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Find the smallest subarray having atleast one duplicate

  • Difficulty Level : Medium
  • Last Updated : 25 May, 2021

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:

 

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++




// C++ program to find
// the smallest subarray having
// atleast one duplicate
#include <bits/stdc++.h>
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




// 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




# 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#




// 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

Javascript




<script>
 
// javascript program to find
// the smallest subarray having
// atleast one duplicate
    var INT_MAX = Number.MAX_VALUE;
 
    // Function to calculate
    // SubArray Length
    function subArrayLength( arr , n) {
 
        var minLen = INT_MAX;
 
        for (var i = 1; i < n; i++) {
            for (var 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
     
        var n = 7;
        var arr = [ 1, 2, 3, 1, 5, 4, 5 ];
 
        var ans = subArrayLength(arr, n);
        document.write(ans);
 
 
// This code contributed by Princi Singh
</script>
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++




// C++ program to find
// the smallest subarray having
// atleast one duplicate
#include <bits/stdc++.h>
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<int, int> 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




// 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<Integer, Integer> last = new HashMap<Integer, Integer>();
 
        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




# 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#




// 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<int, int> last = new Dictionary<int, int>();
 
        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

Javascript




<script>
 
// JavaScript program to find
// the smallest subarray having
// atleast one duplicate
 
// Function to calculate
    // SubArray Length
    function subArrayLength(arr, n)
    {
  
        let minLen = Number.MAX_VALUE;
          
        // Last stores the index of the last
        // occurrence of the corresponding value
        let last = new Map();
  
        for (let i = 0; i < n; i++)
        {
            // If the element has already occurred
            if (last.has(arr[i]) && last.get(arr[i]) != 0)
            {
                minLen =
                Math.min(minLen, i - last.get(arr[i]) + 2);
            }
            last.set(arr[i], i + 1);
        }
        if (minLen == Number.MAX_VALUE)
        {
            return -1;
        }
  
        return minLen;
    }
 
// Driver code
     
      let n = 7;
        let arr = [ 1, 2, 3, 1, 5, 4, 5 ];
  
        let ans = subArrayLength(arr, n);
        document.write(ans);
                                                                                 
</script>
Output: 
3

 

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

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