the subarray arr[i..j]                  // will definitely have a duplicate                  if (arr[i] == arr[j])                  {                      // Update the minimum length                      // obtained so far                      minLen = Math.Min(minLen, i - j + 1);                  }              }          }          if (minLen == INT_MAX)         {              return -1;          }                 return minLen;      }             // Driver Code      public static void Main()     {          int n = 7;          int []arr = { 1, 2, 3, 1, 5, 4, 5 };                 int ans = subArrayLength(arr, n);          Console.WriteLine(ans);                 }  }    // This code is contributed by AnkitRai01

Output:

3

Time Complexity: O(N2)

Efficient Approach:

This problem can be solved in O(N) time and O(N) Auxiliary space using the idea of hashing technique. The idea is to iterate through each element of the array in a linear way and for each element, find its last occurrence using a hashmap and then update the value of min length using the difference of the last occurrence and the current index. Also, update the value of the last occurrence of the element by the value of the current index.

Below is the implementation of the above approach:

 // C++ program to find // the smallest subarray having // atleast one duplicate #include using namespace std;    // Function to calculate // SubArray Length int subArrayLength(int arr[], int n) {        int minLen = INT_MAX;     // Last stores the index of the last     // occurrence of the corresponding value     unordered_map last;        for (int i = 0; i < n; i++) {         // If the element has already occurred         if (last[arr[i]] != 0) {             minLen = min(minLen, i - last[arr[i]] + 2);         }         last[arr[i]] = i + 1;     }     if (minLen == INT_MAX) {         return -1;     }        return minLen; }    // Driver Code int main() {     int n = 7;     int arr[] = { 1, 2, 3, 1, 5, 4, 5 };        int ans = subArrayLength(arr, n);     cout << ans << '\n';        return 0; }

 // Java program to find // the smallest subarray having // atleast one duplicate import java.util.*;    class GFG  {        // Function to calculate     // SubArray Length     static int subArrayLength(int arr[], int n)     {            int minLen = Integer.MAX_VALUE;                    // Last stores the index of the last         // occurrence of the corresponding value         HashMap last = new HashMap();            for (int i = 0; i < n; i++)          {             // If the element has already occurred             if (last.containsKey(arr[i]) && last.get(arr[i]) != 0)              {                 minLen = Math.min(minLen, i - last.get(arr[i]) + 2);             }             last.put(arr[i], i + 1);         }         if (minLen == Integer.MAX_VALUE)         {             return -1;         }            return minLen;     }        // Driver Code     public static void main(String[] args)      {         int n = 7;         int arr[] = { 1, 2, 3, 1, 5, 4, 5 };            int ans = subArrayLength(arr, n);         System.out.print(ans);     } }    // This code is contributed by 29AjayKumar

 # Python program for above approach    n = 7 arr = [1, 2, 3, 1, 5, 4, 5]    last = dict()    minLen = n + 1    for i in range(0, n):     if arr[i] in last:         minLen = min(minLen, i-last[arr[i]]+2)        last[arr[i]]= i + 1           if minLen == n + 1:        print("-1") else:        print(minLen)

 // C# program to find // the smallest subarray having // atleast one duplicate using System; using System.Collections.Generic;    class GFG  {        // Function to calculate     // SubArray Length     static int subArrayLength(int []arr, int n)     {            int minLen = int.MaxValue;                    // Last stores the index of the last         // occurrence of the corresponding value         Dictionary last = new Dictionary();            for (int i = 0; i < n; i++)          {             // If the element has already occurred             if (last.ContainsKey(arr[i]) && last[arr[i]] != 0)              {                 minLen = Math.Min(minLen, i - last[arr[i]] + 2);             }             if(last.ContainsKey(arr[i]))                 last[arr[i]] = i + 1;             else                 last.Add(arr[i], i + 1);         }         if (minLen == int.MaxValue)         {             return -1;         }            return minLen;     }        // Driver Code     public static void Main(String[] args)      {         int n = 7;         int []arr = { 1, 2, 3, 1, 5, 4, 5 };            int ans = subArrayLength(arr, n);         Console.Write(ans);     } }    // This code is contributed by PrinciRaj1992

Output:
3

Time Complexity: O(N), where N is size of array

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